SOLUTION: x+2y-z=3 2x-y+z=22 x+3y-z=4 x,y,z

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Question 1134190: x+2y-z=3
2x-y+z=22
x+3y-z=4
x,y,z

Found 2 solutions by MathLover1, greenestamps:
Answer by MathLover1(20850)   (Show Source): You can put this solution on YOUR website!

.....eq.1
....eq.2
.....eq.3
start with
.....eq.1
.....eq.3
---------------------------------subtract






....eq.2.... plug in

........solve for
........eq.2a
.....eq.3...plug in

.....solve for

....eq.3a
from eq.2a and eq.3a we have




go to ....eq.3a plug in value


solutions: ,,


Answer by greenestamps(13200)   (Show Source): You can put this solution on YOUR website!


There are always a huge number of different ways you could go about solving a system of equations like this. Unless you have a huge amount of experience solving them, you don't know what paths are going to make the solution easy and which will make it a mess.

The solution by tutor @MathLover1 starts by observing that the first and third equations have the same coefficients for x and z, so subtracting one equation from the other immediately allows you to solve for y. But the path from there to the solution turns out to be a bit messy.

Usually, with a system of three linear equations in three variables, you look for ways to eliminate one variable at a time. The easiest variable to eliminate first is nearly always the one that has the "least complicated" coefficients in the three equations.

In your example, with "-z" in two of the equations and "+z" in the other, the path to the solution is probably going to be easiest if we eliminate z first.

So add the first and second equations to eliminate z, and do the same with the second and third equations. The two resulting equations are




It turns out this path makes the solution VERY easy, because subtracting one of these equations from the other gives us y=1.

Then substituting y=1 in either of those two equations gives x=8; then substituting x=8 and y=1 in any of the original equations gives z=7.

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