91Ó°ÊÓ

We have been given equation

$\begin{array}{l}5x+2y+z=5…..\left(1\right)\\ -3x-y+2z=6…..\left(2\right)\\ 2x+3y-3z=5…\left(3\right)\end{array}$

Multiply equation $2$with $2$

$\begin{array}{l}2(-3x-y+2z)=2\left(6\right)\\ -6x-2y+4z=12…\left(4\right)\end{array}$

Lets consider equations $1$and $4$

Add both the equation:

We get:

$-x+5z=17...\left(5\right)$

Multiply equation $2$with $3$

localid="1644476065194" $\begin{array}{l}3(-3x-y+2z)=3\left(6\right)\\ -9x-3y+6z=18..................\left(6\right)\end{array}$

From equations $3$ and $6$

localid="1644476607766" $\begin{array}{c}2x+3y-3z=5\\ \frac{-9x-3y+6z=18}{-7x+3z=23}\\ -7x+3z=23…\left(7\right)\end{array}$

Multiply equation $5$with -$7.$

$\begin{array}{l}-7(-x+5z)=-7\left(17\right)\\ 7x-35z=-119.................\left(8\right)\end{array}$

In order to get opposite coefficients oflocalid="1644476782615" $x$multiply equation 2 with localid="1644476773997" $-7$

Consider equations localid="1645252459553" $7$and localid="1645252391399" $8$

and solve for localid="1645252543588" $z$

localid="1644476797962" $\begin{array}{c}-7x+3z=23\\ 7x-35z=-119\\ -32z=-96\\ 32z=96\\ z=\frac{96}{32}\\ =3\end{array}$

Substitute localid="1644476825823" $z=3$in the equation

localid="1644476816932" $-x+5z=17$and solve for localid="1645252381659" $x$

localid="1645252184150" $$\begin{gathered} -x+5z=17 \\ -x+5\left(3\right)=17 \\ -x+15=17 \\ -x=17-15 \\ -x=2 \\ x=-2 \end{gathered}$$

Now we will substitutelocalid="1645252201161" $x=-2,$localid="1644476807751" $z=3$in the equation localid="1644476791031" $5x+2y+z=5$and solve for localid="1645252443741" $y$

localid="1645252686964" $$\begin{gathered} 5x+2y+z=5 \\ 5(-2)+2y+3=5 \\ -10+2y+3=5 \\ 2y-7=5 \\ 2y=5+7 \\ 2y=12 \\ y=6 \end{gathered}$$

The solution is $(-2,6,3)$