Unit 4 Linear Inequalities And Systems — Unit Plan

We have used equations and the equal sign to represent relationships and constraints in various situations. Not all relationships and constraints involve equality, however. 

In some situations, one quantity is, or needs to be, greater than or less than another. To describe these situations, we can use inequalities and symbols such as $<, \leq, >$, or $\geq$.

When working with inequalities, it helps to remember what the symbol means, in words. For example:

• $100 < a$ means “100 is less than $a$.” 
• $y \le 55$ means “$y$ is less than or equal to 55,” or "$y$ is not more than 55."
• $20 > 18$ means “20 is greater than 18.”
• $t \ge 40$ means “$t$ is greater than or equal to 40,” or "$t$ is at least 40."

These inequalities are fairly straightforward. Each inequality states the relationship between two numbers ($20>18$), or it describes the limit or boundary of a quantity in terms of a number (100<a></a></span>). </p> <p>Inequalities can also express relationships or constraints that are more complex. Here are some examples:</p> <div class="imgrid"> <div class="g--row"> <div class="g--column g--content g--three-fourth"> <ul> <li>The area of a rectangle, <span class="math">\(A, with a length of 4 meters and a width of $w$ meters is no more than 100 square meters.

The equation $\frac12 t = 10$ is an equation in one variable. Its solution is any value of $t$ that makes the equation true. Only $t=20$ meets that requirement, so 20 is the only solution.

The inequality $\frac12t >10$ is an inequality in one variable. Any value of $t$ that makes the inequality true is a solution. For instance, 30 and 48 are both solutions because substituting these values for $t$ produces true inequalities. $\frac12(30) >10$ is true, as is $\frac12(48) >10$. Because the inequality has a range of values that make it true, we sometimes refer to all the solutions as the solution set.

One way to find the solutions to an inequality is by reasoning. For example, to find the solution to $2p<8$, we can reason that if 2 times a value is less than 8, then that value must be less than 4. So a solution to $2p<8$ is any value of $p$ that is less than 4.

Another way to find the solutions to $2p<8$ is to solve the related equation $2p=8$. In this case, dividing each side of the equation by 2 gives $p=4$. This point, where $p$ is 4, is the boundary of the solution to the inequality.

To find out the range of values that make the inequality true, we can try values less than and greater than 4 in our inequality and see which ones make a true statement.

In general, the inequality is false when $p$ is greater than or equal to 4 and true when $p$ is less than 4.

We can represent the solution set to an inequality by writing an inequality, $p<4$, or by graphing on a number line. The ray pointing to the left represents all values less than 4.

Writing and solving inequalities can help us make sense of the constraints in a situation and solve problems. Let's look at an example.

Clare would like to buy a video game system that costs $130 and to have some extra money for games. She has saved $48 so far and plans on saving $5 of her allowance each week. How many weeks, $w$, will it be until she has enough money to buy the system and have some extra money remaining? To represent the constraints, we can write $48 + 5w \geq 130$. Let’s reason about the solutions:

• Because Clare has $48 already and needs to have at least $130 to afford the game system, she needs to save at least $82 more.
• If she saves $5 each week, it will take at least $\frac{82}{5}$ weeks to reach $82.
• $\frac{82}{5}$ is 16.4. Any time shorter than 16.4 weeks won't allow her to save enough.
• Assuming she saves $5 at the end of each week (instead of saving smaller amounts throughout a week), it will be at least 17 weeks before she can afford the game system. 

We can also solve by writing and solving a related equation to find the boundary value for $w$, and then determine whether the solutions are less than or greater than that value.

Sometimes the structure of an inequality can help us see whether the solutions are less than or greater than a boundary value. For example, to find the solutions to $3x > 8x$, we can solve the equation $3x = 8x$, which gives us $x = 0$. Then, instead of testing values on either side of 0, we could reason as follows about the inequality:

• If $x$ is a positive value, then $3x$ would be less than $8x$.
• For $3x$ to be greater than $8x$, $x$ must include negative values.
• For the solutions to include negative values, they must be less than 0, so the solution set would be $x < 0$.

The equation $x+y = 7$ is an equation in two variables. Its solution is any pair of $x$ and $y$ whose sum is 7. The pairs $x=0, y=7$ and $x =\text5, y= 2$ are two examples.

We can represent all the solutions to $x+y = 7$ by graphing the equation on a coordinate plane.

The graph is a line. All the points on the line are solutions to $x+y = 7$.

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The inequality $x+y \leq 7$ is an inequality in two variables. Its solution is any pair of $x$ and $y$ whose sum is 7 or less than 7.

This means it includes all the pairs that are solutions to the equation $x+y=7$, but also many other pairs of $x$ and $y$ that add up to a value less than 7. The pairs $x=4, y=\text-7$ and $x=\text-6, y=0$ are two examples.

On a coordinate plane, the solution to $x+y \leq 7$ includes the line that represents $x+y=7$. If we plot a few other $(x,y)$ pairs that make the inequality true, such as $(4, \text-7)$ and $(\text-6,0)$, we see that these points fall on one side of the line. (In contrast, $(x,y)$ pairs that make the inequality false fall on the other side of the line.)

We can shade that region on one side of the line to indicate that all points in it are solutions.

Inequality graphed on a coordinate plane, origin O. Each axis from negative 8 to 10, by 2’s. Solid line goes through 0 comma 7 and 5 comma 2. The region below the solid line is shaded. The points negative 6 comma 0 and 4 comma negative 7 are labeled.

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What about the inequality $x+y <7$?

The solution is any pair of $x$ and $y$ whose sum is less than 7. This means pairs like $x=0, y=7$ and $x =5, y=2$ are not solutions. 

On a coordinate plane, the solution does not include points on the line that represent $x+y=7$ (because those points are $x$ and $y$ pairs whose sum is 7).

To exclude points on that boundary line, we can use a dashed line. 

All points below that line are $(x,y)$ pairs that make $x+y<7$ true. The region on that side of the line can be shaded to show that it contains the solutions. 

Inequality graphed on a coordinate plane, origin O. Each axis from negative 8 to 10, by 2’s. Dashed line goes through 0 comma 7 and 7 comma 0. The region below the dashed line is shaded.

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Suppose we want to find the solution to  $x - y > 5$. We can start by graphing the related equation $x - y = 5$. 

When identifying the solution region, it is important not to assume that the solution will be above the line because of a “>” symbol or below the line because of a “<” symbol.

Instead, test the points on the line and on either side of the line, and see if they are solutions.​​​​​​

Graph of a line, origin O, with grid. X axis from negative 2 to 12 by 2’s. Y axis from negative 6 to 8 by 2’s. Line passes through approximately 1 comma negative 4, 5 comma 0, and 12 comma 7.

For  $x-y>5$, points on the line and above the line are not solutions to the inequality because the $(x,y)$ pairs make the inequality false. Points that are below the lines are solutions, so we can shade that lower region. 

Graphing technology can help us graph the solution to an inequality in two variables.

Many graphing tools allow us to enter inequalities such as $x-y >5$ and will show the solution region, as shown here.

Some tools, however, may require the inequalities to be in slope-intercept form or another form before displaying the solution region. Be sure to learn how to use the graphing technology available in your classroom.​​​​​​

Inequality graphed on a coordinate plane. X axis from 0 to 10, by 5’s. Y axis from negative 5 to 10, by 5’s. Dashed line goes through 0 comma negative 5, 5 comma 0, and 10 comma 5. The region below the dashed line is shaded.

Although graphing using technology is efficient, we still need to analyze the graph with care. Here are some things to consider: 

• The graphing window. If the graphing window is too small, we may not be able to really see the solution region or the boundary line, as shown here.
• The meaning of solution points in the situation. For example, if $x$ and $y$ represent the lengths of two sides of a rectangle, then only positive values of $x$ and $y$ (or points in the first quadrant) make sense in the situation.

Inequality graphed on a coordinate plane. Each axis from negative 2 to 4, by 2’s. Dashed line goes through 3 comma negative 2, 3 point 5 comma negative 1 point 5, and 4 comma negative 1. The region below the dashed line is shaded.

In this lesson, two linear inequalities in two variables represent the constraints in a situation. Each pair of inequalities forms a system of inequalities.

A solution to a system of inequalities is any $(x,y)$ pair that makes both inequalities true, or any pair of values that simultaneously meet both constraints in the situation. The solution to the system is often best represented by a region on a graph.

Suppose there are two numbers, $x$ and $y$, and there are two things we know about them.

• The value of one number is more than double the value of the other.
• The sum of the two numbers is less than 10.

There are many possible pairs of numbers that meet the first constraint, for example: 1 and 3, or 4 and 9.

The same can be said about the second constraint, for example: 1 and 3, or 2.4 and 7.5.

The pair $x=1$ and $y=3$ meets both constraints, so it is a solution to the system.

The pair $x=4$ and $y=9$ meets the first constraint but not the second ($9 >2(4)$ is a true statement, but $4+9<10$ is not true.)

Remember that graphing is a great way to show all the possible solutions to an inequality, so let’s graph the solution region for each inequality.​​​​​​

A graph of an inequality on a coordinate plane, origin O. Each axis from negative 10 to 5, by 5’s. Dashed line starts below x axis and right of y axis, goes through negative 2 point 5 comma negative 5, 0 comma 0, and 2 point 5 comma 5. The region above the dashed line is shaded.

A graph of an inequality on a coordinate plane, origin O. Each axis from negative 10 to 5, by 5’s. A dashed line starts on the y axis at 10, goes through 5 comma 5, and ends on x axis at 10. The region below the dashed line is shaded.

Because we are looking for a pair of numbers that meet both constraints or make both inequalities true at the same time, we want to find points that are in the solution regions of both graphs.

To do that, we can graph both inequalities on the same coordinate plane.

The solution set to the system of inequalities is represented by the region where the two graphs overlap.

A graph of two intersecting inequalities on a coordinate plane, origin O. Each axis from negative 10 to 5, by 5’s. The first dashed line starts below x axis and right of y axis, goes through negative 2 point 5 comma negative 5, 0 comma 0, and 2 point 5 comma 5. The region above the dashed line is shaded. Second line at on the y axis at 10, goes through 5 comma 5, end on x axis at 10. The region below the dashed line is shaded.