Chapter EP, Problem 3.1.5EP

To determine

Solve the given system of equation by graphing or making table

Answer to Problem 3.1.5EP

As the graph shows, both lines are parallel, So the given system of equation intersects at $\left(5,3\right)$ .

Explanation of Solution

Given information:

System of equation:

  $\begin{array}{l}2x-y=7\\ \frac{2}{5}x-\frac{4}{3}y=-2\end{array}$

Formula used:

A line can be plotted by knowing two points on line. To know these two points any one value from x or y is put on equation and other is obtained.

  $2x-y=7$

Putting $x=0$ ,

  $\begin{array}{l}2x-y=7\\ 2\left(0\right)-y=7\\ -y=7\\ y=-7\end{array}$

First point is $\left(0,-7\right)$

Putting $y=0$ ,

  $\begin{array}{l}2x-y=7\\ 2x-y\left(0\right)=7\\ 2x=7\\ x=\frac{7}{2}\end{array}$

second point is $\left(\frac{7}{2},0\right)$

The line $2x-y=7$ passes from $\left(0,-7\right)$ and $\left(\frac{7}{2},0\right)$ ,

  $\frac{2}{5}x-\frac{4}{3}y=-2$

Putting $x=0$ ,

  $\begin{array}{l}\frac{2}{5}x-\frac{4}{3}y=-2\\ \frac{2}{5}\left(0\right)-\frac{4}{3}y=-2\\ -\frac{4}{3}y=-2\\ y=\frac{3}{2}\end{array}$

First point is $\left(0,\frac{3}{2}\right)$

Putting $y=0$ ,

  $\begin{array}{l}\frac{2}{5}x-\frac{4}{3}y=-2\\ \frac{2}{5}x-\frac{4}{3}\left(0\right)=-2\\ \frac{2}{5}x=-2\\ x=-5\end{array}$

second point is $\left(-5,0\right)$

The line $\frac{2}{5}x-\frac{4}{3}y=-2$ passes from $\left(-5,0\right)$ and $\left(0,\frac{3}{2}\right)$ ,

Now, plotting graphs with the points located on them,

So the given system of equation intersects at $\left(5,3\right)$ .