Q. 15 - Precalculus Enhanced with Graphing Utilities

Question

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

$$\begin{gathered} \left\{\begin{array}{c}\begin{array}{rcl}3x+3y+2z& =& 4\\ x-3y+z& =& 10\end{array}\\ 5x-2y-3z=8\end{array}\right. \\ x=2, y=-2, z=2; \left(2,-2,2\right) \end{gathered}$$

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:

$\begin{array}{rcl}3x+3y+2z& =& 4\\ x-3y+z& =& 10\\ 5x-2y-3z& =& 8\end{array}$

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

Step 2. Check the solution.

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

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

$\begin{array}{rcl}3\left(2\right)+3(-2)+2\left(2\right)& =& 6-6+4\\ & =& 4\end{array}$

So, the first equation is true.

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

$\begin{array}{rcl}\left(2\right)-3(-2)+\left(2\right)& =& 2+6+2\\ & =& 10\end{array}$

So, the second equation is true.

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

$\begin{array}{rcl}5\left(2\right)-2(-2)-3\left(2\right)& =& 10+4-6\\ & =& 8\end{array}$

So, the third equation is true.

Step 3. Conclusion.

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