Q. 47 - Precalculus Enhanced with Graphing Utilities

Question

In Problems $41-60$, use the inverses found in Problems $31-40$ to solve each system of equations.

$\left\{\begin{array}{l}6x+5y=13\\ 2x+2y=5\end{array}\right.$

Step-by-Step Solution

Answer

The solution for the system of equations $\left\{\begin{array}{l}6x+5y=13\\ 2x+2y=5\end{array}\right.$ is, $x=\frac{1}{2},y=2$.

Step 1 . Given information

$\left\{\begin{array}{l}6x+5y=13\\ 2x+2y=5\end{array}\right.$

The inverse found in problem $33$ is, $\left[\begin{array}{cc}1& -\frac{5}{2}\\ -1& 3\end{array}\right]$.

Step 2 . We know that, X = A - 1 . B

From the given information, $B=\left[\begin{array}{c}13\\ 5\end{array}\right]$.

Substitute the known values in the formula.

$$\begin{gathered} \left[\begin{array}{c}x\\ y\end{array}\right]=X \\ X=\left[\begin{array}{cc}1& -\frac{5}{2}\\ -1& 3\end{array}\right]\left[\begin{array}{c}13\\ 5\end{array}\right] \\ =\left[\begin{array}{c}1\left(13\right)-\frac{5}{2}\left(5\right)\\ -1\left(13\right)+3\left(5\right)\end{array}\right] \\ =\left[\begin{array}{c}13-\frac{25}{2}\\ -13+15\end{array}\right] \\ =\left[\begin{array}{c}\frac{1}{2}\\ 2\end{array}\right] \end{gathered}$$

So, the solution is, in ordered pairs $\left(\frac{1}{2},2\right)$ or $x=\frac{1}{2},y=2$.