Q. 10 - Precalculus Enhanced with Graphing Utilities

Question

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

$$\begin{gathered} \left\{\begin{array}{c}2x+\frac{1}{2}y=0\\ 3x-4y=-\frac{19}{2}\end{array}\right. \\ x=-\frac{1}{2}, y=2;\left(-\frac{1}{2},2\right) \end{gathered}$$

Step-by-Step Solution

Answer

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

Step 1. Given information.

The given equations are:

$\begin{array}{rcl}2x+\frac{1}{2}y& =& 0\\ 3x-4y& =& -\frac{19}{2}\end{array}$

The given solution is $x=-\frac{1}{2}, y=2$ or $\left(-\frac{1}{2},2\right)$.

Step 2. Check the solution.

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

Substitute $x=-\frac{1}{2}, y=2$ in the first equation.

$\begin{array}{rcl}2\left(-\frac{1}{2}\right)+\frac{1}{2}\left(2\right)& =& 0\\ -1+1& =& 0\\ 0& =& 0\end{array}$

This statement is true.

Substitute $x=-\frac{1}{2}, y=2$ in the second equation.

$\begin{array}{rcl}3\left(-\frac{1}{2}\right)-4\left(2\right)& =& -\frac{19}{2}\\ -\frac{3}{2}-8& =& -\frac{19}{2}\\ -\frac{19}{2}& =& -\frac{19}{2}\end{array}$

This statement is true.

Step 3. Conclusion.

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