Q18. - Algebra 2

Question

 Solve each system of equations.

$\begin{array}{c}x+z=7\\ 2y-z=-3 \\ -x-3y+2z=11\end{array}$

Step-by-Step Solution

Answer

The solution of the given equation $x=-2,y=3$ is  and $z=9.$

Step 1 – Use elimination method to make a system of two equations in two variables

$\begin{array}{c}x+z=7\\ 2y-z=-3 \\ -x-3y+2z=11\end{array}$

$\begin{array}{r}\mathrm{....}\left(1\right)\\ \mathrm{....}\left(2\right)\\ \mathrm{....}\left(3\right)\end{array}$

Adding (1) and (3), we get

$\begin{array}{c}\left(x+z-7\right)+\left(-x-3y+2z-11\right)=0\\ x+z-7-x-3y+2z-11=0\\ 3z-3y-18=0\\ 3z-3y=18\\ 3(z-y)=18\\ z-y=6\end{array}$

$\mathrm{....}\left(4\right)$

Step 2 – Solve the system of two equations

Adding (2) and (4), we get

$\begin{array}{c}\left(2y-z+3\right)+\left(z-y-6\right)=0\\ 2y-z+3+z-y-6=0\\ y-3=0\\ y=3\end{array}$

$\mathrm{....}\left(5\right)$

Substituting  in equation (4), we get

$\begin{array}{c}z-y=6\\ z-3=6\\ z=9\end{array}$

Step 3 – Evaluate the value of x

Substituting the value of y and z in (1), we get 

$\begin{array}{c}x+z=7\\ x+9=7\\ x=-2\end{array}$

Thus, the values are $x=-2,y=3$ and $z=9.$