Q. 166 - Intermediate Algebra

Question

To solve the system of linear equations by elimination method:

$\begin{array}{l}5x+2y+z=5\\ -3x-y+2z=6\\ 2x+3y-3z=5\end{array}$

Step-by-Step Solution

Answer

The solution is $(-2,6,3)$$(-2,6)$

Step 1. Given information

We have been given equation

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

Step 2. Solving equation

Multiply equation $2$ with $2$

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

Lets consider equations $1$and $4$

Add both the equation:

We get:

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

Multiply equation $2$ with $3$

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

From equations $3$  and $6$

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

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 of$x$ multiply equation 2 with $-7$

Consider equations $7$and  $8$

and solve for $z$

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

Substitute $z=3$in the equation

$-x+5 z=17$ and solve for $x$

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

Now we will substitute$x=-2,$ $z=3$ in the equation $5 x+2 y+z=5$ and solve for $y$

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

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