Chapter 9.5, Problem 45WE

To determine

To find: The value of $x,y\text{ and }z$ by using multiplication with the addition-or-subtraction method.

Answer to Problem 45WE

The value of $x$ is $2$ , the value of $y$ is $3$ , and the value of $z$ is $-1$ .

Explanation of Solution

Given information:

The given equations are,

  $\begin{array}{c}x+y+z=4\\ 2x-y+z=0\\ x-y+2z=-3\end{array}$

Calculation:

Calculate the equations,

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

Add equation $\left(1\right)\&\left(2\right)$ .

  $\begin{array}{c}x+y+z=4\\ 2x-y+z=0\end{array}$

  $3x+2z\text{ = 4}$

Solve the result,

  $3x+2z=\mathrm{4..............................}\left(4\right)$

Add equation $\left(1\right)\&\left(3\right)$ .

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

  $2x+3z\text{ = 1}$

Solve the result,

  $2x+3z=\mathrm{1..............................}\left(5\right)$

Multiply by $3$ in equation $\left(4\right)$ .

  $\begin{array}{c}3\left(3x+2z\right)=3\left(4\right)\\ 9x+6z=\mathrm{12......................}\left(6\right)\end{array}$

Multiply by $2$ in equation $\left(5\right)$ .

  $\begin{array}{c}2\left(2x+3z\right)=2\left(1\right)\\ 4x+6z=\mathrm{2.......................}\left(7\right)\end{array}$

Subtract equation $\left(7\right)$ from equation $\left(6\right)$ .

  $\begin{array}{l}9x+6z=12\\ 4x+6z=2\\ ---\end{array}$

  $5x\text{ = 10}$

Solve the result,

  $x=2$

Substitute the value of $x$ in equation $\left(6\right)or\left(7\right)$ to get $z$ .

  $\begin{array}{c}4\left(2\right)+6z=2\\ z=-1\end{array}$

Substitute the value of $x\&z$ in original equation get $y$ .

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

Therefore,

  $\begin{array}{l}x=2\\ y=3\\ z=-1\end{array}$

Therefore, the value of $x$ is $2$ , the value of $y$ is $3$ , and the value of $z$ is $-1$ .