# 7.7 Graphing linear inequalities in two variables

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This module is from Elementary Algebra by Denny Burzynski and Wade Ellis, Jr. In this chapter the student is shown how graphs provide information that is not always evident from the equation alone. The chapter begins by establishing the relationship between the variables in an equation, the number of coordinate axes necessary to construct its graph, and the spatial dimension of both the coordinate system and the graph. Interpretation of graphs is also emphasized throughout the chapter, beginning with the plotting of points. The slope formula is fully developed, progressing from verbal phrases to mathematical expressions. The expressions are then formed into an equation by explicitly stating that a ratio is a comparison of two quantities of the same type (e.g., distance, weight, or money). This approach benefits students who take future courses that use graphs to display information.The student is shown how to graph lines using the intercept method, the table method, and the slope-intercept method, as well as how to distinguish, by inspection, oblique and horizontal/vertical lines. Objectives of this module: be able to locate solutions to linear inequalities in two variables using graphical techniques.

## Overview

• Location of Solutions
• Method of Graphing

## Location of solutions

In our study of linear equations in two variables, we observed that all the solutions to the equation, and only the solutions to the equation, were located on the graph of the equation. We now wish to determine the location of the solutions to linear inequalities in two variables. Linear inequalities in two variables are inequalities of the forms:

$\begin{array}{cc}ax+by\le c& ax+by\ge c\\ ax+byc\end{array}$

## Half-planes

A straight line drawn through the plane divides the plane into two half-planes .

## Boundary line

The straight line is called the boundary line .

## Solution to an inequality in two variables

Recall that when working with linear equations in two variables, we observed that ordered pairs that produced true statements when substituted into an equation were called solutions to that equation. We can make a similar statement for inequalities in two variables. We say that an inequality in two variables has a solution when a pair of values has been found such that when these values are substituted into the inequality a true statement results.

## The location of solutions in the plane

As with equations, solutions to linear inequalities have particular locations in the plane. All solutions to a linear inequality in two variables are located in one and only in one entire half-plane. For example, consider the inequality

$2x+3y\le 6$

All the solutions to the inequality $2x+3y\le 6$ lie in the shaded half-plane.

Point $A\left(1,\text{\hspace{0.17em}}-1\right)$ is a solution since

$\begin{array}{l}2x+3y\le 6\hfill \\ 2\left(1\right)+3\left(-1\right)\le 6?\hfill \\ 2-3\le 6?\hfill \\ -1\le 6.\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{True}\hfill \end{array}$

Point $B\left(2,\text{\hspace{0.17em}}5\right)$ is not a solution since

$\begin{array}{l}2x+3y\le 6\hfill \\ 2\left(2\right)+3\left(5\right)\le 6?\hfill \\ 4+15\le 6?\hfill \\ 19\le 6.\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{False}\hfill \end{array}$

## Method of graphing

The method of graphing linear inequalities in two variables is as follows:

1. Graph the boundary line (consider the inequality as an equation, that is, replace the inequality sign with an equal sign).
1. If the inequality is $\le$ or $\ge$ , draw the boundary line solid . This means that points on the line are solutions and are part of the graph.
2. If the inequality is $<$ or $>$ , draw the boundary line dotted . This means that points on the line are not solutions and are not part of the graph.
2. Determine which half-plane to shade by choosing a test point.
1. If, when substituted, the test point yields a true statement, shade the half-plane containing it.
2. If, when substituted, the test point yields a false statement, shade the half-plane on the opposite side of the boundary line.

## Sample set a

Graph $3x-2y\ge -\text{\hspace{0.17em}}4$ .

1. Graph the boundary line. The inequality is $\ge$ so we’ll draw the line solid . Consider the inequality as an equation.

$3x-2y=-4$

 $x$ $y$ $\left(x,\text{\hspace{0.17em}}y\right)$ $\begin{array}{r}\hfill 0\\ \hfill \frac{-4}{3}\end{array}$ $\begin{array}{c}2\\ \\ 0\end{array}$ $\begin{array}{l}\left(0,\text{\hspace{0.17em}}2\right)\hfill \\ \left(\frac{-4}{3},\text{\hspace{0.17em}}0\right)\hfill \end{array}$

2. Choose a test point. The easiest one is $\left(0,\text{\hspace{0.17em}}0\right)$ . Substitute $\left(0,\text{\hspace{0.17em}}0\right)$ into the original inequality.

$\begin{array}{l}3x-2y\ge -4\hfill \\ 3\left(0\right)-2\left(0\right)\ge -4?\hfill \\ 0-0\ge -4?\hfill \\ 0\ge -4.\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{True}\hfill \end{array}$
Shade the half-plane containing $\left(0,\text{\hspace{0.17em}}0\right)$ .

Graph $x+y-3<0$ .

1. Graph the boundary line: $x+y-3=0$ . The inequality is $<$ so we’ll draw the line dotted .

2. Choose a test point, say $\left(0,\text{\hspace{0.17em}}0\right)$ .

$\begin{array}{l}x+y-3<0\hfill \\ 0+0-3<0?\hfill \\ -3<0.\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{True}\hfill \end{array}$
Shade the half-plane containing $\left(0,\text{\hspace{0.17em}}0\right)$ .

Graph $y\le 2x$ .

1. Graph the boundary line $y=2x$ . The inequality is $\le$ , so we’ll draw the line solid .

2. Choose a test point, say $\left(0,\text{\hspace{0.17em}}0\right)$ .

$\begin{array}{l}y\le 2x\hfill \\ 0\le 2\left(0\right)?\hfill \\ 0\le 0.\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{True}\hfill \end{array}$

Shade the half-plane containing $\left(0,\text{\hspace{0.17em}}0\right)$ . We can’t! $\left(0,\text{\hspace{0.17em}}0\right)$ is right on the line! Pick another test point, say $\left(1,\text{\hspace{0.17em}}6\right)$ .

$\begin{array}{l}y\le 2x\hfill \\ 6\le 2\left(1\right)?\hfill \\ 6\le 2.\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{False}\hfill \end{array}$

Shade the half-plane on the opposite side of the boundary line.

Graph $y>2$ .

1. Graph the boundary line $y=2$ . The inequality is $>$ so we’ll draw the line dotted .

2. We don’t really need a test point. Where is $y>2?$ Above the line $y=2!$ Any point above the line clearly has a $y\text{-coordinate}$ greater than 2.

## Practice set a

Solve the following inequalities by graphing.

$-3x+2y\le 4$

$x-4y<4$

$3x+y>0$

$x\ge 1$

## Exercises

Solve the inequalities by graphing.

$y

$x+y\le 1$

$-x+2y+4\ge 0$

$-x+5y-10<0$

$-3x+4y>-12$

$2x+5y-15\ge 0$

$y\le 4$

$x\ge 2$

$x\le 0$

$x-y<0$

$x+3y\ge 0$

$-2x+4y>0$

## Exercises for review

( [link] ) Graph the inequality $-3x+5\ge -1$ .

( [link] ) Supply the missing word. The geometric representation (picture) of the solutions to an equation is called the of the equation.

( [link] ) Supply the denominator: $m=\frac{{y}_{2}-{y}_{1}}{?}$ .

$m=\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}$

( [link] ) Graph the equation $y=-3x+2$ .

( [link] ) Write the equation of the line that has slope 4 and passes through the point $\left(-1,\text{\hspace{0.17em}}2\right)$ .

$y=4x+6$

can someone help me with some logarithmic and exponential equations.
20/(×-6^2)
Salomon
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it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
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Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
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