# 6.5: Linear Inequalities in Two Variables

**At Grade**Created by: CK-12

## Learning Objectives

- Graph linear inequalities in one variable on the coordinate plane.
- Graph linear inequalities in two variables.
- Solve real-world problems using linear inequalities.

## Introduction

Yasmeen is selling handmade bracelets for $5 each and necklaces for $7 each. How many of both does she need to sell to make at least $100?

A **linear inequality** in two variables takes the form \begin{align*}y > mx+b\end{align*}

When we graph a line in the coordinate plane, we can see that it divides the plane in half:

The solution to a linear inequality includes all the points in one half of the plane. We can tell which half by looking at the inequality sign:

> The solution set is the half plane above the line.

\begin{align*}\ge\end{align*}

< The solution set is the half plane below the line.

\begin{align*}\le\end{align*}

For a strict inequality, we draw a **dashed line** to show that the points in the line *are not* part of the solution. For an inequality that includes the equals sign, we draw a **solid line** to show that the points on the line *are* part of the solution.

Here are some examples of linear inequality graphs. This is a graph of \begin{align*}y \ge mx + b\end{align*}

This is a graph of \begin{align*}y < mx + b\end{align*}

## Graph Linear Inequalities in One Variable in the Coordinate Plane

In the last few sections we graphed inequalities in one variable on the number line. We can also graph inequalities in one variable on the coordinate plane. We just need to remember that when we graph an equation of the type \begin{align*}x = a\end{align*}

**Example 1**

*Graph the inequality \begin{align*}x > 4\end{align*} x>4 on the coordinate plane.*

**Solution**

First let’s remember what the solution to \begin{align*}x > 4\end{align*} looks like on the number line.

The solution to this inequality is the set of all real numbers \begin{align*}x\end{align*} that are bigger than 4, not including 4. The solution is represented by a line.

In two dimensions, the solution still consists of all the points to the right of \begin{align*}x = 4\end{align*}, but for all possible \begin{align*}y-\end{align*}values as well. This solution is represented by the half plane to the right of \begin{align*}x = 4\end{align*}. (You can think of it as being like the solution graphed on the number line, only stretched out vertically.)

The line \begin{align*}x = 4\end{align*} is dashed because the equals sign is not included in the inequality, meaning that points on the line are not included in the solution.

**Example 2**

*Graph the inequality \begin{align*}|x| \ge 2\end{align*}.*

**Solution**

The absolute value inequality \begin{align*}|x| \ge 2\end{align*} can be re-written as a compound inequality:

\begin{align*}x \le -2 \quad \text{or} \quad x \ge 2\end{align*}

In other words, the solution is all the coordinate points for which the value of \begin{align*}x\end{align*} is smaller than or equal to -2 **or** greater than or equal to 2. The solution is represented by the plane to the left of the vertical line \begin{align*}x = -2\end{align*} and the plane to the right of line \begin{align*}x = 2\end{align*}.

Both vertical lines are solid because points on the lines are included in the solution.

**Example 3**

Graph the inequality \begin{align*}|y| < 5\end{align*}

**Solution**

The absolute value inequality \begin{align*}|y| < 5\end{align*} can be re-written as \begin{align*}-5 < y < 5\end{align*}. This is a compound inequality which can be expressed as

\begin{align*}y > -5 \quad \text{and} \quad y < 5\end{align*}

In other words, the solution is all the coordinate points for which the value of \begin{align*}y\end{align*} is larger than -5 **and** smaller than 5. The solution is represented by the plane between the horizontal lines \begin{align*}y = -5\end{align*} and \begin{align*}y = 5\end{align*}.

Both horizontal lines are dashed because points on the lines are not included in the solution.

## Graph Linear Inequalities in Two Variables

The general procedure for graphing inequalities in two variables is as follows:

- Re-write the inequality in slope-intercept form: \begin{align*}y=mx+b\end{align*}. Writing the inequality in this form lets you know the direction of the inequality.
- Graph the line of the equation \begin{align*}y=mx+b\end{align*} using your favorite method (plotting two points, using slope and \begin{align*}y-\end{align*}intercept, using \begin{align*}y-\end{align*}intercept and another point, or whatever is easiest). Draw the line as a dashed line if the equals sign is not included and a solid line if the equals sign is included.
- Shade the half plane above the line if the inequality is “greater than.” Shade the half plane under the line if the inequality is “less than.”

**Example 4**

*Graph the inequality \begin{align*}y \ge 2x-3\end{align*}.*

**Solution**

The inequality is already written in slope-intercept form, so it’s easy to graph. First we graph the line \begin{align*}y=2x-3\end{align*}; then we shade the half-plane above the line. The line is solid because the inequality includes the equals sign.

**Example 5**

*Graph the inequality \begin{align*}5x-2y>4\end{align*}.*

**Solution**

First we need to rewrite the inequality in slope-intercept form:

\begin{align*} -2y & > -5x +4\\ y & < \frac{5}{2}x - 2\end{align*}

Notice that the inequality sign changed direction because we divided by a negative number.

To graph the equation, we can make a table of values:

\begin{align*}x\end{align*} | \begin{align*}y\end{align*} |
---|---|

-2 | \begin{align*}\frac{5}{2}(-2)-2=-7\end{align*} |

0 | \begin{align*}\frac{5}{2}(0)-2=-2\end{align*} |

2 | \begin{align*}\frac{5}{2}(2)-2=3\end{align*} |

After graphing the line, we shade the plane **below** the line because the inequality in slope-intercept form is **less than**. The line is dashed because the inequality does not include an equals sign.

## Solve Real-World Problems Using Linear Inequalities

In this section, we see how linear inequalities can be used to solve real-world applications.

**Example 8**

A retailer sells two types of coffee beans. One type costs $9 per pound and the other type costs $7 per pound. Find all the possible amounts of the two different coffee beans that can be mixed together to get a quantity of coffee beans costing $8.50 or less.

**Solution**

Let \begin{align*}x =\end{align*} weight of $9 per pound coffee beans in pounds.

Let \begin{align*}y =\end{align*} weight of $7 per pound coffee beans in pounds.

The cost of a pound of coffee blend is given by \begin{align*}9x + 7y\end{align*}.

We are looking for the mixtures that cost $8.50 or less. We write the inequality \begin{align*}9x+7y \le 8.50\end{align*}.

Since this inequality is in standard form, it’s easiest to graph it by finding the \begin{align*}x-\end{align*} and \begin{align*}y-\end{align*}intercepts. When \begin{align*}x=0\end{align*}, we have \begin{align*}7y=8.50\end{align*} or \begin{align*}y=\frac{8.50}{7} \approx 1.21\end{align*}. When \begin{align*}y=0\end{align*}, we have \begin{align*}9x=8.50\end{align*} or \begin{align*}x=\frac{8.50}{9} \approx 0.94\end{align*}. We can then graph the line that includes those two points.

Now we have to figure out which side of the line to shade. In \begin{align*}y-\end{align*}intercept form, we shade the area **below** the line when the inequality is “less than.” But in standard form that’s not always true. We could convert the inequality to \begin{align*}y-\end{align*}intercept form to find out which side to shade, but there is another way that can be easier.

The other method, which works for any linear inequality in any form, is to plug a random point into the inequality and see if it makes the inequality true. Any point that’s not on the line will do; the point (0, 0) is usually the most convenient.

In this case, plugging in 0 for \begin{align*}x\end{align*} and \begin{align*}y\end{align*} would give us \begin{align*}9(0)+7(0) \le 8.50\end{align*}, which is true. That means we should shade the half of the plane that includes (0, 0). If plugging in (0, 0) gave us a false inequality, that would mean that the solution set is the part of the plane that does *not* contain (0, 0).

Notice also that in this graph we show only the first quadrant of the coordinate plane. That’s because weight values in the real world are always nonnegative, so points outside the first quadrant don’t represent real-world solutions to this problem.

**Example 9**

*Julius has a job as an appliance salesman. He earns a commission of $60 for each washing machine he sells and $130 for each refrigerator he sells. How many washing machines and refrigerators must Julius sell in order to make $1000 or more in commissions?*

**Solution**

Let \begin{align*}x =\end{align*} number of washing machines Julius sells.

Let \begin{align*}y =\end{align*} number of refrigerators Julius sells.

The total commission is \begin{align*}60x + 130y\end{align*}.

We’re looking for a total commission of $1000 or more, so we write the inequality \begin{align*}60x+130y \ge 1000\end{align*}.

Once again, we can do this most easily by finding the \begin{align*}x-\end{align*} and \begin{align*}y-\end{align*}intercepts. When \begin{align*}x=0\end{align*}, we have \begin{align*}130y=1000\end{align*}, or \begin{align*}y=\frac{1000}{30} \approx 7.69\end{align*}. When \begin{align*}y=0\end{align*}, we have \begin{align*}60x=1000\end{align*}, or \begin{align*}x=\frac{1000}{60} \approx 16.67\end{align*}.

We draw a solid line connecting those points, and shade above the line because the inequality is “greater than.” We can check this by plugging in the point (0, 0): selling 0 washing machines and 0 refrigerators would give Julius a commission of $0, which is *not* greater than or equal to $1000, so the point (0, 0) is *not* part of the solution; instead, we want to shade the side of the line that does *not* include it.

Notice also that we show only the first quadrant of the coordinate plane, because Julius’s commission should be nonnegative.

The video at http://www.youtube.com/watch?v=7629PsZLP1A&feature=related contains more examples of real-world problems using inequalities in two variables.

## Review Questions

Graph the following inequalities on the coordinate plane.

- \begin{align*}x < 20\end{align*}
- \begin{align*}y \ge -5\end{align*}
- \begin{align*}|x| > 10\end{align*}
- \begin{align*}|y| \le 7\end{align*}
- \begin{align*}y \le 4x+3\end{align*}
- \begin{align*}y > -\frac{x}{2}-6\end{align*}
- \begin{align*}3x-4y \ge 12\end{align*}
- \begin{align*}x+7y < 5\end{align*}
- \begin{align*}6x+5y>1\end{align*}
- \begin{align*}y+5 \le -4x+10\end{align*}
- \begin{align*}x-\frac{1}{2}y \ge 5\end{align*}
- \begin{align*}6x+y < 20\end{align*}
- \begin{align*}30x+5y < 100\end{align*}
- Remember what you learned in the last chapter about families of lines.
- What do the graphs of \begin{align*}y > x+2\end{align*} and \begin{align*}y < x+5\end{align*} have in common?
- What do you think the graph of \begin{align*}x+2 < y < x+5\end{align*} would look like?

- How would the answer to problem 6 change if you subtracted 2 from the right-hand side of the inequality?
- How would the answer to problem 7 change if you added 12 to the right-hand side?
- How would the answer to problem 8 change if you flipped the inequality sign?
- A phone company charges 50 cents per minute during the daytime and 10 cents per minute at night. How many daytime minutes and nighttime minutes could you use in one week if you wanted to pay less than $20?
- Suppose you are graphing the inequality \begin{align*}y > 5x\end{align*}.
- Why can’t you plug in the point (0, 0) to tell you which side of the line to shade?
- What happens if you do plug it in?
- Try plugging in the point (0, 1) instead. Now which side of the line should you shade?

- A theater wants to take in at least $2000 for a certain matinee. Children’s tickets cost $5 each and adult tickets cost $10 each.
- If \begin{align*}x\end{align*} represents the number of adult tickets sold and \begin{align*}y\end{align*} represents the number of children’s tickets, write an inequality describing the number of tickets that will allow the theater to meet their minimum take.
- If 100 children’s tickets and 100 adult tickets have already been sold, what inequality describes how many
*more*tickets of both types the theater needs to sell? - If the theater has only 300 seats (so only 100 are still available), what inequality describes the
*maximum*number of additional tickets of both types the theater can sell?

## Texas Instruments Resources

*In the CK-12 Texas Instruments Algebra I FlexBook, there are graphing calculator activities designed to supplement the objectives for some of the lessons in this chapter. See http://www.ck12.org/flexr/chapter/9616.*

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