x

x

x

## Ask a Question

• Ask a Question
• Scan a Question
• Post MCQ
• Note: File extension must be of jpg, jpeg, png, bmp format and file size must not exceed 5 MB
x

Hire a Tutor

# Answers and Solutions

## Question:7x-2y=3,11x-3/2y=8 then(x,y)=

Posted by: A.Vijay on: 30 Mar, 2023

(x,y)=(1,2)

7x – 2y = 3 -----  Eq 1

11x- 3y/2 = 8 ----- Eq 2

To solve these kind of two variable equations, we need to eliminate one variable

Coefficient of y in Eq 1 is -2. Coefficient of y in Eq 2 is -3/2

We have make the coefficient of y same in both the equations.

Looking at the equations, Multiply

Eq 1 with 3 such that coefficient of y becomes -6

Eq 2 with 4 such that coefficient of y also becomes -6

So Eq 1 and Eq 2 can be written as,

3(7x – 2y) = 3x3

• 21x – 6y = 9 ----- Eq 3

4(11x – 3y/2) = 4x8

• 44x – 6y = 32 ----- Eq 4

Now subtract Eq 3 from Eq 4 to eliminate y term

44x – 6y = 32

-21x + 6y = 9

---------------------

23x = 23   =>  x = 1

Now, Substitute value of x in Eq 1

Eq 1:   7(1) – 2y = 3

• 7 – 2y = 3
• 2y = 4
• y = 2

Therefore (x,y) = (1,2)

Another way is to eliminate x term.

This can be done by multiplying Eq 1 with 11 and Eq 2 with 7 so that coefficient of x becomes same in both the equations

Eq 1 becomes  11(7x – 2y) = 11 x 3

• 77x – 22y = 33 ----- Eq 5

Eq 2 becomes  7(11x- 3y/2) = 7 x 8

• 77x – 21y/2 = 56 ----- Eq 6

Subtracting Eq 5 from Eq 6

77x – 21y/2 = 56

-77x + 22y = 33

------------------------

22y – 21y/2  = 23

23y = 2 x 23 => y = 2

Substituting y in Eq 1

7x – 2(2) = 3

7x = 7 => x = 1

(x,y) = (1,2)

Verifying x and y values in Eq 1and Eq 2

Eq 1 LHS: 7x – 2y Substitute x = 1 and y = 2

LHS: 7(1) – 2(2)

LHS: 3

RHS = 3

Therefore, for Eq 1, LHS = RHS for (x,y) = (1,2)

Eq 2 LHS: 11x – 3y/2 = 8

LHS: 11(1) – 3(2)/2

LHS: 11-3 = 8

RHS = 8

Therefore, for Eq 2 also LHS = RHS for (x,y) = (1,2)

Hence (x,y) = (1,2) is the correct solution for Eq 1 and Eq 2

Ans. (x, y) = (1, 2)

Multiplying 1st equation by 3 and 2nd equation by 4, and then subtracting the former fro the latter, we get, 23x = 23 .  => x = 1. Then from 1st equation, 2y = 4 .  => y = 2.

7x -  2y       =  3 ............ ....A

11x - (3/2)*y =   8 ................B

Multiplying equation 'B' with 4  and equation ' A' with 3 we get

44x -  6 y  =   32  ..............C

21x -  6y   =      9................D

Subtracting  equation 'D' from equation 'C' ,we get

23x =   23.Therfore  x =  1

putting this value in equation 'A' ,we get  7-2y =  3  or  2y= 4 and y = 2

Therfore x = 1 and y =2

### Post Answer and Earn Credit Points

Get 5 credit points for each correct answer. The best one gets 25 in all.