Q. 46 - Precalculus Enhanced with Graphing Utilities

Question

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

$\left\{\begin{array}{l}-4x+y=0\\ 6x-2y=14\end{array}\right.$

Step-by-Step Solution

Answer

The solution for the system of equations $\left\{\begin{array}{l}-4x+y=0\\ 6x-2y=14\end{array}\right.$is, $x=-7,y=-28$.

Step 1 . Given information

$\left\{\begin{array}{l}-4x+y=0\\ 6x-2y=14\end{array}\right.$

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, $B=\left[\begin{array}{c}0\\ 14\end{array}\right]$.

Substitute the known values in the formula.

$$\begin{gathered} \left[\begin{array}{c}x\\ y\end{array}\right]=X \\ X=\left[\begin{array}{cc}-1& \frac{-1}{2}\\ -3& -2\end{array}\right]\left[\begin{array}{c}0\\ 14\end{array}\right] \\ =\left[\begin{array}{c}-1\left(0\right)-\frac{1}{2}\left(14\right)\\ -3\left(0\right)-2\left(14\right)\end{array}\right] \\ =\left[\begin{array}{c}0-7\\ 0-28\end{array}\right] \\ =\left[\begin{array}{c}-7\\ -28\end{array}\right] \end{gathered}$$

So, the solution is, in ordered pairs $\left(-7,-28\right)$ or $x=-7,y=-28$.