Q. 76 - Precalculus Enhanced with Graphing Utilities

Question

In Problems 75– 82, algebraically solve each system of equations using any method you wish.

$\left\{\begin{array}{l}2x+8y=-8\\ x+7y=-13\end{array}\right.$

Step-by-Step Solution

Answer

The values of unknown variables are $x=8 and y=-3$

Step 1. Given information

The given equations $\left\{\begin{array}{l}2x+8y=-8\\ x+7y=-13\end{array}\right.$

Step 2. Finding the values of x , y

Consider the equations $2x+8y=-8$ and $x+7y=-13$.To solve these equations,we first convert them into matrix form and then transform it into reduced echelon form

$$\begin{gathered} \left[\begin{array}{cc}2& 8\\ 1& 7\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}-8\\ -13\end{array}\right] \left\{R_1=\frac{R_1}{2}\right\} \\ \left[\begin{array}{cc}1& 4\\ 1& 7\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}-4\\ -13\end{array}\right] \left\{R_2=R_2-R_1\right\} \\ \left[\begin{array}{cc}1& 4\\ 0& 3\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}-4\\ -9\end{array}\right] \left\{R_2=\frac{R_2}{3}\right\} \\ \left[\begin{array}{cc}1& 4\\ 0& 1\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}-4\\ -3\end{array}\right] \left\{R_1=R_1-4R_2\right\} \\ \left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}8\\ -3\end{array}\right] \\ \left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}8\\ -3\end{array}\right] \end{gathered}$$

Thus,we have solved the equations and the values of unknown variables are $x=8 and y=-3$