```Question 622962
Hi, there--
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Here is how to check your answer. If you got it right, the values you found for x, y, and z will make every equation in the system true.
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Just substitute -2 for x, 3 for y, and -4 for z in each equation.

x+y+z=-3
(-2)+(3)+(-4)=-3
-3=-3
Check!!
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x-y+3z=-13
(-2)-(3)+3(-4)=-13
-2-3-12=-13
-17=-13
Oops...Your solution doesn't work in this equation.
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x=-2,y=3,z=-4
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We can see that there is a problem with your calculations because the solution must work in every equation.
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Let's start with the elimination method. Multiply every term in  equation {A} by -4.
{A} x+y+z=-3  ---------------> -4x-4y-4z=12
{B} x-y+3z=-13
{C} 4x+4y+4z=-15 -----------> 4x+4y+4z=-15
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-4x-4y-4z=12
4x+4y+4z=-15
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Since the variables all cancel out (-4x+4x=0, etc.) we have
0=-3
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This result tells us that these two equations have no points in common. In 2-D, this is analogous to parallel lines. Your system has no single point (x,y,z) solution.
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Here are some pictures of what solutions to systems of equations with three variables can look like:
http://www.mathwarehouse.com/algebra/planes/systems/three-variable-equations.php
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