Chapter 3.2, Problem 3CYU

To determine

To graph: The inequalities $y>-2x+4$ and $y\le -3x-3$ .

Explanation of Solution

Given information: Graph the linear inequalities $y>-2x+4$ and $y\le -3x-3$ .

Method used: Use slope and y - intercept to graph the lines and find the common region for both inequalities which would be the solution region. Then use $slope=\frac{rise}{run}$ .

Graph:

Consider the inequalities as equation

  $\begin{array}{l}y=-2x+4\\ y=-3x-3\end{array}$

Graph them by finding slope and y intercept by comparing it with $y=mx+b$ .

For the line $y=-2x+4$

  $\begin{array}{l}slope=-2\\ y-\mathrm{int}ercept=4\end{array}$

Firstly represent the y-intercept (0,4)on xy plane then move down 2 units and right 1 unit to get another point because slope is -2 this could be written as $\frac{-2}{1}$ . Then join the points and extend the line on both ends.

This line is dotted line because the inequality has no equal sign (no small line under inequality sign) with it.

Like so graph the second line.

For the line $y=-3x-3$

  $\begin{array}{l}slope=-3\\ y-\mathrm{int}ercept=-3\end{array}$

Now, represent (0,-3) on the xy plane and then move down 3 units and right 1 unit to get another point because slope is -3 which could be written as $\frac{-3}{1}$ .

Then join the points and extend the line on both ends. This line is a solid line because there is a small line under the inequality sign. It means it is less than or equal to.

  

Interpretation: The dark area of the graph is solution region to the inequalities. Ignore the Purple and black areas. Just the combination of these two colors area is solution region.

That region would satisfy the linear inequalities.