What if you were given a linear inequality like \begin{align*}2x - 3y \le 5\end{align*}? How could you graph that inequality in the coordinate plane? After completing this Concept, you'll be able to graph linear inequalities in two variables like this one.

### Watch This

CK-12 Foundation: 0612S Linear Inequalities in Two Variables (H264)

### Guidance

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 A

*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 B

*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 C

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.

Watch this video for help with the Examples above.

CK-12 Foundation: Linear Inequalities in Two Variables

### Guided Practice

*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 non-negative.

### Explore More

Graph the following inequalities on the coordinate plane.

- \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. Sketch a graph showing 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?

### Answers for Explore More Problems

To view the Explore More answers, open this PDF file and look for section 6.12.

### Texas Instruments Resources

*In the CK-12 Texas Instruments Algebra I FlexBook® resource, 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.*