The Art of Solving Linear Equations

Solving linear equations is an essential skill in mathematics that requires attention to detail and a systematic approach. In this article, we will explore two different examples of solving linear equations using various methods.

**Example 1: Solving Two Linear Equations**

In the first example, we are given two linear equations:

10m + 2n = 4

15m - 5 = -2

The goal is to solve for the variables m and n. To do this, we need to simplify the equations by equating the coefficients of n.

We multiply the second equation by 2 and the first equation by 5 to equate the coefficients of n:

50m + 10n = 20

30m - 10 = -4

Now that the coefficients of n are equal, we can cancel them out. This leaves us with two new equations:

50m = 16

30m = 2

Dividing both sides of each equation by their respective coefficients, we get:

m = 16/80

m = 1/5

So, the value of m is 1/5.

Now that we have the value of m, we can substitute it into one of the original equations to solve for n. Let's use the first equation:

10(1/5) + 2n = 4

Simplifying the equation, we get:

2 + 2n = 4

2n = 2

n = 1

So, the value of n is 1.

Finally, we can substitute the values of m and n back into one of the original equations to find x and y. Let's use the equation "1/x + y = f". We know that f = 3 and x + y = 5. Substituting these values in, we get:

1/x + y = 3

x + y = 5

Subtracting the second equation from the first, we get:

-1/x = -2

x = 1/2

Now that we have the value of x, we can substitute it into one of the original equations to solve for y. Let's use the equation "x + y = 5":

1/2 + y = 5

y = 9/2

So, the values of x and y are 1/2 and 9/2, respectively.

**The Caution of Speed**

In the second example, we are given a problem involving speed, distance, and time. We need to find the speed of rowing in still water and the speed of the current.

Let's assume that the speed of rowing in still water is x km/h and the speed of the current is y km/h.

When rowing downstream, the distance traveled is 20 kilometers and the time taken is 2 hours. The speed of the boat is the sum of the speed of rowing in still water and the speed of the current:

x + y = 10

Similarly, when rowing upstream, the distance traveled is 4 kilometers and the time taken is 2 hours. The speed of the boat is the difference between the speed of rowing in still water and the speed of the current:

x - y = 2

Now we have two linear equations with two variables. We can solve for x and y using various methods, including substitution or elimination.

Using the elimination method, we add the two equations to eliminate one variable:

(x + y) + (x - y) = 10 + 2

2x = 12

x = 6

Now that we have the value of x, we can substitute it into one of the original equations to solve for y. Let's use the equation "x + y = 10":

6 + y = 10

y = 4

So, the values of x and y are 6 and 4, respectively.

In conclusion, solving linear equations requires attention to detail and a systematic approach. By using various methods such as substitution or elimination, we can solve for the variables in a system of linear equations. Additionally, by applying these methods to real-world problems like speed, distance, and time, we can find the values of unknown variables with precision.

"WEBVTTKind: captionsLanguage: enthat is the pair of linear equation in two variables so first of all a linear equation equation having degree one is called the linear equation and next is two variable two variable means that equation involves two variable that i means x and y and the pair the pair means we are having such two pair of two equations like 5x plus 3y equal to 9 and 7x minus 2y is equal to 6 so look at here this is one of the example of pair of linear equation in two variable so the first of all here there are two equation equation number one and equation number two so this is a pair of linear equation look at here the degree of the equation is 1 the degree is 1 the power of x is 1 power of y is 1 similarly in the second equation also equation the quality signs are there and 2 variable means they are two variable x and y so this is one of the example of pair of linear equation in two variable now in this step chapter we have to deal with such type of equations the next one is solution what is the solution of a pair of linear equation in two variable so a point m comma 1 n is a solution of pair of linear equation in two variable a 1 com x plus b 1 y is equal to c 1 a 2 x plus b 2 y is equal to c 2 if it will satisfy both the such equations simultaneously now what is the meaning of this that a point m comma n is a point that is a solution of the this type of equation if it will satisfy both those equations simultaneously what is the meaning of simultaneously means whatever the solution is that is satisfy both the equations at a time so let's take example if 3 comma 1 solution of x plus y is equal to 4 and x minus y is equal to 2 means we have to check that is 3 comma 1 is the solution of this pair of linear equations yes or no so let's try if if this point is a solution of this system of solution equations that it will satisfy both the equation so let's try this is equation number one and this is equation number two first of all we will try for the question number one equation number one is x plus y is equal to 4 now point is 3 comma 1 so replace x by 3 and y by 1 left hand side we will get 4 and that is equal to right hand side so yes this point satisfy the equation number one now we will try for the question number two so equation number two is x minus y equal to two and the point again point is 3 comma 1 so try for this 3 minus 1 that is 2 and this is also equal to right hand side so that means 3 comma 1 point satisfy both the equation equation number one and equation number two so it is a solution of both the equation so we will say that three comma one is the solution of the pair of linear equation in two variable the next example is is 2 comma 1 solution of 3x minus 2y is equal to 4 and 2x plus 4y is equal to again the one example is here we have to check that is this given point solution of pair of linear question intuitive yes or no so again we will try for the both the equations this is equation number one and this is equation number two that's for first equation 3x minus 2y equal to 4 and the point is 2 comma 1 so easily x is replaced by 2 and the y is replaced by 1 this is 3 into 2 minus 2 into 1 that is 6 minus 2 that is equal to 4 and this is equal to right hand side so that means 2 comma 1 is the solution of first linear equation in 2 variable now let's check for the second equation also that is 2x plus 4y is equal to 2 again replace x by 2 and y by 1 so this is 2 2s of 4 4 was the 4 that is 8 this is not equal to right hand side so 2 comma 1 is solution of first equation but is it is not a solution of second equation so overall 2 comma 1 is not a solution of such pair of linear equation in two variables means for the solution of pair of linear equation in two variable that given point has to satisfy both the equations simultaneously in as in this example it will satisfy first equation but it will not satisfy the second equation so overall it is not a solution of pair of linear equation in two variables so this is how we have to find or we have to check the the given point is solution of pair of linear equation two variable yes or no the next one is now in grade nine as we know that a single linear equation in two variable represents a straight line on the graph so similarly if we are having a pair of linear equation two variable so simply it will represents two straight lines on the graph so so pair of linear equation in two variable represents two straight lines of the graph and there are three possible cases related to the two lines straight lines so first case is at the the two lines intersect each other at a point so these are the two straight lines the line l1 and line l2 and this will intersect at a given point so look at here in this chapter as we know that a single point is a solution if it will satisfy both the equations so that means or in graphically the solution of pair of linear equation two variable is a point that is common to both the lines so look at here in the first case how many common points are there there is only a single common point a one point in common so there is a one solution the solution in a pair of linear equation represents a point that is common to both the equation if there are two common points there is two solutions five common points five solution one common point one solution no common point means there isn't no solution so similarly if there are only three possible cases in the first case when the lines are intersecting to each other there is only one common point so we will say that there is only one common point that why it is having a unique solution means it is having a single solution and the system is called consistent system that means solution is possible for the second case when the lines are parallel when the lines are parallel now look at here between these two lines there is no comma point there is no compromise between these two lines so there is no common point so we will say that there is a no solution because i will repeat solution is the common point to both the equation or the both the straight lines as there is no comma point so there is a no solution and the system is called inconsistent inconsistent system where solution is not possible the third and the last case is when one line is just upon the second line means a client just loosely link is just over here both the line coincides with each other now every point is a common point every point of both the lines is a common point so they are infinite number of points in common so we will say that there are infinite many solution because solution is again a common point so between these two lines there are infinite points are common so they are infinitely many solution again the system is called consistent system because the solutions are possible so i will repeat so what is the solution graphically graphically a solution is a common point to both the straight lines if we are having two lines there are only possible three cases case one case two and case three in first case lines are intersecting where only one common point is there so we are having only unique solution in second case we only no point is common so there is a no solution in third case every point is a common point to both the lines so there are infrared common points so we will say that there are infinitely many solutions the next one is how to find the solution the next ways as we know from the last slide what is the solution solution is a common point to both the equations or the or the straight lines now the content is how to find such solutions so look at here there are four methods according to ncrt grade 10 we have to find the solution of pair of linear equation in two variable the first method is substitution method second is graphical method third is elimination method and the fourth is cos multiply method in this video we will discuss the first two methods that are substitution method and the graphical method so the first method is substitution method so sol x plus y is equal to 5 and 2x minus 3y is equal to 4 using the substitution method so this is a simply this is a pair of linear equation in two variable this is the first equation and this one is the second equation and we have to solve this system of equation using the substitution method so very easy look at here the first equation that is x plus y is equal to 5 and the second equation is 2x minus 3y is equal to 4 what is the first step look at here the first step is we have to find the value of x or y from the equation number one so simply the value of x is equal to 5 minus y this is the value of x from the first equation now we have to put this value in the equation number 2. look at here 2 x is replaced by its value that is 5 minus y minus 3 y is equal to 4 now so this two 5's are 10 2 y's are 2y minus 3y equal to 4 that is 10 minus minus plus that is minus 5y equals equal to 4 10 minus 4 is equal to 5y that is 6 is equal to 5 y and the value of y is 6 upon 5 that is the value of y so what is the first step i will repeat the first step is from first equation we have to find the value of x or y if when we are having the value of x or y we will substitute that value in the question number 2 from there we will get the value of one variable we will get here we will get the value of y that is 6 by 5 again put that value in this equation so we will get x is equal to 5 minus y that is 6 upon 5 on calculation we will get 25 minus 6 x is equal to 19 upon 5 so the overall answer is 19 upon 5 comma 6 0.5 is the solution of this pair of linear equation in two variable now after getting the solution we will easily verify that is the solution correct or not we simply have to substitute these solution in the equations if it will satisfy both the equations this solution is correct so substitute here x plus y is equal to 5 first equation x plus y x is 19 upon 5 plus y that is 6 upon 5 and the answer is 19 plus 6 upon 5 that is 25 upon 5 that is equal to 5 is equal to right hand side our solution will satisfy the first equation easily now let's try for the second equation that is 2x minus 3y is equal to 4 2 into x that is 19 upon 5 minus 3 into y that is 6 upon 5 that is 38 upon 5 minus 18 upon 5 that is equal to 20 upon 5 and that is 4 is equal to right hand side so a solution will also satisfy the question number two so that means our solutions satisfy both the equations simultaneously so this is the correct answer for this system of solution now we are having a single solution so there is this system having the unique solution and the lines are intersecting lines because we are having a single solution so this system is represents a intersecting straight lines now the second question is 3x plus 4y equal to 10 and 2x minus 2y equal to 2 again this is the first equation 3x plus 4y is equal to 10 from here we have to find the value of x or y for the first equation so value of x is that is 10 minus 4 y upon 3 this is the value of x from the first equation now we will substitute this value in the second equation that is 2 into x that is 10 minus 4 y upon 3 minus 2 2y equal to 2 now calculating 20 minus 8y upon 3 minus 6y is equal to 2 i will take calcium here so this is equal to 20 minus 14 y is equal to 6 that is 20 minus 6 is equal to 14 y 14 is equal to 14y we will get y is equal to 1 now we are having the value of one variable that y is equal to 1 we will substitute that value in the equation this one x is equal to 10 minus 4 into 1 upon 3 x x is equal to x is equal to 10 minus 4 that is 6 upon 3 so x is equal to 2 so 2 comma 1 is the solution of this system of equations now again we will check is this is the solution correct so we will set trying to satisfy these both the equations the first equation 3 into x that is 3 into 2 6 plus 4 into y that is 4 into 1 4 the answer is 10 and that is equal to right hand side yes the first equation is satisfy this by the solution try for the second 2x this is 2 into x that is 2 into 2 minus 2 into y that is 1 the answer is 4 minus 2 and the answer is 2 that is equal to right hand side so this solution will satisfy both the equation so our answer is correct that means 2 comma 1 is the solution of this pair of linear equation in two variables so this is the substitution method again i will repeat in substitution method the first step is we have to find the value of x or y from the first equation that is like this then we have to substitute that value of that very value of that variable in the second equation from there after calculating we will get the value of one variable and after substituting that variable in again in this equation we will get the second variable also so this is the overall method that is called substitution method the next one is if we add this is a statement problem if we add 1 to the numerator and subtract 1 from the denominator a fraction becomes 1 it also becomes half if we only add 1 to the denominator what is the fraction so this is a question related to fraction so we have to find the fraction let the nominator is let denominator of the fraction is x and denominator of the fraction is y so overall the fraction is x upon y this is the very basic thing in a fraction there are two parts denominator and denominator as we have to find the fraction that denominator is x and the denominator is y so overall fraction is x upon y now what is the first condition if we add 1 to the numerator if we add 1 to the numerator and subtract 1 from the denominator subtract 1 from the denominator fraction becomes 1 the fraction becomes 1 this is the first condition so just cross multiplying x plus 1 is equal to y minus 1 that is x minus y is equal to minus 2 this is our first equation now what is the second condition it becomes half the fraction become half if we only add 1 to the denominator means there is no change in the denominator there is only we have to add 1 in the denominator the equation becomes half again cross multiplying 2x is equal to y plus 1 and 2x minus y is equal to 1 this is equation number 2. now we are having the a pair of linear equation in 2 variable again using the substitution method we will solve these equations from here we will get the value one variable that is minus 2 plus y now substitute this value in the this is step one step two substitute this value here in the question number 2 that is 2 into x that is minus 2 plus y minus y equal to 1 that is minus 4 plus 2y minus y equal to 1 this is 2y minus y y 4 get added that side is equal to 5 so value of y is 5 again substitute that value here x is equal to minus 2 plus y and y is 5 so x is equal to 3 so value of x is 3 and y is 4 so our fraction is 3 upon 5 so fraction is 3 upon 5 if we want to check the solution we will go through the conditions look at here the first condition is add 1 to the numerator that is 3 plus 1 4 subtract 1 from the denominator 5 minus 1 4 the answer is 1 yes the answer is 1 first condition satisfied the second condition is if there is no change in denominator means 3 is as this and if we add 1 to the denominator means 5 plus 1 6 it becomes half yes it becomes half so uh answer that is 3 upon 5 will satisfy both the conditions so 3.5 is the right correct answer for this statement problem now the next one is five years ago nuri was thrice as old as sonu 10 years later nori will be twice as old as sunu how old the newly and so no now this is a statement problem related to edges so for that we will divide the question in three parts the first is percentages that is future edges and the past ages as we have to find the present edges let the present age of lurie is axiors and the present age of sonu is y years so this is the x and y is the present edges of sonu and nuri respectively now 10 years later 10 years later nori will be twice as old as so no but this is the second condition the first condition is five years ago five years ago means in the past five years in the past so nuri's age five years ago is x minus five five years ago is y minus five the nuri was thrice as old as sonu the nori is as thrice as old as solu means lori is elder than as compared to sonu so the according to condition x minus 5 is equal to as thrice old as sonu this is 3 times y minus 5 because the nuri is already elder then so no to equate them we have to multiply the age of 7 by 3 so from here we will get the equation x minus 5 is equal to 3y minus 15 so that is x minus 3y is equal to minus 10 this is our equation number one from first equation condition future in future 10 years later means means 10 years in the future so lurie's age is x plus 10 and sonu's age is y plus 10 nori will be twice as old as sonu twice as compared to sonu means x plus 10 is equal to 2 times y plus 10 so that becomes x plus 10 is equal to 2y plus 20 that is x minus 2y is equal to 10 this is equation number 2 now we are having the pair of linear equation in two variable so we have to satisfy we have to solve this equation using the substitution method so again from first equation this is step one from first equation we will find the value of x that is minus 10 plus 3 y and we have to substitute that value in the equation number 2 that becomes minus 10 plus 3y minus 2 y is equal to 10 that becomes y is equal to 20 the y is equal to 20 now again substitute that by here x is equal to minus 10 plus 3 into 20 that is equal to minus 10 plus 60 and x is equal to 50 so that means present age of lurie is 50 and the present age of sunday is 20. these are the present edges of nori so no again we have to verify our solution so simply look at here five years in past that age is fifty minus four five that is forty five and here order's age is fifteen and that lures ages three times as solute this is the condition thrice yes first condition fulfilled in 10 years future here nuri is 50 plus 10 that is 68 20 plus 10 30 h here nuri is twice twice as old as so no yes second condition also satisfy means our solution is correct so this is the way if there is a statement question related to edges you have to divide the question like this presentages passages and future ranges and you have to find the conditions the next one is mina went to bank to withdraw rupees 2000 she asked the cashier to give me the notes of 50 and 100 only meena got 25 notes in all find how many notes of 5800 she received means this is a question that a girl meena goes to bank to withdraw 2000 means she total amount she has to withdraw is 2 000 but she asked the cashier that give me the notes of 5800 only and the cashier gave him her 25 notes overall we have to find how many nodes of 50 and 100 each given by cashier to the mean now so let rupees 50 nodes are x and rupees hundred nodes are y nodes and look at here if we are having two nodes of rupees 50 means overall amount is 100 rupees if we are having three nodes of rupees 50 means overall amount is 150 rupees similarly this time we are having x nuts of rupees 50 so amount is 50 x similarly we are having y nodes of 100 rupees so amount is 100 y now the first condition is the cashier gave 25 dots to meter but according to uh there are only x plus y number of nodes because x naughts of rupees 50 and y dots of rupees 100 so we now get x plus y notes but mean now go 25 nodes overall so first condition is x plus y is equal to 25 this is equation number one now look at the amount the amount mina got is 50 x plus 100 y but according to meera she got 2 000 rupees so that means 50x plus 100 y is equal to 2000 this is equation number two now we are having two linear equation two variable so we will solve this using substitution method first of all the equation number 2 is simply divided by 50 that is 50 is a common factor that becomes x plus 2y is equal to 40 this is actually question number two so again from step one we will find the value of x or y from the first equation that x is equal to 25 minus y and substitute that value in the step 2 in the equation number 2 that is x 25 minus y plus 2 y is equal to 40 that is 25 plus y is equal to 40 and y is equal to 40 minus 25 that y is equal to 50 so value 5 is y is 50 to put that value by here x is equal to 25 minus 50 x is equal to 10 so what does this mean this means that she got 10 nodes of rupees 50 and 15 knots of rupees 100 so look at here there are 15 plus 10 25 nodes are there first condition full field now the amount is 15 10 500 this is 1500 and the amount is 2 000 yes second condition also fulfill that means y15 and extra is the correct solution for this statement problem the next one is a landing library has fixed charges per day charges for the first three days is at additional charges for every day thereafter arushi paid rupees 27 for the book kept for seven days and the fixed charges four piece x per day and the additional charges four piece y per day write the linear question representing the above information so this is a question where we have to convert the statement in the linear equation so what what actually is this question about that there are some fixed charges and some charges are that are per day according to per day so let the fixed charges are lacked fixed charges are actually piece and variable charges are variable charges are y rupees per day so look at here arushi paid 27 piece she kept the book for seven days now she kept in the book for seven days but the fixed charges are for first three days fixed charges are for first three days that means for first three days she has to please pay rupees x now after and total number of days are seven and she has to pay repeats x for first three days now the remaining days are four days so for remaining four days she has to pay five rupees per day so the amount is 4 y this is the total amount for seven days that is x rupees for first three days and one piece for each day after three days that means remaining four days the charges are four why so but according to her the total paid she paid that is 27 she paid total 27 rupees for seven days so this is the linear equation representing the this situation again if one more situation is there if she rent a book for five days she paid 21 rupees for five days now how to convert it again for the first three days she will pay extra rupees for first three days now the remaining days that are out of five if three days there are two remaining days for again for remaining two days she have to pay wider piece according to per day that charges are 2y for wire piece per day for many two days she has to pay 2y and the total overall pay she's 21. this is a second equation so here this question is mainly here for how to convert such situation in the pair of linear equation in two variables now you can you are having the two situations and two equations you will easily solve these equations using the substitution method the next one is how to find the solution of pair of linear equation in two variables so basically there are four methods the first method is substitution method second graphical third elimination method and the fourth is cross multiplied method the first two methods are already discussed in the first part of this chapter so now in this video we will discuss about the third and fourth method that are elimination method and the cross multiply method so the third method that is elimination method sol x plus y is equal to 5 and 2x minus y 3y is equal to 4 using the elimination method for this the first of all equations are x plus y is equal to 5 and second equation is 2x minus 3y is equal to 4. now for the elimination method we have to equate two variables of either x or y equal to each other so we want to make the coefficient of x is equal to 2 here so for that we multiply the first equation by 2 and second equation by 1 so that coefficient of x are equal 2x minus 3y is equal to 4 now the equations become 2x plus 2y is equal to 10 and the second equation 2x minus 3y is equal to 4. now look at here coefficient of xr equal 2 and 2. so change the sign of the second equation to solve both plus 2x minus 2x get cancel out plus 2y plus 3y that is 5y 10 minus 4 that is equal to 6 so value of y is equal to 6 upon 5 this is the value of y variable now to find the value of second variable that is x we will substitute that value in any of the equation to get the second value so again x plus y is equal to 5 so put the value of y here that is x plus 6 upon 5 is equal to 5 that is equal to x is equal to 5 minus 6 upon 5 that is equal to 25 minus 6 upon 5 that is equal to 19 upon 5 so solution is 19 upon 5 comma 6 upon 5 so this is the third method that is elimination method to use that method we need that the coefficient of x or the coefficient of y are equal to each other so for that we will multiply the first equation by 2 to get the coefficient of x are equal to each other now for the second question 3x plus 4y is equal to 10 and 2x minus 2y is equal to 2 again we want to equate the coefficients of x so multiply the first equation by 2 that is coefficient of x in second equation and multiply the second equation by coefficient of x in first equation that is 3 into 2x minus 2y is equal to 2. the equations become 6x plus 8y is equal to 20 and the second equation becomes 6x minus 6y is equal to 6. again the change the sign minus plus minus 6 positive 6 x negative cancel out 8 by positive 6 by positive 14 y positive is equal to 20 minus 6 that is again 14 so y is equal to 14 upon 14 that is equal to y is equal to 1 now again to get the value of x we will substitute that value of y in the first or second equation so 3 x plus 4 y that is replaced by 1 is equal to 10 so that is 3x plus 4 is equal to 10 and 3x is equal to 6. so x is equal to 6 upon 3 that x is equal to 2 so 2 comma 1 is the solution of second question using the elimination method so this is how we can find the solution of the pair of linear equation using the elimination method the next one is the sum of the digits of a two digit number is nine if nine times the number is equal to twice the number obtained by reversing the order of the digits find the number so this is a question related to two digit numbers so if we know that in two zero number there is a one that is ruled place and the tens place so let digit at one's place is x and the digit at tens place is equal to y so now what is the first condition the sum of the digits sum of the digits is nine the first condition is x plus y is equal to 9 this is the first condition now there is second condition that the 9 times of a number is equal to twice the numbers by obtained by reversing the order now there are two things first of all what is the number number is 10 into tens plus ones that is 10 into tens y plus once this is actual number where one's place is x and ten space is y now if we reverse the order means now at one's place y is there and tens place x is there the new number is the new number is again 10 into tens plus once this is a new number now according to the condition nine times of this number nine times of this number this one old number is equal to twice is equal to twice two times of the number obtained by reversing obtained by reversing the order that is 10 x plus y from here we will get the second equation that is 19 y plus 9 x is equal to 20x plus 2y the equation is that is equal to 11x 2y minus 90y that is equal to minus 88y is equal to 0 now dividing by 11 we will get that equation is x minus 8 y this is the second equation now we are having the two equation we will solve these two questions using the elimination method to get the value of x and y so x plus y is equal to 9 and x minus 8y is equal to 0 so look at here the coefficients of x are already equal to each other so no need to multiply equation by any number so just change the signs so plus x is cancelled by minus x plus y plus 8y that is 9y and 9 minus 0 is 9 so value of y is 1 because 9 upon 9 is 1. now substitute that value by in the this equation that is x plus y that is 1 is equal to 9 we will get x is equal to 8 so x is 8 and y is 1 so the number is 18 because 1 is at 10's place and 8 that is x is on the first place so number is 18. we will also verify our answer using the constraints the first constraint is the sum of digit is 9 look at here 1 plus 8 is 9 the first constraint is ok the number is 18 and the number by reversing is 18 1 now constraint is 9 times of this is equal to 2 times of that look at here we will check 9 times of this is equal to 2 times of this one this is simply 162 this is also 162. yes second constraint is also satisfied here so this means the solution is so this is how we can solve this question the next is taxi charges in a city consist of fixed charges together with the charges of the distance covered find the distance of the 10 kilometer that charge pay 105 rupees and for the journey of 15 kilometer the charge paid is 155 rupees how much does the person have so there are two charges the first is fixed charges and let the fixed charges are fixed charges are x rupees and the variable charges are variable charges are y rupees according to per kilometer okay so now what is the first condition it has to pay for 10 millimeter has to pay 105 rupees now for 10 kilometer simply access the fixed charges that we must have to pay for 10 kilometers the charges are 10 y because for one kilometer the chances are y rupees for 10 kilometer chances are 10 y and the total amount is 1 0 5 rupees this is the first equation now for the second we have to pay 155 rupees for 15 kilometer so again x are the fixed charges and for 15 kilometer charger of 15y and the total amount is 155 this is the second equation now we will solve these two equations using the elimination method so look at here already coefficient so far x are equal to each other so no need to multiply the an equation by any number so simply change the sign minus minus and minus plus nx minus x plus cancel out plus 10 and minus 15 that is minus 5y 105 positive and 155 negative that is equal to 50 negative so value of 5 is minus cancel out with minus 50 upon 5 that is y is equal to 10 now substitute the value of 10 in first equation we will get the value of x so x plus 10 y that is 10 into 10 100 is equal to 1 0 5 so x is equal to 5 that means the fixed charges are 5 rupees and the variable charges are 10 rupees according to per kilometer now this is answer so let's verify the solution for the first 10 kilometers we have to pay one zero five please look at here the fixed charges are five and for 10 kilometers we have to pay 10 tens of 100 and that is equal to one zero five this is correct and for 15 kilometer five ps are fixed charges and for 15 millimeter we have to pay 150 and that is 155 the second condition is also fulfilled so answer is correct now there is a next statement how much we have to pay for the 25 kilometer now we have to travel 25 kilometer for that distance how much we have to pay simply fixed charges x and plus 25k coding 25i that are the variable charges now put the value x is 5 plus 25 into y is 10 that is 5 plus 2 50 that is 255 four 25 kilometer maze for 25 kilometer one has to pay 255 rupees to cover that 25 kilometer this is the how this question you have to deal with the next one is cross multiply method now the fourth and the last method of this chapter is cross multiply method so 2 x plus 3y is equal to 2 and 2x minus y is equal to 4. now this is the first question that we have to solve with the cross multiply method so look at here for that one we have to turn all the terms including constant terms towards left hand side so 2x plus 3y minus 2 is equal to 0 and 2x minus y minus 4 equal to 0 so take all the constant terms towards left hand side and this is the new equations that is 2x plus 3y minus 2 equal to 0 and 2x minus y minus 4 equal to 0 now we have to use the rule 2 3 1 2 this is a rule remember it 2 3 1 2 two means two means coefficient of y second terms three means third term that are the constant terms one means the first term that is the coefficient of x and again two means the coefficient of second terms that are coefficients of y so first of all second terms that are three and minus one that is third term that are the constant terms minus two and minus four i will write it here two three one two second terms this one second terms third terms then first terms two and two and again the second terms that are the three and minus one so this is the pattern how will arrange the term according to 2 1 3 rule now look at here we will do this step x upon something is equal to y or something is equal to 1 upon something now how to use what to use here look at here first of all 3 is multiplied by minus 4 that is equal to minus 12 now we have to put a minus sign here then the cross multiplication is there minus 1 into minus 2 again this minus 2 into 2 then you have to put a minus sign here then the cross multiplication minus 4 into 2 for the third one also same process 2 into minus 1 you have to put minus sign here then cross product that is 3 into 2 and that is equal to minus 12 minus minus plus 2 and this is minus 2 that is equal to y upon this is minus 2 and minus 2 minus 2 into 2 that is minus 4 and this is minus minus plus that is 8 is equal to 1 upon minus 2 minus 6 so x over that is minus 2 minus 12 minus 14 that is minus 4 plus 8 that is 4 and this is 1 upon 1 upon minus 2 minus 6 that is minus 8 again now to get the value of x we will equate first and the third term so look at here to get the value of x we will equate first term that is x upon minus 14 is equal to third of that is 1 upon 8 minus so from here we will get x is equal to 14 by 8 that is 7 by 4 now to get the value of y we will equate second and third term so y upon 4 is equal to minus 1 upon 8 so y is equal to minus 4 upon 8 that is minus 1 by 2 so the solution is 7 by 4 comma minus 1 by 2 this is a required solution now i will repeat the whole process first step you have to bring all the terms constant terms including constant terms towards the left means you have to make the right hand side 0. now second step is you have to arrange the terms according to 2 3 1 2 rule that means here 2 represents the second terms that are coefficients of y 3 represents the third term that are the constant terms one represents the coefficient of x that are first terms and again two represents the second term that are the coefficients of y now after arrangement we will write x x by something is equal to y by up is equal to 1 by something now we have to fill these parts so for that first 3 is multiplied by minus 4 and then minus sign is substituted here then minus 1 is multiplied by minus 2 again minus 2 is multiplied by minus 2 here minus sign we have to insert then minus 4 is multiplied by 2 again repeat the process 2 into multiply by minus 1 we have to insert a minus sign then 2 is multiplied by 3. so the calculation is x upon minus 12 to minus 2 that is x upon minus 14 y upon minus 4 plus 8 that is y upon 4 and the 1 upon minus 2 minus 6 that is 1 upon minus 8. now to get the value of x we will equate first and third member and to get the value of 5 we will equate second and third member and cross multiplying we will get the values of x and y so this is the cross multiply method and the very important method now we will solve the second question using the same method so 3x minus 4y minus 10 equal to 0 and the second is 2x plus 5y minus 2 equal to 0 now we will arrange the term using the rule 2 3 1 2 rule 2 means second terms that is minus 4 and 5. 3 means third term that is minus 10 and minus 2 1 means the first offset of x first from 3 and 2 again 2 means second terms now x upon cross multiplication of this minus 4 multiplied by minus 2 and we have to insert a minus sign here than the cross multiplication 5 is multiplied by minus 10 that is equal to y upon minus 10 is multiplied by 2 and we have to insert a minus sign here then minus 2 is multiplied by 3 that is equal to 1 upon minus that that 3 is multiply by 5 insert minus sign here and 2 is multiply by minus 4 so on calculating look at here this is 8 plus 50 is equal to y upon minus 20 plus 6 is equal to 1 upon 50 plus 8 so x over 58 is equal to y upon minus 14 is equal to 1 upon 23 so to get the value of x we will equate first and third member that is x upon 58 is equal to 1 upon 23 so x is equal to 58 upon 23 this is a solution for value of x now to get the value of y we will equate second and third member that is y upon minus 14 is equal to 1 upon 23 so value of y is minus 14 upon 23 so this is the value of y so this is how we can use the cross multiply method to get the solution for the pair of linear equation in two variable the next one is the coach of cricket team by seven bats and six balls for rupees 3800 and later he buys three bats and five balls four rupees one seven five zero find the cost of each bat and bowl so simple very simple let the cost of batteries extra rupees next cost of bat is rupees x and the cost of one ball is rupees y now according to the first condition he buys seven bags that is seven x because cost of one battery is the extra piece and cost of seven bats is seven x rupees and six balls that is plus six y and the total bill is 3800 rupees and according to the second situation three bats and three x plus five volts that is 5 i and the cost is 7 1 7 5 0 rupees now this is a pair of linear equation 2 variable let's solve this equation using the cross multiply method so that is equal to 7x plus 6y minus 3800 is equal to 0 and the second is 3x plus five i minus one seven five zero is equal to zero now arrange the terms according to two three one two rule the second terms are 6 5 the third terms are minus 38 100 upon that is minus 750 1750 and the first terms are 7 3 and again the second terms are 6 5 so x upon 6 is multiplied by minus 750 that is minus ten thousand five hundred then minus sign is there minus minus plus five is multiplied by thirty eight hundred that is nineteen 000 that is equal to y upon minus 38 in this multiplied by 3 that is 11 hundred then minus sign is there that this that is seven is multiplied by minus r50 minus minus plus that is equal to 12 250 and that is equal to 1 upon 7 5 35 minus that is 6 3's are 80. so look at here this becomes x upon minus 10 500 plus 19 000 that is 8 500 positive that is equal to y upon minus eleven thousand four hundred and twelve thousand two hundred fifty that is eight hundred fifty positive and is equal to one upon 70 now to get the value of x we will equate the first and the third member so look at here x upon 8500 is equal to 1 upon 70 so x is 8500 upon 17 that is x is equal to 500 and so forget the value of y we will equate the second and third member so y is equal to 850 upon 17 that is equal to 50 that means cost of the bat is 500 rupees and cost of the ball is 50 rupees now we will easily verify the answer as cost of seven bats look at here 70 to 500 3500 plus six more six fives are 300 that is 3800 correct three bears that is 1500 plus 5 volts that is 250 rupees and that is equal to 7 15 17 15 rupees yes the answer is correct so this is how we will first convert the statement into two equations that we have to solve the equation using the cross multiply method you can also solve these equations using the elimination method and the substitution method of graphical method you will get the same answer the next one is so 2x plus 3y equal to 11 and 2x minus 4y equal to minus 24 and hence find the value of m such that y is equal to mx plus 3. now very interesting question and look at here this is a question is actually related to three lines that three lines pass each other passing through a single point and the first line is 2x plus 3y is equal to 11 and the second line is 2x minus 4y is equal to minus 24 and the third line is y is equal to m x plus three now this this common point is the solution for all the three lines so now what we have to do from first two lines we will find the solution and we will put that solution in the third line because that solution also satisfied the third equation so that is 2x plus 3y minus 11 is equal to 0 and 2x minus 4y plus 24 equal to 0. now again for the cross multiply method we will arrange the terms according to 2 3 1 2 rule that are the second terms second terms means three and minus four third terms minus 11 and 24 first terms two and two and again second term 3x minus 4 now x upon 24 theta 72 minus insert then minus 1 is plus plus minus minus that is 44. again that is equal to y minus 11 to minus 20 to that minus sign you have we have to insert then 24 2's are 40 8 that is equal to 1 upon this is minus 8 minus sine we have to insert that this is 6 so this is x upon 28 that is y is equal to minus minus plus sine of minus 70 that is minus minus plus so to get the value of a we will equate first and third member that x is equal to 28 upon minus 14 that is equal to minus 2 so value of x is minus 2 to get the value of equa y we will equate second and third member so y is equal to minus 70 upon minus 14 that is equal to 5 so the solution is minus 2 comma 5 this common point is minus 2 comma 5 now this point also satisfy the third equation so we will substitute or substitute value of x by 2 minus 2 and the value of y by 5 so here 5 is equal to m into x that is minus 2 plus 3 that is 5 is equal to minus 2 m plus 3 3 gets subtracted that becomes 2 is equal to minus 2 f this minus 2 is divided we will get m is equal to minus 1 so this is the way how we can solve or we will find the value fab because three lines are passing through a single point so solution of any two is also the solution of third one so using the first two members first two equations and using the cross multiply method we will get the solution of the two equation and that solution also satisfy the third equation and using that we will get m is equal to minus 1 the next one is equations reducible to pair of linear equation in two variables now look at here these two equations are little different from the previous one because the variables are in the denominator so first of all we have to convert this equation in the simple one and then solve so look at here i will write 4 this equation as 4 into 1 by under root x minus 9 into 1 by under root y is equal to minus 1 and this equation has 2 into 1 by under root x plus 3 into 1 by under root y is equal to 2 now if i will put 1 by under root x is equal to m and 1 by under y is equal to n these two equations become 4 m minus 9 and is equal to minus 1 and this equation become 2 m plus 3 and is equal to 2. now look at here these two equations convert into a simple one equations now we have to solve this equation using any of the method so let's try the elimination method 4m minus 3 9 n is equal to minus 1 and 2 m plus 3 n is equal to 2 so as we will try to make the coefficient of m equal so multiplying the second equation by 2 that is equal to 4 m minus nine n is equal to minus one two to the four m plus six n is equal to four now coefficients of m are equal so change the size of the second equation plus four m minus four and cancel out minus 9 minus 6 that is minus 15 n and 1 minus 1 minus 4 that is minus 5 so value of n is 1 by 3 value of n is 1 by 3 here now put that value fed in any of the first equation or second you will get the value of m so 4 m minus 9 n that is 1 by 3 is equal to minus 1 that is 4 m minus 3 is equal to minus 1 and 4 m is equal to minus 1 plus 3 that is m is equal to 2 upon 4 so m is equal to 1 upon 2 now this the solution of these two equation is n is equal to 1 by 3 and m is equal to 1 by 2 but our question is in the variable of x and y so we have to find the final solution in terms of x and y only so these are the this is the solution for m and n now we have to convert this solution in the form of x and y so look at here 1 by under root x that is equal to m here and m is equal to 1 by 2 so 1 by under root x is equal to 1 by 2 now cross multiplying 2 is equal to under root x and squaring that 4 is equal to x so value of x is 4 similarly 1 by under root y that is equal to n and n is 1 by 3 here so this is 1 by 3 cross multiplying 3 is equal to under root y and squaring we will get y is equal to 9 so this is how if if the variable terms are in the denominator this is how we have to deal with such questions we have to convert that question in the simple equations by putting as mu by root x is equal to m and one by under root y is equal to n so this is the method so here this solution happened and for the these equations not for the actual equations of x and y we have to convert that solution in the verb of x and y also at the last the next question is 10 over x plus y plus 2 over x minus y is equal to 4 15 over x plus y minus 5 over x minus y equal to minus 2 this is again the question of same type we have to convert this question in the simple form first of all so we will write it as 10 into 1 over x plus y plus 2 into 1 over x minus y is equal to 4 and this one has 15 into 1 over x plus y minus 5 into 1 over x minus y is equal to minus 2 so put x plus y 1 over x minus y is equal to m and 1 over x minus y is equal to n these two equation reduces to 10 m plus 2 and is equal to 4 and the second equation reduces to 15 m minus 5 and is equal to minus 2. now the questions in variable x and y converts into simple equation with the variable m and so solve this equation using any of the method so let's multiply second equation by two and first by five first by five and second by two to equate the coefficients of and so multiply this by 5 and second equation by 2 the equations become 50 the equation become 50 m plus 10 n is equal to 20 and the second equation become 30 m minus 10 and is equal to minus 4. now look at here coefficients of n are equal and they are also of opposite side so no need to change the size this is directly cancel out this is 80 m is equal to 16 so m is equal to 16 upon 80 that is 1 upon 5 the value of m is 1 upon 5 now substitute that value in the first equation to get the value of n so 10 into 1 by 5 plus 2 n is equal to 4 that will give 2 plus 2 n is equal to 4 2 n is equal to 2 at n is equal to 1 the value of n is 1 and value of m is 1 by 5. now substitute that values here look at here value of m that is 1 by 5 and value of n is 1 so 1 over x plus y is equal to 1 by 5 cross multiplying we will get x plus y is equal to 5 we will get another equation and for the second one 1 over x minus y that is equal to n that is equal to y so we will get x minus y is equal to 1 so we get another pair of equations now we have to solve these pair of equations also x plus y is equal to 5 and x minus y is equal to 1 again coefficients of y are equal and they are opposite side so directly cancel out x plus x 2x and 5 plus 1 that is 6 so x is equal to 3 the value of x is 3 and from first one this is 3 plus y is equal to 5 so y is equal to 2 now the value of x is 3 and value of y is 2 for the this collision look at here carefully the first we will convert the equation into simple equations just putting 1 by x plus y is equal to f and 1 by x minus y is equal to n now from all solving these equations we'll get another pair of equations look at here x plus y is equal to 5 and x minus y is equal to 1 on solving the first pair of equation we will get another pair of equations so only that solving that pair we will get the actual values of x y so this is the how to solve this type of questions the next one is ritu can row downstream 20 kilometer in two hours at upstream four kilometer in towards also find her speed of rowing and still water and the speed of current so this is basically caution of speed and vote water and more so let the speed of growing or still water is x kilometer per hour and the speed of current that is speed of water is y kilometer now there are two cases the one is for downstream and second is for upstream now in downstream the distance is 20 kilometer time is two hours and the speed is x plus y and in case of upstream the distance is four kilometer the time is again two hours and the speed is x minus y so as we know that speed is equal to distance upon time so x plus y is equal to 20 upon 2 that is 10 and here again speed is equal to distance upon time so x minus y is equal to 4 upon 2 that is 2. now there are two pair of linear equation 2 variable let's solve it this is x plus y is equal to 10 and x minus y is equal to 2 plus y and minus y cancel out using the elimination method x plus x that is 2x is equal to 12 so x is equal to 6 now put the value of x6 here 6 minus y is equal to 2 so 6 minus 2 is equal to y y is equal to 4 that means the speed of the speed drawing in still water is 6 km per hour and the speed of the current is 4 kilometer per hour so dear students this is all about the chapter thanks for watching the video please like share and subscribe the channel to watch more videos thank youthat is the pair of linear equation in two variables so first of all a linear equation equation having degree one is called the linear equation and next is two variable two variable means that equation involves two variable that i means x and y and the pair the pair means we are having such two pair of two equations like 5x plus 3y equal to 9 and 7x minus 2y is equal to 6 so look at here this is one of the example of pair of linear equation in two variable so the first of all here there are two equation equation number one and equation number two so this is a pair of linear equation look at here the degree of the equation is 1 the degree is 1 the power of x is 1 power of y is 1 similarly in the second equation also equation the quality signs are there and 2 variable means they are two variable x and y so this is one of the example of pair of linear equation in two variable now in this step chapter we have to deal with such type of equations the next one is solution what is the solution of a pair of linear equation in two variable so a point m comma 1 n is a solution of pair of linear equation in two variable a 1 com x plus b 1 y is equal to c 1 a 2 x plus b 2 y is equal to c 2 if it will satisfy both the such equations simultaneously now what is the meaning of this that a point m comma n is a point that is a solution of the this type of equation if it will satisfy both those equations simultaneously what is the meaning of simultaneously means whatever the solution is that is satisfy both the equations at a time so let's take example if 3 comma 1 solution of x plus y is equal to 4 and x minus y is equal to 2 means we have to check that is 3 comma 1 is the solution of this pair of linear equations yes or no so let's try if if this point is a solution of this system of solution equations that it will satisfy both the equation so let's try this is equation number one and this is equation number two first of all we will try for the question number one equation number one is x plus y is equal to 4 now point is 3 comma 1 so replace x by 3 and y by 1 left hand side we will get 4 and that is equal to right hand side so yes this point satisfy the equation number one now we will try for the question number two so equation number two is x minus y equal to two and the point again point is 3 comma 1 so try for this 3 minus 1 that is 2 and this is also equal to right hand side so that means 3 comma 1 point satisfy both the equation equation number one and equation number two so it is a solution of both the equation so we will say that three comma one is the solution of the pair of linear equation in two variable the next example is is 2 comma 1 solution of 3x minus 2y is equal to 4 and 2x plus 4y is equal to again the one example is here we have to check that is this given point solution of pair of linear question intuitive yes or no so again we will try for the both the equations this is equation number one and this is equation number two that's for first equation 3x minus 2y equal to 4 and the point is 2 comma 1 so easily x is replaced by 2 and the y is replaced by 1 this is 3 into 2 minus 2 into 1 that is 6 minus 2 that is equal to 4 and this is equal to right hand side so that means 2 comma 1 is the solution of first linear equation in 2 variable now let's check for the second equation also that is 2x plus 4y is equal to 2 again replace x by 2 and y by 1 so this is 2 2s of 4 4 was the 4 that is 8 this is not equal to right hand side so 2 comma 1 is solution of first equation but is it is not a solution of second equation so overall 2 comma 1 is not a solution of such pair of linear equation in two variables means for the solution of pair of linear equation in two variable that given point has to satisfy both the equations simultaneously in as in this example it will satisfy first equation but it will not satisfy the second equation so overall it is not a solution of pair of linear equation in two variables so this is how we have to find or we have to check the the given point is solution of pair of linear equation two variable yes or no the next one is now in grade nine as we know that a single linear equation in two variable represents a straight line on the graph so similarly if we are having a pair of linear equation two variable so simply it will represents two straight lines on the graph so so pair of linear equation in two variable represents two straight lines of the graph and there are three possible cases related to the two lines straight lines so first case is at the the two lines intersect each other at a point so these are the two straight lines the line l1 and line l2 and this will intersect at a given point so look at here in this chapter as we know that a single point is a solution if it will satisfy both the equations so that means or in graphically the solution of pair of linear equation two variable is a point that is common to both the lines so look at here in the first case how many common points are there there is only a single common point a one point in common so there is a one solution the solution in a pair of linear equation represents a point that is common to both the equation if there are two common points there is two solutions five common points five solution one common point one solution no common point means there isn't no solution so similarly if there are only three possible cases in the first case when the lines are intersecting to each other there is only one common point so we will say that there is only one common point that why it is having a unique solution means it is having a single solution and the system is called consistent system that means solution is possible for the second case when the lines are parallel when the lines are parallel now look at here between these two lines there is no comma point there is no compromise between these two lines so there is no common point so we will say that there is a no solution because i will repeat solution is the common point to both the equation or the both the straight lines as there is no comma point so there is a no solution and the system is called inconsistent inconsistent system where solution is not possible the third and the last case is when one line is just upon the second line means a client just loosely link is just over here both the line coincides with each other now every point is a common point every point of both the lines is a common point so they are infinite number of points in common so we will say that there are infinite many solution because solution is again a common point so between these two lines there are infinite points are common so they are infinitely many solution again the system is called consistent system because the solutions are possible so i will repeat so what is the solution graphically graphically a solution is a common point to both the straight lines if we are having two lines there are only possible three cases case one case two and case three in first case lines are intersecting where only one common point is there so we are having only unique solution in second case we only no point is common so there is a no solution in third case every point is a common point to both the lines so there are infrared common points so we will say that there are infinitely many solutions the next one is how to find the solution the next ways as we know from the last slide what is the solution solution is a common point to both the equations or the or the straight lines now the content is how to find such solutions so look at here there are four methods according to ncrt grade 10 we have to find the solution of pair of linear equation in two variable the first method is substitution method second is graphical method third is elimination method and the fourth is cos multiply method in this video we will discuss the first two methods that are substitution method and the graphical method so the first method is substitution method so sol x plus y is equal to 5 and 2x minus 3y is equal to 4 using the substitution method so this is a simply this is a pair of linear equation in two variable this is the first equation and this one is the second equation and we have to solve this system of equation using the substitution method so very easy look at here the first equation that is x plus y is equal to 5 and the second equation is 2x minus 3y is equal to 4 what is the first step look at here the first step is we have to find the value of x or y from the equation number one so simply the value of x is equal to 5 minus y this is the value of x from the first equation now we have to put this value in the equation number 2. look at here 2 x is replaced by its value that is 5 minus y minus 3 y is equal to 4 now so this two 5's are 10 2 y's are 2y minus 3y equal to 4 that is 10 minus minus plus that is minus 5y equals equal to 4 10 minus 4 is equal to 5y that is 6 is equal to 5 y and the value of y is 6 upon 5 that is the value of y so what is the first step i will repeat the first step is from first equation we have to find the value of x or y if when we are having the value of x or y we will substitute that value in the question number 2 from there we will get the value of one variable we will get here we will get the value of y that is 6 by 5 again put that value in this equation so we will get x is equal to 5 minus y that is 6 upon 5 on calculation we will get 25 minus 6 x is equal to 19 upon 5 so the overall answer is 19 upon 5 comma 6 0.5 is the solution of this pair of linear equation in two variable now after getting the solution we will easily verify that is the solution correct or not we simply have to substitute these solution in the equations if it will satisfy both the equations this solution is correct so substitute here x plus y is equal to 5 first equation x plus y x is 19 upon 5 plus y that is 6 upon 5 and the answer is 19 plus 6 upon 5 that is 25 upon 5 that is equal to 5 is equal to right hand side our solution will satisfy the first equation easily now let's try for the second equation that is 2x minus 3y is equal to 4 2 into x that is 19 upon 5 minus 3 into y that is 6 upon 5 that is 38 upon 5 minus 18 upon 5 that is equal to 20 upon 5 and that is 4 is equal to right hand side so a solution will also satisfy the question number two so that means our solutions satisfy both the equations simultaneously so this is the correct answer for this system of solution now we are having a single solution so there is this system having the unique solution and the lines are intersecting lines because we are having a single solution so this system is represents a intersecting straight lines now the second question is 3x plus 4y equal to 10 and 2x minus 2y equal to 2 again this is the first equation 3x plus 4y is equal to 10 from here we have to find the value of x or y for the first equation so value of x is that is 10 minus 4 y upon 3 this is the value of x from the first equation now we will substitute this value in the second equation that is 2 into x that is 10 minus 4 y upon 3 minus 2 2y equal to 2 now calculating 20 minus 8y upon 3 minus 6y is equal to 2 i will take calcium here so this is equal to 20 minus 14 y is equal to 6 that is 20 minus 6 is equal to 14 y 14 is equal to 14y we will get y is equal to 1 now we are having the value of one variable that y is equal to 1 we will substitute that value in the equation this one x is equal to 10 minus 4 into 1 upon 3 x x is equal to x is equal to 10 minus 4 that is 6 upon 3 so x is equal to 2 so 2 comma 1 is the solution of this system of equations now again we will check is this is the solution correct so we will set trying to satisfy these both the equations the first equation 3 into x that is 3 into 2 6 plus 4 into y that is 4 into 1 4 the answer is 10 and that is equal to right hand side yes the first equation is satisfy this by the solution try for the second 2x this is 2 into x that is 2 into 2 minus 2 into y that is 1 the answer is 4 minus 2 and the answer is 2 that is equal to right hand side so this solution will satisfy both the equation so our answer is correct that means 2 comma 1 is the solution of this pair of linear equation in two variables so this is the substitution method again i will repeat in substitution method the first step is we have to find the value of x or y from the first equation that is like this then we have to substitute that value of that very value of that variable in the second equation from there after calculating we will get the value of one variable and after substituting that variable in again in this equation we will get the second variable also so this is the overall method that is called substitution method the next one is if we add this is a statement problem if we add 1 to the numerator and subtract 1 from the denominator a fraction becomes 1 it also becomes half if we only add 1 to the denominator what is the fraction so this is a question related to fraction so we have to find the fraction let the nominator is let denominator of the fraction is x and denominator of the fraction is y so overall the fraction is x upon y this is the very basic thing in a fraction there are two parts denominator and denominator as we have to find the fraction that denominator is x and the denominator is y so overall fraction is x upon y now what is the first condition if we add 1 to the numerator if we add 1 to the numerator and subtract 1 from the denominator subtract 1 from the denominator fraction becomes 1 the fraction becomes 1 this is the first condition so just cross multiplying x plus 1 is equal to y minus 1 that is x minus y is equal to minus 2 this is our first equation now what is the second condition it becomes half the fraction become half if we only add 1 to the denominator means there is no change in the denominator there is only we have to add 1 in the denominator the equation becomes half again cross multiplying 2x is equal to y plus 1 and 2x minus y is equal to 1 this is equation number 2. now we are having the a pair of linear equation in 2 variable again using the substitution method we will solve these equations from here we will get the value one variable that is minus 2 plus y now substitute this value in the this is step one step two substitute this value here in the question number 2 that is 2 into x that is minus 2 plus y minus y equal to 1 that is minus 4 plus 2y minus y equal to 1 this is 2y minus y y 4 get added that side is equal to 5 so value of y is 5 again substitute that value here x is equal to minus 2 plus y and y is 5 so x is equal to 3 so value of x is 3 and y is 4 so our fraction is 3 upon 5 so fraction is 3 upon 5 if we want to check the solution we will go through the conditions look at here the first condition is add 1 to the numerator that is 3 plus 1 4 subtract 1 from the denominator 5 minus 1 4 the answer is 1 yes the answer is 1 first condition satisfied the second condition is if there is no change in denominator means 3 is as this and if we add 1 to the denominator means 5 plus 1 6 it becomes half yes it becomes half so uh answer that is 3 upon 5 will satisfy both the conditions so 3.5 is the right correct answer for this statement problem now the next one is five years ago nuri was thrice as old as sonu 10 years later nori will be twice as old as sunu how old the newly and so no now this is a statement problem related to edges so for that we will divide the question in three parts the first is percentages that is future edges and the past ages as we have to find the present edges let the present age of lurie is axiors and the present age of sonu is y years so this is the x and y is the present edges of sonu and nuri respectively now 10 years later 10 years later nori will be twice as old as so no but this is the second condition the first condition is five years ago five years ago means in the past five years in the past so nuri's age five years ago is x minus five five years ago is y minus five the nuri was thrice as old as sonu the nori is as thrice as old as solu means lori is elder than as compared to sonu so the according to condition x minus 5 is equal to as thrice old as sonu this is 3 times y minus 5 because the nuri is already elder then so no to equate them we have to multiply the age of 7 by 3 so from here we will get the equation x minus 5 is equal to 3y minus 15 so that is x minus 3y is equal to minus 10 this is our equation number one from first equation condition future in future 10 years later means means 10 years in the future so lurie's age is x plus 10 and sonu's age is y plus 10 nori will be twice as old as sonu twice as compared to sonu means x plus 10 is equal to 2 times y plus 10 so that becomes x plus 10 is equal to 2y plus 20 that is x minus 2y is equal to 10 this is equation number 2 now we are having the pair of linear equation in two variable so we have to satisfy we have to solve this equation using the substitution method so again from first equation this is step one from first equation we will find the value of x that is minus 10 plus 3 y and we have to substitute that value in the equation number 2 that becomes minus 10 plus 3y minus 2 y is equal to 10 that becomes y is equal to 20 the y is equal to 20 now again substitute that by here x is equal to minus 10 plus 3 into 20 that is equal to minus 10 plus 60 and x is equal to 50 so that means present age of lurie is 50 and the present age of sunday is 20. these are the present edges of nori so no again we have to verify our solution so simply look at here five years in past that age is fifty minus four five that is forty five and here order's age is fifteen and that lures ages three times as solute this is the condition thrice yes first condition fulfilled in 10 years future here nuri is 50 plus 10 that is 68 20 plus 10 30 h here nuri is twice twice as old as so no yes second condition also satisfy means our solution is correct so this is the way if there is a statement question related to edges you have to divide the question like this presentages passages and future ranges and you have to find the conditions the next one is mina went to bank to withdraw rupees 2000 she asked the cashier to give me the notes of 50 and 100 only meena got 25 notes in all find how many notes of 5800 she received means this is a question that a girl meena goes to bank to withdraw 2000 means she total amount she has to withdraw is 2 000 but she asked the cashier that give me the notes of 5800 only and the cashier gave him her 25 notes overall we have to find how many nodes of 50 and 100 each given by cashier to the mean now so let rupees 50 nodes are x and rupees hundred nodes are y nodes and look at here if we are having two nodes of rupees 50 means overall amount is 100 rupees if we are having three nodes of rupees 50 means overall amount is 150 rupees similarly this time we are having x nuts of rupees 50 so amount is 50 x similarly we are having y nodes of 100 rupees so amount is 100 y now the first condition is the cashier gave 25 dots to meter but according to uh there are only x plus y number of nodes because x naughts of rupees 50 and y dots of rupees 100 so we now get x plus y notes but mean now go 25 nodes overall so first condition is x plus y is equal to 25 this is equation number one now look at the amount the amount mina got is 50 x plus 100 y but according to meera she got 2 000 rupees so that means 50x plus 100 y is equal to 2000 this is equation number two now we are having two linear equation two variable so we will solve this using substitution method first of all the equation number 2 is simply divided by 50 that is 50 is a common factor that becomes x plus 2y is equal to 40 this is actually question number two so again from step one we will find the value of x or y from the first equation that x is equal to 25 minus y and substitute that value in the step 2 in the equation number 2 that is x 25 minus y plus 2 y is equal to 40 that is 25 plus y is equal to 40 and y is equal to 40 minus 25 that y is equal to 50 so value 5 is y is 50 to put that value by here x is equal to 25 minus 50 x is equal to 10 so what does this mean this means that she got 10 nodes of rupees 50 and 15 knots of rupees 100 so look at here there are 15 plus 10 25 nodes are there first condition full field now the amount is 15 10 500 this is 1500 and the amount is 2 000 yes second condition also fulfill that means y15 and extra is the correct solution for this statement problem the next one is a landing library has fixed charges per day charges for the first three days is at additional charges for every day thereafter arushi paid rupees 27 for the book kept for seven days and the fixed charges four piece x per day and the additional charges four piece y per day write the linear question representing the above information so this is a question where we have to convert the statement in the linear equation so what what actually is this question about that there are some fixed charges and some charges are that are per day according to per day so let the fixed charges are lacked fixed charges are actually piece and variable charges are variable charges are y rupees per day so look at here arushi paid 27 piece she kept the book for seven days now she kept in the book for seven days but the fixed charges are for first three days fixed charges are for first three days that means for first three days she has to please pay rupees x now after and total number of days are seven and she has to pay repeats x for first three days now the remaining days are four days so for remaining four days she has to pay five rupees per day so the amount is 4 y this is the total amount for seven days that is x rupees for first three days and one piece for each day after three days that means remaining four days the charges are four why so but according to her the total paid she paid that is 27 she paid total 27 rupees for seven days so this is the linear equation representing the this situation again if one more situation is there if she rent a book for five days she paid 21 rupees for five days now how to convert it again for the first three days she will pay extra rupees for first three days now the remaining days that are out of five if three days there are two remaining days for again for remaining two days she have to pay wider piece according to per day that charges are 2y for wire piece per day for many two days she has to pay 2y and the total overall pay she's 21. this is a second equation so here this question is mainly here for how to convert such situation in the pair of linear equation in two variables now you can you are having the two situations and two equations you will easily solve these equations using the substitution method the next one is how to find the solution of pair of linear equation in two variables so basically there are four methods the first method is substitution method second graphical third elimination method and the fourth is cross multiplied method the first two methods are already discussed in the first part of this chapter so now in this video we will discuss about the third and fourth method that are elimination method and the cross multiply method so the third method that is elimination method sol x plus y is equal to 5 and 2x minus y 3y is equal to 4 using the elimination method for this the first of all equations are x plus y is equal to 5 and second equation is 2x minus 3y is equal to 4. now for the elimination method we have to equate two variables of either x or y equal to each other so we want to make the coefficient of x is equal to 2 here so for that we multiply the first equation by 2 and second equation by 1 so that coefficient of x are equal 2x minus 3y is equal to 4 now the equations become 2x plus 2y is equal to 10 and the second equation 2x minus 3y is equal to 4. now look at here coefficient of xr equal 2 and 2. so change the sign of the second equation to solve both plus 2x minus 2x get cancel out plus 2y plus 3y that is 5y 10 minus 4 that is equal to 6 so value of y is equal to 6 upon 5 this is the value of y variable now to find the value of second variable that is x we will substitute that value in any of the equation to get the second value so again x plus y is equal to 5 so put the value of y here that is x plus 6 upon 5 is equal to 5 that is equal to x is equal to 5 minus 6 upon 5 that is equal to 25 minus 6 upon 5 that is equal to 19 upon 5 so solution is 19 upon 5 comma 6 upon 5 so this is the third method that is elimination method to use that method we need that the coefficient of x or the coefficient of y are equal to each other so for that we will multiply the first equation by 2 to get the coefficient of x are equal to each other now for the second question 3x plus 4y is equal to 10 and 2x minus 2y is equal to 2 again we want to equate the coefficients of x so multiply the first equation by 2 that is coefficient of x in second equation and multiply the second equation by coefficient of x in first equation that is 3 into 2x minus 2y is equal to 2. the equations become 6x plus 8y is equal to 20 and the second equation becomes 6x minus 6y is equal to 6. again the change the sign minus plus minus 6 positive 6 x negative cancel out 8 by positive 6 by positive 14 y positive is equal to 20 minus 6 that is again 14 so y is equal to 14 upon 14 that is equal to y is equal to 1 now again to get the value of x we will substitute that value of y in the first or second equation so 3 x plus 4 y that is replaced by 1 is equal to 10 so that is 3x plus 4 is equal to 10 and 3x is equal to 6. so x is equal to 6 upon 3 that x is equal to 2 so 2 comma 1 is the solution of second question using the elimination method so this is how we can find the solution of the pair of linear equation using the elimination method the next one is the sum of the digits of a two digit number is nine if nine times the number is equal to twice the number obtained by reversing the order of the digits find the number so this is a question related to two digit numbers so if we know that in two zero number there is a one that is ruled place and the tens place so let digit at one's place is x and the digit at tens place is equal to y so now what is the first condition the sum of the digits sum of the digits is nine the first condition is x plus y is equal to 9 this is the first condition now there is second condition that the 9 times of a number is equal to twice the numbers by obtained by reversing the order now there are two things first of all what is the number number is 10 into tens plus ones that is 10 into tens y plus once this is actual number where one's place is x and ten space is y now if we reverse the order means now at one's place y is there and tens place x is there the new number is the new number is again 10 into tens plus once this is a new number now according to the condition nine times of this number nine times of this number this one old number is equal to twice is equal to twice two times of the number obtained by reversing obtained by reversing the order that is 10 x plus y from here we will get the second equation that is 19 y plus 9 x is equal to 20x plus 2y the equation is that is equal to 11x 2y minus 90y that is equal to minus 88y is equal to 0 now dividing by 11 we will get that equation is x minus 8 y this is the second equation now we are having the two equation we will solve these two questions using the elimination method to get the value of x and y so x plus y is equal to 9 and x minus 8y is equal to 0 so look at here the coefficients of x are already equal to each other so no need to multiply equation by any number so just change the signs so plus x is cancelled by minus x plus y plus 8y that is 9y and 9 minus 0 is 9 so value of y is 1 because 9 upon 9 is 1. now substitute that value by in the this equation that is x plus y that is 1 is equal to 9 we will get x is equal to 8 so x is 8 and y is 1 so the number is 18 because 1 is at 10's place and 8 that is x is on the first place so number is 18. we will also verify our answer using the constraints the first constraint is the sum of digit is 9 look at here 1 plus 8 is 9 the first constraint is ok the number is 18 and the number by reversing is 18 1 now constraint is 9 times of this is equal to 2 times of that look at here we will check 9 times of this is equal to 2 times of this one this is simply 162 this is also 162. yes second constraint is also satisfied here so this means the solution is so this is how we can solve this question the next is taxi charges in a city consist of fixed charges together with the charges of the distance covered find the distance of the 10 kilometer that charge pay 105 rupees and for the journey of 15 kilometer the charge paid is 155 rupees how much does the person have so there are two charges the first is fixed charges and let the fixed charges are fixed charges are x rupees and the variable charges are variable charges are y rupees according to per kilometer okay so now what is the first condition it has to pay for 10 millimeter has to pay 105 rupees now for 10 kilometer simply access the fixed charges that we must have to pay for 10 kilometers the charges are 10 y because for one kilometer the chances are y rupees for 10 kilometer chances are 10 y and the total amount is 1 0 5 rupees this is the first equation now for the second we have to pay 155 rupees for 15 kilometer so again x are the fixed charges and for 15 kilometer charger of 15y and the total amount is 155 this is the second equation now we will solve these two equations using the elimination method so look at here already coefficient so far x are equal to each other so no need to multiply the an equation by any number so simply change the sign minus minus and minus plus nx minus x plus cancel out plus 10 and minus 15 that is minus 5y 105 positive and 155 negative that is equal to 50 negative so value of 5 is minus cancel out with minus 50 upon 5 that is y is equal to 10 now substitute the value of 10 in first equation we will get the value of x so x plus 10 y that is 10 into 10 100 is equal to 1 0 5 so x is equal to 5 that means the fixed charges are 5 rupees and the variable charges are 10 rupees according to per kilometer now this is answer so let's verify the solution for the first 10 kilometers we have to pay one zero five please look at here the fixed charges are five and for 10 kilometers we have to pay 10 tens of 100 and that is equal to one zero five this is correct and for 15 kilometer five ps are fixed charges and for 15 millimeter we have to pay 150 and that is 155 the second condition is also fulfilled so answer is correct now there is a next statement how much we have to pay for the 25 kilometer now we have to travel 25 kilometer for that distance how much we have to pay simply fixed charges x and plus 25k coding 25i that are the variable charges now put the value x is 5 plus 25 into y is 10 that is 5 plus 2 50 that is 255 four 25 kilometer maze for 25 kilometer one has to pay 255 rupees to cover that 25 kilometer this is the how this question you have to deal with the next one is cross multiply method now the fourth and the last method of this chapter is cross multiply method so 2 x plus 3y is equal to 2 and 2x minus y is equal to 4. now this is the first question that we have to solve with the cross multiply method so look at here for that one we have to turn all the terms including constant terms towards left hand side so 2x plus 3y minus 2 is equal to 0 and 2x minus y minus 4 equal to 0 so take all the constant terms towards left hand side and this is the new equations that is 2x plus 3y minus 2 equal to 0 and 2x minus y minus 4 equal to 0 now we have to use the rule 2 3 1 2 this is a rule remember it 2 3 1 2 two means two means coefficient of y second terms three means third term that are the constant terms one means the first term that is the coefficient of x and again two means the coefficient of second terms that are coefficients of y so first of all second terms that are three and minus one that is third term that are the constant terms minus two and minus four i will write it here two three one two second terms this one second terms third terms then first terms two and two and again the second terms that are the three and minus one so this is the pattern how will arrange the term according to 2 1 3 rule now look at here we will do this step x upon something is equal to y or something is equal to 1 upon something now how to use what to use here look at here first of all 3 is multiplied by minus 4 that is equal to minus 12 now we have to put a minus sign here then the cross multiplication is there minus 1 into minus 2 again this minus 2 into 2 then you have to put a minus sign here then the cross multiplication minus 4 into 2 for the third one also same process 2 into minus 1 you have to put minus sign here then cross product that is 3 into 2 and that is equal to minus 12 minus minus plus 2 and this is minus 2 that is equal to y upon this is minus 2 and minus 2 minus 2 into 2 that is minus 4 and this is minus minus plus that is 8 is equal to 1 upon minus 2 minus 6 so x over that is minus 2 minus 12 minus 14 that is minus 4 plus 8 that is 4 and this is 1 upon 1 upon minus 2 minus 6 that is minus 8 again now to get the value of x we will equate first and the third term so look at here to get the value of x we will equate first term that is x upon minus 14 is equal to third of that is 1 upon 8 minus so from here we will get x is equal to 14 by 8 that is 7 by 4 now to get the value of y we will equate second and third term so y upon 4 is equal to minus 1 upon 8 so y is equal to minus 4 upon 8 that is minus 1 by 2 so the solution is 7 by 4 comma minus 1 by 2 this is a required solution now i will repeat the whole process first step you have to bring all the terms constant terms including constant terms towards the left means you have to make the right hand side 0. now second step is you have to arrange the terms according to 2 3 1 2 rule that means here 2 represents the second terms that are coefficients of y 3 represents the third term that are the constant terms one represents the coefficient of x that are first terms and again two represents the second term that are the coefficients of y now after arrangement we will write x x by something is equal to y by up is equal to 1 by something now we have to fill these parts so for that first 3 is multiplied by minus 4 and then minus sign is substituted here then minus 1 is multiplied by minus 2 again minus 2 is multiplied by minus 2 here minus sign we have to insert then minus 4 is multiplied by 2 again repeat the process 2 into multiply by minus 1 we have to insert a minus sign then 2 is multiplied by 3. so the calculation is x upon minus 12 to minus 2 that is x upon minus 14 y upon minus 4 plus 8 that is y upon 4 and the 1 upon minus 2 minus 6 that is 1 upon minus 8. now to get the value of x we will equate first and third member and to get the value of 5 we will equate second and third member and cross multiplying we will get the values of x and y so this is the cross multiply method and the very important method now we will solve the second question using the same method so 3x minus 4y minus 10 equal to 0 and the second is 2x plus 5y minus 2 equal to 0 now we will arrange the term using the rule 2 3 1 2 rule 2 means second terms that is minus 4 and 5. 3 means third term that is minus 10 and minus 2 1 means the first offset of x first from 3 and 2 again 2 means second terms now x upon cross multiplication of this minus 4 multiplied by minus 2 and we have to insert a minus sign here than the cross multiplication 5 is multiplied by minus 10 that is equal to y upon minus 10 is multiplied by 2 and we have to insert a minus sign here then minus 2 is multiplied by 3 that is equal to 1 upon minus that that 3 is multiply by 5 insert minus sign here and 2 is multiply by minus 4 so on calculating look at here this is 8 plus 50 is equal to y upon minus 20 plus 6 is equal to 1 upon 50 plus 8 so x over 58 is equal to y upon minus 14 is equal to 1 upon 23 so to get the value of x we will equate first and third member that is x upon 58 is equal to 1 upon 23 so x is equal to 58 upon 23 this is a solution for value of x now to get the value of y we will equate second and third member that is y upon minus 14 is equal to 1 upon 23 so value of y is minus 14 upon 23 so this is the value of y so this is how we can use the cross multiply method to get the solution for the pair of linear equation in two variable the next one is the coach of cricket team by seven bats and six balls for rupees 3800 and later he buys three bats and five balls four rupees one seven five zero find the cost of each bat and bowl so simple very simple let the cost of batteries extra rupees next cost of bat is rupees x and the cost of one ball is rupees y now according to the first condition he buys seven bags that is seven x because cost of one battery is the extra piece and cost of seven bats is seven x rupees and six balls that is plus six y and the total bill is 3800 rupees and according to the second situation three bats and three x plus five volts that is 5 i and the cost is 7 1 7 5 0 rupees now this is a pair of linear equation 2 variable let's solve this equation using the cross multiply method so that is equal to 7x plus 6y minus 3800 is equal to 0 and the second is 3x plus five i minus one seven five zero is equal to zero now arrange the terms according to two three one two rule the second terms are 6 5 the third terms are minus 38 100 upon that is minus 750 1750 and the first terms are 7 3 and again the second terms are 6 5 so x upon 6 is multiplied by minus 750 that is minus ten thousand five hundred then minus sign is there minus minus plus five is multiplied by thirty eight hundred that is nineteen 000 that is equal to y upon minus 38 in this multiplied by 3 that is 11 hundred then minus sign is there that this that is seven is multiplied by minus r50 minus minus plus that is equal to 12 250 and that is equal to 1 upon 7 5 35 minus that is 6 3's are 80. so look at here this becomes x upon minus 10 500 plus 19 000 that is 8 500 positive that is equal to y upon minus eleven thousand four hundred and twelve thousand two hundred fifty that is eight hundred fifty positive and is equal to one upon 70 now to get the value of x we will equate the first and the third member so look at here x upon 8500 is equal to 1 upon 70 so x is 8500 upon 17 that is x is equal to 500 and so forget the value of y we will equate the second and third member so y is equal to 850 upon 17 that is equal to 50 that means cost of the bat is 500 rupees and cost of the ball is 50 rupees now we will easily verify the answer as cost of seven bats look at here 70 to 500 3500 plus six more six fives are 300 that is 3800 correct three bears that is 1500 plus 5 volts that is 250 rupees and that is equal to 7 15 17 15 rupees yes the answer is correct so this is how we will first convert the statement into two equations that we have to solve the equation using the cross multiply method you can also solve these equations using the elimination method and the substitution method of graphical method you will get the same answer the next one is so 2x plus 3y equal to 11 and 2x minus 4y equal to minus 24 and hence find the value of m such that y is equal to mx plus 3. now very interesting question and look at here this is a question is actually related to three lines that three lines pass each other passing through a single point and the first line is 2x plus 3y is equal to 11 and the second line is 2x minus 4y is equal to minus 24 and the third line is y is equal to m x plus three now this this common point is the solution for all the three lines so now what we have to do from first two lines we will find the solution and we will put that solution in the third line because that solution also satisfied the third equation so that is 2x plus 3y minus 11 is equal to 0 and 2x minus 4y plus 24 equal to 0. now again for the cross multiply method we will arrange the terms according to 2 3 1 2 rule that are the second terms second terms means three and minus four third terms minus 11 and 24 first terms two and two and again second term 3x minus 4 now x upon 24 theta 72 minus insert then minus 1 is plus plus minus minus that is 44. again that is equal to y minus 11 to minus 20 to that minus sign you have we have to insert then 24 2's are 40 8 that is equal to 1 upon this is minus 8 minus sine we have to insert that this is 6 so this is x upon 28 that is y is equal to minus minus plus sine of minus 70 that is minus minus plus so to get the value of a we will equate first and third member that x is equal to 28 upon minus 14 that is equal to minus 2 so value of x is minus 2 to get the value of equa y we will equate second and third member so y is equal to minus 70 upon minus 14 that is equal to 5 so the solution is minus 2 comma 5 this common point is minus 2 comma 5 now this point also satisfy the third equation so we will substitute or substitute value of x by 2 minus 2 and the value of y by 5 so here 5 is equal to m into x that is minus 2 plus 3 that is 5 is equal to minus 2 m plus 3 3 gets subtracted that becomes 2 is equal to minus 2 f this minus 2 is divided we will get m is equal to minus 1 so this is the way how we can solve or we will find the value fab because three lines are passing through a single point so solution of any two is also the solution of third one so using the first two members first two equations and using the cross multiply method we will get the solution of the two equation and that solution also satisfy the third equation and using that we will get m is equal to minus 1 the next one is equations reducible to pair of linear equation in two variables now look at here these two equations are little different from the previous one because the variables are in the denominator so first of all we have to convert this equation in the simple one and then solve so look at here i will write 4 this equation as 4 into 1 by under root x minus 9 into 1 by under root y is equal to minus 1 and this equation has 2 into 1 by under root x plus 3 into 1 by under root y is equal to 2 now if i will put 1 by under root x is equal to m and 1 by under y is equal to n these two equations become 4 m minus 9 and is equal to minus 1 and this equation become 2 m plus 3 and is equal to 2. now look at here these two equations convert into a simple one equations now we have to solve this equation using any of the method so let's try the elimination method 4m minus 3 9 n is equal to minus 1 and 2 m plus 3 n is equal to 2 so as we will try to make the coefficient of m equal so multiplying the second equation by 2 that is equal to 4 m minus nine n is equal to minus one two to the four m plus six n is equal to four now coefficients of m are equal so change the size of the second equation plus four m minus four and cancel out minus 9 minus 6 that is minus 15 n and 1 minus 1 minus 4 that is minus 5 so value of n is 1 by 3 value of n is 1 by 3 here now put that value fed in any of the first equation or second you will get the value of m so 4 m minus 9 n that is 1 by 3 is equal to minus 1 that is 4 m minus 3 is equal to minus 1 and 4 m is equal to minus 1 plus 3 that is m is equal to 2 upon 4 so m is equal to 1 upon 2 now this the solution of these two equation is n is equal to 1 by 3 and m is equal to 1 by 2 but our question is in the variable of x and y so we have to find the final solution in terms of x and y only so these are the this is the solution for m and n now we have to convert this solution in the form of x and y so look at here 1 by under root x that is equal to m here and m is equal to 1 by 2 so 1 by under root x is equal to 1 by 2 now cross multiplying 2 is equal to under root x and squaring that 4 is equal to x so value of x is 4 similarly 1 by under root y that is equal to n and n is 1 by 3 here so this is 1 by 3 cross multiplying 3 is equal to under root y and squaring we will get y is equal to 9 so this is how if if the variable terms are in the denominator this is how we have to deal with such questions we have to convert that question in the simple equations by putting as mu by root x is equal to m and one by under root y is equal to n so this is the method so here this solution happened and for the these equations not for the actual equations of x and y we have to convert that solution in the verb of x and y also at the last the next question is 10 over x plus y plus 2 over x minus y is equal to 4 15 over x plus y minus 5 over x minus y equal to minus 2 this is again the question of same type we have to convert this question in the simple form first of all so we will write it as 10 into 1 over x plus y plus 2 into 1 over x minus y is equal to 4 and this one has 15 into 1 over x plus y minus 5 into 1 over x minus y is equal to minus 2 so put x plus y 1 over x minus y is equal to m and 1 over x minus y is equal to n these two equation reduces to 10 m plus 2 and is equal to 4 and the second equation reduces to 15 m minus 5 and is equal to minus 2. now the questions in variable x and y converts into simple equation with the variable m and so solve this equation using any of the method so let's multiply second equation by two and first by five first by five and second by two to equate the coefficients of and so multiply this by 5 and second equation by 2 the equations become 50 the equation become 50 m plus 10 n is equal to 20 and the second equation become 30 m minus 10 and is equal to minus 4. now look at here coefficients of n are equal and they are also of opposite side so no need to change the size this is directly cancel out this is 80 m is equal to 16 so m is equal to 16 upon 80 that is 1 upon 5 the value of m is 1 upon 5 now substitute that value in the first equation to get the value of n so 10 into 1 by 5 plus 2 n is equal to 4 that will give 2 plus 2 n is equal to 4 2 n is equal to 2 at n is equal to 1 the value of n is 1 and value of m is 1 by 5. now substitute that values here look at here value of m that is 1 by 5 and value of n is 1 so 1 over x plus y is equal to 1 by 5 cross multiplying we will get x plus y is equal to 5 we will get another equation and for the second one 1 over x minus y that is equal to n that is equal to y so we will get x minus y is equal to 1 so we get another pair of equations now we have to solve these pair of equations also x plus y is equal to 5 and x minus y is equal to 1 again coefficients of y are equal and they are opposite side so directly cancel out x plus x 2x and 5 plus 1 that is 6 so x is equal to 3 the value of x is 3 and from first one this is 3 plus y is equal to 5 so y is equal to 2 now the value of x is 3 and value of y is 2 for the this collision look at here carefully the first we will convert the equation into simple equations just putting 1 by x plus y is equal to f and 1 by x minus y is equal to n now from all solving these equations we'll get another pair of equations look at here x plus y is equal to 5 and x minus y is equal to 1 on solving the first pair of equation we will get another pair of equations so only that solving that pair we will get the actual values of x y so this is the how to solve this type of questions the next one is ritu can row downstream 20 kilometer in two hours at upstream four kilometer in towards also find her speed of rowing and still water and the speed of current so this is basically caution of speed and vote water and more so let the speed of growing or still water is x kilometer per hour and the speed of current that is speed of water is y kilometer now there are two cases the one is for downstream and second is for upstream now in downstream the distance is 20 kilometer time is two hours and the speed is x plus y and in case of upstream the distance is four kilometer the time is again two hours and the speed is x minus y so as we know that speed is equal to distance upon time so x plus y is equal to 20 upon 2 that is 10 and here again speed is equal to distance upon time so x minus y is equal to 4 upon 2 that is 2. now there are two pair of linear equation 2 variable let's solve it this is x plus y is equal to 10 and x minus y is equal to 2 plus y and minus y cancel out using the elimination method x plus x that is 2x is equal to 12 so x is equal to 6 now put the value of x6 here 6 minus y is equal to 2 so 6 minus 2 is equal to y y is equal to 4 that means the speed of the speed drawing in still water is 6 km per hour and the speed of the current is 4 kilometer per hour so dear students this is all about the chapter thanks for watching the video please like share and subscribe the channel to watch more videos thank you\n"