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4. Systems of Equations & Inequalities
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Algebra 1

4.04 Two variable linear inequalities

Lesson
Practice

Two variable linear inequalities

If a linear inequality involves two variables, we can represent it as a region on a coordinate plane rather than an interval on a number line.

Linear inequality in two variables

An inequality whose solution is a set of ordered pairs represented by a region of the coordinate plane on one side of a boundary line.

Example:

2y > 4 - 3x

Exploration

Drag the blue point to the different regions of the graph (shaded, unshaded, boundary line) and note what happens.

Use the dropdown to change the inequality symbol and note what happens.

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  1. What do you notice about the label when the point is in the shaded vs. the unshaded region? Why do you think that happens?
  2. What do you notice about the label when the point is on the boundary line? Why do you think that happens?
  3. How does the inequality symbol affect the graph and the label on the point? Why do you think that happens?

The solution to an inequality in two variables includes many solutions, similarly to inequalities in one variable. Instead of shading a number line to show a solution set, we now have a region that represents the solution to a linear inequality. Instead of an unfilled circle to show the boundary of a solution set that is excluded, we use a dashed line. Instead of a filled circle to show the boundary of a solution set that is included, we use a solid line.

Boundary line

A line which divides the coordinate plane into two regions. A boundary line of a linear inequality is solid if it is included in the solution set, and dashed if it is not.

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Depending on the inequality sign, the boundary line will be solid or dashed, and region shaded will be above or below the boundary line.

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y \geq x
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y \leq x

We can also use linear inequalities to represent real-world situations. Recall that a viable solution is a valid solution that makes sense within the context of the question or problem while a non-viable solution is an algebraically valid solution that does not make sense within the context of the question or problem.

We were previously introduced to constraints on inequalities that present a limitation or restriction of the possible x-values when solving inequalities in one variable. We should also consider constraints in context when solving two variable linear inequalities.

Examples

Example 1

Consider the inequality

y \leq -2x + 5

a

Graph the boundary line y=-2x+5.

Worked Solution
Create a strategy

We can graph the equation of the boundary line using the y-intercept and slope.

Apply the idea
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b

Determine whether the point (4,\,2) is a solution to the inequality y \leq -2x+5.

Worked Solution
Create a strategy

We will substitute the ordered pair into the inequality to determine whether the statement is true.

Apply the idea
\displaystyle y\displaystyle \leq\displaystyle -2x+5Given inequality
\displaystyle 2\displaystyle \leq\displaystyle -2(4) + 5Substitute \left(4,\,2\right)
\displaystyle 2\displaystyle \leq\displaystyle -3Evaluate the multiplication and addition

Since 2 is not less than or equal to -3 the statement is false and the point (4,\,2) is not a solution to the inequality y \leq -2x+5.

c

Determine whether the point (0,\,0) is a solution to the inequality y \leq -2x+5.

Worked Solution
Create a strategy

We will substitute the ordered pair into the inequality to determine whether the statement is true.

Apply the idea
\displaystyle y\displaystyle \leq\displaystyle -2x+5Given inequality
\displaystyle 0\displaystyle \leq\displaystyle -2(0) + 5Substitute \left(0,\,0\right)
\displaystyle 0\displaystyle \leq\displaystyle 5Evaluate the multiplication and addition

Since the statement is true, the point (0,\,0) is a solution to the inequality y \leq -2x+5.

d

Shade the side of the graph that includes the point from part (b) or part (c) that is a solution to the inequality.

Worked Solution
Create a strategy

Since the point (0,\,0) is a solution to the inequality, we can shade the side of the graph that includes this test point to show that any point in that region is in the solution set to the inequality.

Apply the idea
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Note that the line of the graph remains a solid line because the inequality includes "or equal to" meaning solutions on the boundary line are included in the solution set.

Reflect and check

When an inequality is written in slope-intercept form, we can determine the solution set by reading the inequality and it will inform us where to shade on the graph.

  • If the inequality is in the form y < mx +b or y \leq mx +b, the solution set is below the line
  • If the inequality is in the form y > mx +b or y \geq mx +b, the solution set is above the line

Example 2

Consider the linear inequality 5x + 3y > 15

a

Graph the inequality.

Worked Solution
Create a strategy

First, find the intercepts of the graph of the equation 5x+3y=15 and sketch the line. Then, choose test points to determine the region of the solution set and change the line to a dashed line to show that the boundary line is excluded from the solution set.

Apply the idea

Find the x-intercept:

\displaystyle 5x+3y\displaystyle =\displaystyle 15Equation of boundary line
\displaystyle 5x+3(0)\displaystyle =\displaystyle 15Substitute y=0
\displaystyle 5x\displaystyle =\displaystyle 15Evaluate the multiplication
\displaystyle x\displaystyle =\displaystyle 3Division property of equality

Find the y-intercept:

\displaystyle 5x+3y\displaystyle =\displaystyle 15Equation of boundary line
\displaystyle 5(0)+3y\displaystyle =\displaystyle 15Substitute x=0
\displaystyle 3y\displaystyle =\displaystyle 15Evaluate the multiplication
\displaystyle y\displaystyle =\displaystyle 5Division property of equality

The intercepts are (3,\,0) and (0,\,5).

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The point (0,\,0) can be used as a test point.

\displaystyle 5x+3y\displaystyle >\displaystyle 15Original inequality
\displaystyle 5(0)+3(0)\displaystyle >\displaystyle 15Substitute test point \left(0,\,0\right)
\displaystyle 0\displaystyle >\displaystyle 15Evaluate the multiplication and addition

Since 0 is not greater than 15 the statement is false and (0,\,0) is not in the solution set of the inequality and the shaded region is located on the opposite side of the boundary line.

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Reflect and check

By converting the boundary line inequality to slope-intercept form, we can determine that the shading is above the boundary line, as shown:

\displaystyle 5x+3y\displaystyle >\displaystyle 15Original inequality
\displaystyle 3y\displaystyle >\displaystyle -5x+15Subtract 5x from both sides
\displaystyle y\displaystyle >\displaystyle \dfrac{-5}{3}x+5Divide both sides by 3

Because we are left with an inequality that states y is greater than, we can use that information to shade all of the points above the boundary line.

b

Is the point (0,\,5) in the solution set of the inequality?

Worked Solution
Create a strategy

Substitute the ordered pair in the inequality and determine whether the math statement is true.

Apply the idea
\displaystyle 5x+3y\displaystyle >\displaystyle 15Original inequality
\displaystyle 5(0)+3(5)\displaystyle >\displaystyle 15Substitute the point \left(0,\,5\right)
\displaystyle 15\displaystyle >\displaystyle 15Evaluate the multiplication and addition

This is a false statement, so (0,\,5) is not a solution to the inequality.

Reflect and check

By looking at the graph of the linear inequality, we can see that the point (0,\,5) lies on the boundary line. Since the line is dashed, we know that any points on the boundary line are excluded from the solution set.

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Example 3

Write the inequality that describes the region shaded on the given coordinate plane.

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Worked Solution
Create a strategy

First, we want to determine the equation of the boundary line. Then we need to determine what inequality sign to use.

Apply the idea
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To find the equation of the boundary line in the form y=mx+b:

  • The line crosses the y-axis at (0,\,-3)
  • The rise is 3
  • The run is 1

So, b=-3 and m=3.

Now we have y=3x-3 as the boundary line.

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To determine the inequality:

  • The boundary line is solid, so it is \leq or \geq.
  • Any point in the shaded region would be above the boundary line.

This means that for every x-value, the y-value for any point in the region is greater than the y-value of the point on the line.

So our linear inequality is y \geq 3x-3.

Reflect and check

It is a good idea to check our answer by substituting in a point in the region to make sure it satisfies the inequality. For example, the origin, \left(0,\,0\right), is in the region, so should satisfy the inequality.

\displaystyle y\displaystyle \geq\displaystyle 3x-3Written inequality
\displaystyle 0\displaystyle \geq\displaystyle 3\left(0\right)-3Substitute the point \left(0,0\right)
\displaystyle 0\displaystyle \geq\displaystyle -3Evaluate the multiplication

It does satisfy the inequality, so we have selected the correct inequality sign.

Example 4

A pick-up truck has a maximum weight capacity of 3000 pounds. Each box of oranges weighs 8 pounds and each box of grapefruits weighs 12 pounds.

a

Write an inequality to represent the number of boxes of oranges and grapefruit that can be in the truck.

Worked Solution
Create a strategy

Our unknown values are the number of boxes of oranges and the number of boxes of grapefruits. Let x represent the number of boxes of oranges in the truck. Let y represent the number of boxes of grapefruit in the truck.

We should consider the structure of the inequality. The maximum weight capacity is 3000 pounds, so we know that the total weight of the boxes \leq 3000. From there, we can use the defined variables and numbers from the problem to build the other part of the inequality.

Apply the idea

The weight of all the orange boxes is the product of the weight of one box, 8, and the number of boxes, x.

Weight of orange boxes: 8x pounds

The weight of all the grapefruit boxes is the product of the weight of one box, 12, and the number of boxes, y.

Weight of grapefruit boxes: 12y pounds

The total weight is the sum of the weights of orange boxes and grapefruit boxes.

Total weight: 8x+12y pounds

The truck can carry at most 3000 pounds, so we get our inequality:

8x+12y \leq 3000

b

Create a graph of the region containing the points corresponding to all the different numbers of orange and grapefruit boxes that can be loaded into the truck.

Worked Solution
Create a strategy

We will graph the region representing 8x+12y \leq 3000.

We need to identify our boundary line, then graph it. Then, we need to decide which side of the boundary line to shade.

Apply the idea

The boundary line is 8x+12y = 3000.

This is a line in standard form, so we can graph it by finding the intercepts.

Find the x-intercept by setting y=0 and solving:

\displaystyle 8x+12y\displaystyle =\displaystyle 3000Given equation
\displaystyle 8x-12\left(0\right)\displaystyle =\displaystyle 3000Substitute y=0
\displaystyle 8x\displaystyle =\displaystyle 3000Evaluate the multiplication
\displaystyle x\displaystyle =\displaystyle 375Divide both sides by 8

Find the y-intercept by setting x=0 and solving:

\displaystyle 8x+12y\displaystyle =\displaystyle 3000Given equation
\displaystyle 8\left(0\right)+12y\displaystyle =\displaystyle 3000Substitute x=0
\displaystyle 12y\displaystyle =\displaystyle 3000Evaluate the multiplication
\displaystyle y\displaystyle =\displaystyle 250Divide both sides by 12

The boundary line will be solid because we have \leq as the sign.

We will shade below the line as the point \left(0,\,0\right) satisfies the inequality.

Pick-up truck capacity in pounds
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\text{Number of orange boxes, }x
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\text{Number of grapefruit boxes, }y
Reflect and check

The actual possible solutions are the positive integer coordinates since we are working with whole fruits and have both x\geq 0 and y \geq 0. This region shows the constraints on the problem and where all those possible solutions can be.

c

Explain why the point (325,\,-100) is a non-viable solution to the inequality.

Worked Solution
Create a strategy

We can use the given ordered pair to interpret its meaning in context and explain why it is non-viable.

Apply the idea

The point (325,\,-100) means that 325 boxes of oranges and -100 boxes of grapefruit meet the weight capacity of the pick-up truck.

\displaystyle 8x+12y\displaystyle \leq\displaystyle 3000Given inequality
\displaystyle 8(325)+12(-100)\displaystyle \leq\displaystyle 3000Substitute \left(325,\,-100\right)
\displaystyle 2600-1200\displaystyle \leq\displaystyle 3000Evaluate the multiplication
\displaystyle 1400\displaystyle \leq\displaystyle 3000Evaluate the subtraction

Negative weight is not possible so we know that this solution, while it satisfies the inequality that models the situation, is non-viable because it is unreasonable in context.

Idea summary

We can use test points or the inequality in slope-intercept form to determine the region of the solution set on the coordinate plane:

  • y > mx+b indicates that the solution set is above the boundary line
  • y < mx+b indicates that the solution set is below the boundary line

The constraints of a context are important to consider when writing linear inequalities and interpreting viable solutions.

Outcomes

A.EI.2

The student will represent, solve, explain, and interpret the solution to a system of two linear equations, a linear inequality in two variables, or a system of two linear inequalities in two variables.

A.EI.2d

Create a linear inequality in two variables to represent a contextual situation.

A.EI.2e

Represent the solution of a linear inequality in two variables graphically on a coordinate plane.

A.EI.2h

Verify possible solution(s) to a system of two linear equations, a linear inequality in two variable, or a system of two linear inequalities algebraically, graphically, and with technology to justify the reasonableness of the answer(s). Explain the solution method and interpret solutions for problems given in context.

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