Chapter 6.5, Problem 16PPS

To determine

To find: System of linear equations and solution.

Answer to Problem 16PPS

Linear equations are

  $\begin{array}{l}2x-3y=18\\ 8x-3y+12=0\end{array}$

solution is $\left(-5,-\frac{28}{3}\right)$

Explanation of Solution

From the given table taking two $\left(x,y_1\right)$ points

We get

  $\left(-6,-10\right)$ And $\left(0,-6\right)$

Lines formed by above points are

  $\begin{array}{l}y-\left(-10\right)=\frac{-6-\left(-10\right)}{0-\left(-6\right)}\left(x-\left(-6\right)\right)\\ \Rightarrow y+10=\frac{10-6}{0+6}\left(x+6\right)\\ \Rightarrow y+10=\frac{4}{6}\left(x+6\right)\\ \Rightarrow 3y+30=2x+12\\ \Rightarrow 2x-3y=18\end{array}$

From the given table taking two $\left(x,y_1\right)$ points

We get

  $\left(-6,-12\right)$ And $\left(0,4\right)$

Lines formed by above points are

  $\begin{array}{l}y-\left(-12\right)=\frac{4-\left(-12\right)}{0-\left(-6\right)}\left(x-\left(-6\right)\right)\\ \Rightarrow y+12=\frac{4+12}{0+6}\left(x+6\right)\\ \Rightarrow y+12=\frac{16}{6}\left(x+6\right)\\ \Rightarrow 3y+36=8x+48\\ \Rightarrow 8x-3y+12=0\end{array}$

Therefore,

Linear equations are

  $\begin{array}{l}2x-3y=18\\ 8x-3y+12=0\end{array}$

On graphing both linear equations

We get

  

Here blue line represents the graph of $8x-3y+12=0$ and red line represents the graph of $2x-3y=18$ .

Therefore,

Point of intersection of both lines is the solution that is $\left(-5,-\frac{28}{3}\right)$