## Elimination method questions 3

In this page elimination method questions 3 we are going to see solution of third problem from the worksheet elimination method worksheet.

Question 3

Solve the following system of linear equations by elimination-method

x + (y/2) = 4

(x/3) + 2 y = 5

First we have to make the given equations in the form of a x + by = c.For that we have to take L.C.M

2 x + y = 4 x 2

2 x + y = 8

x + 6 y = 5 x 3

x + 6 y = 15

2 x + y = 8    --------- (1)

x + 6 y = 15   --------- (2)

There are two unknowns in the given equations. By solving these equations we have to find the values of x and y. For that let us consider the coefficients of x and y in both equation. In the first equation we have + y and in the second equation also we have + 6y and the symbols are  same so we have to subtract them for eliminating the variable y.

For that we have to multiply the first equation by 6 =>12x+6y=48

Subtract the second equation from first equation

12 x + 6 y = 48

x + 6 y = 15

(-)     (-)    (-)

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

11 x = 33

x = 33/ 11

x = 3

now we have to apply the value of x in either given equations to get the value of another variable y

Substitute x = 3 in the second equation we get

3 + 6 y = 15

6 y = 15 - 3

6 y = 12

Y =12/6

y =2

Solution:

x = 3

y = 2

verification:

2 x  + y = 8

2(3) + 2 = 8

6 + 2 = 8

8 = 8   elimination method questions 3  elimination method questions 3

## Need to try other questions

Solve each of the following system of equations by elimination-method.

• x + 2 y = 7 , x – 2 y = 1     Solution
• 3 x + y = 8 , 5 x + y = 10    Solution
• 11 x - 7 y = x y , 9 x - 4 y = 6 x y   Solution
• 3/y + 5/x = 20/x y , 2/x + 5/ y = 15/x y   Solution
• 8 x – 3 y = 5 x y , 6 x – 5 y = -  2 x y    Solution
• 13 x+ 11 y = 70 ,11 x + 13 y = 74     Solution
• 65 x – 336 y = 97 , 33 x – 65 y = 1     Solution
• 15/ x + 2/y = 17 , 1/x + 1/y = 36/5  Solution
• 2/x + 2/3y  = 1/6,3/x + 2/y = 0    Solution