### Home > CCA2 > Chapter 6 > Lesson 6.1.4 > Problem6-51

6-51.

Use the algebraic strategies you developed in today's lesson to solve the system of equations below. Be sure to check your solution.

$2x+y−3z=−12\\ 5x−y+z=11\\ x+3y−2z=−13$

Choose a pair of equations that can be added together to eliminate a variable.

$\ \ \ {2x+y-3z=-12}\\ \underline{+5x-y+\ \ z=\ \ \ 11}\\ \ \ \ 7x\ \ \ \ \ \ \ -2z=-\ \ 1$

Choose a different pair of equations to add together to eliminate the same variable.
In this case, you will have to multiply each equation by a constant.

$\ \ \ {-3(2x+\ \ y-3z=-12)}\\ \underline{\ \ \ \ \ \ \ \ \ \ \ \ x+3y-2z=-13}\\ \ \ \ \ \ -5x\ \ \ \ \ \ \ \ \ +7z=\ \ \ 23$

Solve the system of equations you created in steps $1\text{ and}\ 2$ for either $x\ \text{or}\ z$.

$\ \ \ {5(\ \ \ 7x-2z=-1)}\rightarrow\ \ \ \ 35x-10z=-5\\ \underline{+7(-5x+7z=\ 23)}\rightarrow\ \underline{-35x+49z=161}\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad 39z=156\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad z=4$

Solve for $x$ using the two-variable equation you created in either step 1 or step 2.
Then substitute the values for $x\ \text{and}\ z$ into one of the original equations to solve for $y$.

Check your solution in a different equation than the one you used to solve for $y$.