Chapter 8.1, Problem 63E

To determine

To Solve: The system of equations and check the solution.

Answer to Problem 63E

  $x=3,y=1,z=1$

Explanation of Solution

Given:

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

Given:

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

Isolate

  $\begin{array}{l}2x-2y+z=5\\ x=\frac{5+2y-z}{2}\end{array}$

Substitute $x=\frac{5+2y-z}{2}$ in second equation

We get

  $\begin{array}{l}-2x+3y+2z=-1\\ -2\left(\frac{5+2y-z}{2}\right)+3y+2z=-1\\ \frac{5+2y-z}{2}-4y+5z=4\\ y+3z-5=-1\\ \frac{-6y+9z+5}{2}=4\end{array}$

Isolate $y+3z-5=-1$

We get

  $y=-3z+4$

Substitute $y=-3z+4$ in $\frac{-6y+9z+5}{2}=4$

  $\begin{array}{l}\frac{-6(-3z+4)+9z+5}{2}=4\\ \frac{27z-19}{2}=4\\ 27z-19=8\\ 27z=8+19\\ 27z=27\\ z=1\end{array}$

Now

For $y=-3z+4$

Substitute $z=1$

  $\begin{array}{l}y=-3z+4\\ y=-3(1)+4\\ y=1\end{array}$

For $x=\frac{5+2y-z}{2}$

Substitute $y=1$

  $\begin{array}{l}x=\frac{5+2(1)-1}{2}\\ x=3\end{array}$

Thus

The solutions of the system are

  $x=3,y=1,z=1$

Check:

Substitute $x=3,y=1,z=1$ in equations

  $\begin{array}{l}(i)2x-2y+z=5\\ 2(3)-2(1)+1=5\\ 6-2+1=5\\ 5=5\\ (ii)-2x+3y+2z=-1\\ -2(3)+3(1)+2(1)=-1\\ -6+3+2=-1\\ -1=-1\\ (iii)x-4y+5z=4\\ 3-4(1)+5(1)=4\\ 3-4+5=4\\ 4=4\end{array}$

Hence, $x=3,y=1,z=1$