Q. 14 - Precalculus Enhanced with Graphing Utilities

Question

Verify that the values of the variables listed are solutions of the system of equations.

$$\begin{gathered} \left\{\begin{array}{r}\begin{array}{rcl}4x-z& =& 7\\ 8x+5y-z& =& 0\end{array}\\ -x-y+5z=6\end{array}\right. \\ x=2, y=-3, z=1;\left(2,-3,1\right) \end{gathered}$$

Step-by-Step Solution

Answer

The given values are the solutions because all equations are true for the given values.

Step 1. Given information.

The given equations are:

$\begin{array}{rcl}4x-z& =& 7\\ 8x+5y-z& =& 0\\ -x-y+5z& =& 6\end{array}$

The given solution is $x=2, y=-3, z=1$ or $\left(2,-3,1\right)$.

Step 2. Check the solution.

If all equations are true for the given values, then the values represent the solution.

Substitute $x=2, y=-3, z=1$ in the first equation.

$\begin{array}{rcl}4\left(2\right)-\left(1\right)& =& 8-1\\ & =& 7\end{array}$

So, the first equation is true.

Substitute $x=2, y=-3, z=1$ in the second equation.

$\begin{array}{rcl}8\left(2\right)+5(-3)-\left(1\right)& =& 16-15-1\\ & =& 0\end{array}$

So, the second equation is true.

Substitute $x=2, y=-3, z=1$ in the third equation.

$\begin{array}{rcl}-\left(2\right)-(-3)+5\left(1\right)& =& -2+3+5\\ & =& 6\end{array}$

So, the third equation is true.

Step 3. Conclusion.

The given values are the solutions because all equations are true for the given values.