Q. 49 - Precalculus Enhanced with Graphing Utilities

Question

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

$\left\{\begin{array}{l}2x+y=-3\\ ax+ay=-a\end{array}\right. a\ne 0$.

Step-by-Step Solution

Answer

The solution for the system of equations $\left\{\begin{array}{l}2x+y=-3\\ ax+ay=-a\end{array}\right.$is, $x=-2,y=1$.

Step 1 . Given information

$\left\{\begin{array}{l}2x+y=-3\\ ax+ay=-a\end{array}\right. a\ne 0$

The inverse found in problem $35$ is, $\left[\begin{array}{cc}1& -\frac{1}{a}\\ -1& \frac{2}{a}\end{array}\right]$.

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

From the given information, $B=\left[\begin{array}{c}-3\\ -a\end{array}\right]$.

Substitute the known values in the formula.

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

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