Q. 75 - Precalculus Enhanced with Graphing Utilities

Question

Algebraically solve each system of equations using any method you wish.

$\left\{\begin{array}{l}2x+3y=11\\ 5x+7y=24\end{array}\right.$

Step-by-Step Solution

Answer

The values of x and y are $-5 \& 7$

Step 1.Given information

Given equation are $2x+3y=11 \& 5x+7y=24$

Step 2.Conversion of equations

Matrix form $\left[\begin{array}{cc}2& 3\\ 5& 7\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}11\\ 24\end{array}\right]$

Step 3.Reduction of matrix

Reducing the matrix

$$\begin{gathered} \{R_1\to \frac{R_1}{2}\} \left[\begin{array}{cc}1& 1.5\\ 5& 7\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}5.5\\ 24\end{array}\right] \\ \{R_2\to R_2-5R_1\}\left[\begin{array}{cc}1& 1.5\\ 0& -0.5\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}5.5\\ -3.5\end{array}\right] \\ \{R_2\to -2R_2\}\left[\begin{array}{cc}1& 1.5\\ 0& 1\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}5.5\\ 7\end{array}\right] \\ \{R_1\to R_1-5R_2\}\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}-5\\ 7\end{array}\right] \\ \left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}-5\\ 7\end{array}\right] \end{gathered}$$

Step 4.The solution

The values of x and y are $-5 \& 7$