Question

Determine the values for x,y and z in the following system: 2x+3y-z=5 ; 4x-y-z=-1 ; x+4y+z=12.

Answer 1

x = 1

y = 2

z = 3

Explanation:

Given:

2x + 3y - z = 5

or

z = 2x + 3y - 5 ............(1)

4x - y - z = -1 ..........(2)

x + 4y + z = 12 ...........(3)

substituting 1 in 2, we get

4x - y - ( 2x + 3y - 5 ) = -1

or

4x - y - 2x - 3y + 5 = -1

2x - 4y + 6 = 0

or

x - 2y + 3 = 0

or

x = 2y - 3

substituting x in equation 1 , we get

z = 2( 2y - 3 ) + 3y - 5

or

z = 4y - 6 + 3y - 5

z = 7y - 11

substituting x and z in 3 we get

( 2y - 3 ) + 4y + ( 7y - 11 ) = 12

or

-14 + 13y = 12

or

y = 2

substituting y in z = 7y - 11, we get

z = 7 × 2 - 11

or

z = 3

substituting y in x = 2y - 3

we get

x = 2 × 2 - 3

or

x = 1

thus,

x = 1

y = 2

z = 3