Chapter 3.1, Problem 45E

To determine

To find: the solutions for the given system of equations.

Answer to Problem 45E

  $x=-2,y=-3\text{ and }z=7$

Explanation of Solution

Given information:

The given system of equations is as follows:

  $2x+y+z=0$..........(1)

  $3x-2y-3z=-21$..........(2)

  $4x+5y+3z=-2$..........(3)

Calculation:

Multiply equation (1) by 3 and add the resultant equation with equation (2) to eliminate z.

  $\begin{array}{l}\left(2x+y+z=0\right)3\to \text{ 6}x+3y+3z=0\\ \\ 6x+3y+3z=0\\ 3x-2y-3z=-21\\ \overline{\begin{array}{l}9x+y=-21\\ y=-21-9x\end{array}}\end{array}$

Multiply equation (1) by $-3$ and add the resultant equation with equation (3) to eliminate z.

  $\begin{array}{l}\left(2x+y+z=0\right)\left(-3\right)\to -6x-3y-3z=0\\ \\ -6x-3y-3z=0\\ 4x+5y+3z=-2\\ \overline{\begin{array}{l}-2x+2y=-2\\ -x+y=-1\\ y=x-1\end{array}}\end{array}$

Compare both the values of y and solve x as follows:

  $\begin{array}{c}x-1=-21-9x\\ x+9x=-21+1\\ 10x=-20\\ x=-2\end{array}$

Put the values of x in $y=x-1$ and evaluate as follows:

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

Put the values of x and y in equation (1) and evaluate as follows:

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

Thus, the solution $\left(x,y,z\right)$ is $\left(-2,-3,7\right)$ .