Chapter 4.2, Problem 48STP

To determine

To find: The solution for the system of equations.

Answer to Problem 48STP

The solution of the system is $x=3,y=-1,z=4$ .

Explanation of Solution

Given:

The given system of equations is,

  $\begin{array}{l}2x+3y-z=-1\\ 5x+y+4z=30\\ -8x-2y+5z=-2\end{array}$

Calculation:

Consider the given inequalities as,

  $\begin{array}{l}2x+3y-z=-1\\ 5x+y+4z=30\\ -8x-2y+5z=-2\end{array}$

Multiply the first equation by 4and secondequation by 4 and then add both the equation as,

  $\begin{array}{l}8x+12y-4z+5x+y+4z=30-4\\ 13x+13y=26\\ x=2-y\end{array}$

Multiply the first equation by 5and add it to the third equation as,

  $\begin{array}{l}10x+15y-5z-8x-2y+5z=-5-2\\ 2x+13y=-7\end{array}$

Then, from $x=2-y$ ,

  $\begin{array}{l}2\left(2-y\right)+13y=-7\\ 4-2y+13y=-11\\ y=-1\end{array}$

Then the value of $x=2-y$ is,

  $\begin{array}{c}x=2-\left(-1\right)\\ =3\end{array}$

Then consider the equation $29x+26y=98$ as,

  $\begin{array}{l}29\left(6y-8\right)+26y=98\\ 174-232+26y=330\\ y=\frac{33}{20}\end{array}$

Consider $x=6y-8$ as,

  $\begin{array}{c}x=6\left(\frac{33}{20}\right)-8\\ =\frac{19}{10}\end{array}$

Consider the equation $2x+3y-z=-1$ as,

  $\begin{array}{l}2\left(3\right)+3\left(-1\right)-z=-1\\ 6-3-z=-1\\ z=4\end{array}$

Thus, the solution of the system is $x=3,y=-1,z=4$ .