Chapter 2.2, Problem 4CFU

To determine

To solve: the system of equations.

Answer to Problem 4CFU

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

Explanation of Solution

Given information:

  $\begin{array}{c}4x+2y+z=7\mathrm{...}\text{(1)}\\ 2x+2y-4z=-4\mathrm{...}\text{(2)}\\ x+3y-2z=-8\mathrm{...}\text{(3)}\end{array}$

Calculation:

Divide equation (2) by 2 to get,

  $\begin{array}{l}\frac{2x+2y-4z}{2}=\frac{-4}{2}\\ x+y-2z=-2\mathrm{...}\left(4\right)\end{array}$

Use elimination method on equation (3) and (4) to eliminate x as follows:

  $\begin{array}{l}x+y-2z=-2\\ x+3y-2z=-8\\ --\text{ + + }\\ \overline{\text{ 0}-2y+0=6}\\ \\ -2y=6\\ y=\frac{6}{-2}\\ =-3\end{array}$

Multiply equation (2) by 2 to get,

  $\begin{array}{l}\left(2x+2y-4z=-4\right)2\\ 4x+4y-8z=-8\mathrm{...}\left(5\right)\end{array}$

Use elimination method with equation (5) and (1) to eliminate x as follows:

  $\begin{array}{l}4x+4y-8z=-8\\ 4x+2y+z=7\\ ----\\ \overline{\text{ 0 +}2y-9z=-15}\\ \\ 2y-9z=-15\\ 2\left(-3\right)-9z=-15\left(\text{Substitute value of }y\right)\\ -6-9z=-15\\ -9z=-15+6\\ -9z=-9\\ z=\frac{-9}{-9}\\ =1\end{array}$

Substitute the value of y and z in equation (3) to get,

  $\begin{array}{l}x+3y-2z=-8\\ x+3\left(-3\right)-2\left(1\right)=-8\left(\text{Substitute the value of }y\text{ and }z\right)\\ x-9-2=-8\\ x-11=-8\\ x=11-8\\ x=3\end{array}$

The values of x, y and z are $3,-3$ and $1$ respectively.