Q. 48 - Precalculus Enhanced with Graphing Utilities

Question

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

$\left\{\begin{array}{l}-4x+y=5\\ 6x-2y=-9\end{array}\right.$

Step-by-Step Solution

Answer

The solution for the system of equations $\left\{\begin{array}{l}-4x+y=5\\ 6x-2y=-9\end{array}\right.$ is, $x=-\frac{1}{2},y=3$.

Step 1 . Given information

$\left\{\begin{array}{l}-4x+y=5\\ 6x-2y=-9\end{array}\right.$

The inverse found in problem $34$ is, $\left[\begin{array}{cc}-1& -\frac{1}{2}\\ -3& -2\end{array}\right]$.

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

From the given information, $B=\left[\begin{array}{c}5\\ -9\end{array}\right]$.

Substitute the known values in the formula.

$$\begin{gathered} \left[\begin{array}{c}x\\ y\end{array}\right]=X \\ X=\left[\begin{array}{cc}-1& -\frac{1}{2}\\ -3& -2\end{array}\right]\left[\begin{array}{c}5\\ -9\end{array}\right] \\ =\left[\begin{array}{c}-1\left(5\right)-\frac{1}{2}\left(-9\right)\\ -3\left(5\right)-2\left(-9\right)\end{array}\right] \\ =\left[\begin{array}{c}-5+\frac{9}{2}\\ -15+18\end{array}\right] \\ =\left[\begin{array}{c}-\frac{1}{2}\\ 3\end{array}\right] \end{gathered}$$

So, the solution is, $x=-\frac{1}{2},y=3$ or in ordered pairs $\left(\frac{-1}{2},3\right)$.