Q. 50 - Precalculus Enhanced with Graphing Utilities

Question

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

$\left\{\begin{array}{l}bx+3y=2b+3\\ bx+2y=2b+2\end{array} b\ne 0\right.$

Step-by-Step Solution

Answer

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

Step 1 . Given information

$\left\{\begin{array}{l}bx+3y=2b+3\\ bx+2y=2b+2\end{array}\right.$

The inverse found in problem $36$ is, $\left[\begin{array}{cc}\frac{-2}{b}& \frac{3}{b}\\ 1& -1\end{array}\right]$.

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

From the given information, $B=\left[\begin{array}{c}2b+3\\ 2b+2\end{array}\right]$.

Substitute the known values in the formula.

$$\begin{gathered} \left[\begin{array}{c}x\\ y\end{array}\right]=X \\ X=\left[\begin{array}{cc}\frac{-2}{b}& \frac{3}{b}\\ 1& -1\end{array}\right]\left[\begin{array}{c}2b+3\\ 2b+2\end{array}\right] \\ =\left[\begin{array}{c}\frac{-2}{b}\left(2b+3\right)+\frac{3}{b}\left(2b+2\right)\\ 1\left(2b+3\right)-1\left(2b+2\right)\end{array}\right] \\ =\left[\begin{array}{c}-4-\frac{6}{b}+6+\frac{6}{b}\\ 2b+3-2b-2\end{array}\right] \\ =\left[\begin{array}{c}2\\ 1\end{array}\right] \end{gathered}$$

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