Q17. - Algebra 2

Question

Solve each system of equations.

$\begin{array}{c}x+y+z=-1\\ 2x+4y+z=1 \\ x+ 2y-3z=-3\end{array}$

Step-by-Step Solution

Answer

The solution of the given equation is $x=-4,y=2$and $z=1.$

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

$\begin{array}{c}x+y+z=-1\\ 2x+4y+z=1 \\ x+ 2y-3z=-3\end{array}$

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

Multiplying equation (1) by 2, we get

$2x+2y+2z=-2$

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

Subtracting (1) from (4), we get

$\begin{array}{c}\left(2x+2y+2z+2\right)-\left(2x+4y+z-1\right)=0\\ 2x+2y+2z+2-2x-4y-z+1=0\\ -2y+z+3=0\\ 2y-z-3=0\\ 2y-z=3\end{array}$

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

Subtracting (3) from (1), we get

$\begin{array}{c}\left(x+y+z+1\right)-\left(x+2y-3z+3\right)=0\\ x+y+z+1-x-2y+3z-3=0\\ -y+4z-2=0\\ y-4z+2=0\\ y-4z=-2\end{array}$

       $\mathrm{....}\left(6\right)$

Step 2 – Solve the system of two equations

Multiplying (6) by 2, we get

$2y-8z=-4$

$\mathrm{....}\left(7\right)$

Subtracting (7) from (5), we get

$\begin{array}{c}\left(2y-z-3\right)-\left(2y-8z+4\right)=0\\ 2y-z-3-2y+8z-4=0\\ 7z-7=0\\ 7z=7\\ z=1\end{array}$

Substituting the value of in (5), we get   

$\begin{array}{c}2y-z=3\\ 2y-1=3\\ 2y=3+1\\ 2y=4\\ y=2\end{array}$

Step 3 – Evaluate the value of x

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

$\begin{array}{c}x+y+z=-1\\ x+2+1=-1\\ x+3=-1\\ x=-1-3\\ x=-4\end{array}$

Thus, the values are $x=-4,y=2$and $z=1.$