Chapter 11.1, Problem 41AYU

To determine

To determine: The solution of the system of linear equations $\{\begin{array}{l}x+y=9\hfill \\ 2x-z=13\hfill \\ 3y+2z=7\hfill \end{array}$ .

Answer to Problem 41AYU

Solution:

The solution of the system is $x=6,y=3,$ and $z=-1$

Explanation of Solution

Given information:

The system of linear equations $\{\begin{array}{ccccc}x& +& y& =& 9\\ 2x& -& z& =& 13\\ 3y& +& 2z& =& 7\end{array}$

The given system of linear equations is

  $\{\begin{array}{ccccc}x& +& y& =& 9\\ 2x& -& z& =& 13\\ 3y& +& 2z& =& 7\end{array}$ .

To solve this system using elimination method, first multiply equation $x+y=9$ by $2.$

Therefore, the equation becomes $2x+2y=18$

Subtract the equation $2x+2y=18$ from $2x-z=13$ to eliminate the variable.

  $\begin{array}{cccccc}& 2x& -& z& =& 13\\ -& 2x& +& 2y& =& 18\\ & -2y& -& z& =& -5\end{array}$

Then, multiply the equation $-2y-z=-5$ by $3$ and $3y+2z=7$ by $2$

Therefore, the equations become $-6y-3z=-15$ and $6y+4z=14$

Add the equations $-6y-3z=-15$ and $6y+4z=14$ to eliminate the variable $z.$

  $\begin{array}{cccccc}& -6y& -& 3z& =& -15\\ +& 6y& +& 4z& =& 14\\ & & & z& =& -1\end{array}$

Plug $z=-1$ in equation $-6y-3z=-15$ to get the variable $y$ .

Therefore, the equation becomes $-6y-3(-1)=-15$

  $\Rightarrow y=3$

And plug $y=3$ in equation $2x+2y=18$ to get the variable $x.$

Therefore, the equation becomes $2x+2(3)=18$

  $\Rightarrow x=6$ .

Therefore, the solution of the system of equation $\{\begin{array}{ccccc}x& +& y& =& 9\\ 2x& -& z& =& 13\\ 3y& +& 2z& =& 7\end{array}$ is $x=6,y=3,z=-1.$