How do you solve this system of equations: x- y = 6 , 2y + 5z = 1, and 3x - 4z = 8?

2 Answers
Nov 3, 2017

x=4, y=-2 and z=1

Explanation:

We first eliminate z from the latter two equations. This can be done by multiplying second equation by 4 and third equation by 5 and adding the two together.

4xx(2y+5z)+5x(3x-4z)=4xx1+5xx8

or 8y+20z+15x-20z=4+40

or 15x+8y=44 ..........................(A)

Multiplying x-y=6 by 8 we get

8x-8y=48 ..........................(B)

Adding A and B, we get

23x=92 i..e. x=4

and putting this in first equation 4-y=6 i.e. y=-2

and putting value of y in 2y+5z=1, we get

-4+5z=1 or 5z=5 i.e. z=1

The solutions are x=4,y=(-2),z=1

Explanation:

the given equations are
x-y=6->(1)
2y+5z=1->(2)
3x-4z=8->(3)
from (1)
rArry=x-6
put y=x-6 in (2) we get
2x+5z=13->(4)
multiply (4) by 3 and (3) by 2
we get 6x+15z=39->(5) and 6x-8z=16->(6)
solving (5) and (6) we get z=1 put z=1 in (4) we get x=4
put x=4 in (1) we get y=-2