How do you solve the system 17x−y+2z=−9, x+y−4z=8, 3x−2y−12z=24?

1 Answer
Shwetank Mauria
Feb 11, 2016

Solution is x=−27, y=0 and z=−2914

Explanation:

These are three linear equations in three variables. To solve them pick two equations and eliminate one variable and then choose another two equations and eliminate the same variable again. This will give us two linear equations in two variables, which can then be solved. The process in case of these equations could be as follows:

Adding first two equations, we can eliminate y as we get

17x−y+2z+x+y−4z=−9+8 i.e.

18x−2z=−1 .................. Equation (A)

We can also eliminate in second and third equation by adding double of second equation in third equation, as follows.

2⋅(x+y−4z)+3x−2y−12z=2⋅8+24 or 5x−20z=40 or dividing by 4,

we get x−4z=8 .................. Equation (B)

To eliminate z, multiply equation (A) by 2 and subtracting (B) from it, which gives us

2⋅(18x−2z)−(x−4z)=−2−8 or 35x=−10 i.e. x=−1035or−27

Putting this in equation (B) we get −27−4z=8 or −4z=8+27=587. This gives z=−2914

Putting values in x and z in equation x+y−4z=8, we can get y, as y=8−x+4z i.e.

y=8−(−27)+4(−2914)=8+27−587

i.e. y=56+2−587 or y=0

Hence solution is x=−27, y=0 and z=−2914

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