Find the values of x, y and z, so as to satisfy the equations given below:
x + y + z = 12; x + y – z = 6; x – y + z = 4

Correct option is C

Given:

​x + y + z = 12; x + y – z = 6; x – y + z = 4

Concept Used:

Elimination method

Solution:

​x + y + z = 12  (Eq. 1)

x + y – z = 6  (Eq. 2)

x – y + z = 4  (Eq. 3)

Subtracting (2) from (1), we get

x + y + z - x - y + z = 12 - 6

2z = 6

z = $$\frac{6}{2}=3$$​​

Now, subtracting (3) from (1), we get

x + y + z - x + y - z = 12 - 4

2y = 8

y = $$\frac{8}{2}=4$$​​

Putting the values of y and z directly, we get

x + 3 + 4 = 12

x = 12 - 7 = 5

So, x = 5, y = 4, and z = 3.