6-6 solving linear inequalities a linear inequality is similar to a linear equation, but the equal...

6-6 Solving Linear Inequalities

A linear inequality is similar to a linear equation, but the equal sign is replaced with an inequality symbol.

A solution of a linear inequality is any ordered pair that makes the inequality true.


Tell whether the ordered pair is a solution of the inequality.

Additional Example 1A: Identifying Solutions of Inequalities

(–2, 4); y < 2x + 1

Substitute (–2, 4) for (x, y).

y < 2x + 1

4 2(–2) + 14 –4 + 14 –3<

(–2, 4) is not a solution.


Tell whether the ordered pair is a solution of the inequality.

Additional Example 1B: Identifying Solutions of Inequalities

(3, 1); y > x – 4

Substitute (3, 1) for (x, y).

y > x − 4

1 3 – 4

1 – 1>

(3, 1) is a solution.


A linear inequality describes a region of a coordinate plane called a half-plane.

All points in the half-plane region are solutions of the linear inequality.

The boundary line of the region is the graph of the related equation.


Graphing Linear Inequalities

Step 1 Solve the inequality for y (slope-intercept form).

Step 2Graph the boundary line. Use a solid line for ≤ or ≥. Use a dashed line for < or >.

Step 3Shade the half-plane above the line for y > or ≥. Shade the half-plane below the line for y < or y ≤. Check your answer.


Graph the solutions of the linear inequality. Check your answer.

Example 2A: Graphing Linear Inequalities in Two Variables

y 2x – 3

Step 1 The inequality is already solved for y.

Step 2 Graph the boundary line y = 2x – 3. Use a solid line for .

Step 3 The inequality is , so shade below the line.

6-6 Solving Linear InequalitiesAdditional Example 2A Continued

Substitute (0, 0) for (x, y) because it is not on the boundary line.Check y 2x – 3

0 2(0) – 3

0 –3 A false statement means

that the half-plane containing (0, 0) should NOT be shaded. (0, 0) is not one of the solutions, so the graph is shaded correctly.


y 2x – 3


Use the “test point” method shown in Example 2 to check your answers to linear inequalities. The point (0, 0) is a good test point to use if it does not lie on the boundary line. However, this method will check only that your shading is correct. It will not check the boundary line.

Helpful Hint



Example 2B: Graphing Linear Inequalities in Two Variables

5x + 2y > –8

Step 1 Solve the inequality for y.

5x + 2y > –8 –5x –5x

2y > –5x – 8

y > x – 4

Step 2 Graph the boundary line Use a dashed line for >.

y = x – 4.


Step 3 The inequality is >, so shade above the line.

Additional Example 2B Continued


5x + 2y > –8

6-6 Solving Linear InequalitiesAdditional Example 2B Continued

Substitute ( 0, 0) for (x, y) because it is not on the boundary line.

The point (0, 0) satisfies the inequality, so the graph is correctly shaded.

Check

y > x – 4

0 (0) – 4

0 –40 –4>


5x + 2y > –8



Example 2C: Graphing Linear Inequalities in two Variables

4x – y + 2 ≤ 0

Step 1 Solve the inequality for y.

4x – y + 2 ≤ 0

–y ≤ –4x – 2

–1 –1

y ≥ 4x + 2

Step 2 Graph the boundary line y = 4x + 2. Use a solid line for ≥.


Step 3 The inequality is ≥, so shade above the line.

Additional Example 2C Continued


6-6 Solving Linear InequalitiesAdditional Example 2C Continued

Substitute ( –3, 3) for (x, y) because it is not on the boundary line.

The point (–3, 3) satisfies the inequality, so the graph is correctly shaded.

Check

3 4(–3)+ 2 3 –12 + 2

3 ≥ –10

y ≥ 4x + 2


Write an inequality to represent the graph.

Additional Example 4A: Writing an Inequality from a Graph

y-intercept: 1; slope:

Write an equation in slope-intercept form.

The graph is shaded above a dashed boundary line.

Replace = with > to write the inequality


Ada has at most 285 beads to make jewelry. A necklace requires 40 beads, and a bracelet requires 15 beads.

Additional Example 3a: Application

a. Write a linear inequality to describe the situation.

Let x represent the number of necklaces and y the number of bracelets.

Write an inequality. Use ≤ for “at most.”

6-6 Solving Linear InequalitiesAdditional Example 3a Continued

Necklacebeads

braceletbeadsplus

is atmost

285beads.

40x + 15y ≤ 285

Solve the inequality for y.

40x + 15y ≤ 285–40x –40x

15y ≤ –40x + 285Subtract 40x from

both sides.

Divide both sides by 15.

6-6 Solving Linear InequalitiesAdditional Example 3b

b. Graph the solutions.

=

Step 1 Since Ada cannot make a

negative amount of jewelry, the

system is graphed only in

Quadrant I. Graph the boundary

line . Use a solid line

for ≤.


b. Graph the solutions.

Step 2 Shade below the line. Ada can only make whole numbers of jewelry. All points on or below the line with whole number coordinates are the different combinations of bracelets and necklaces that Ada can make.

Additional Example 3b Continued


c. Give two combinations of necklaces and bracelets that Ada could make.

Additional Example 3c

Two different combinations of jewelry that Ada could make with 285 beads could be 2 necklaces and 8 bracelets or 5 necklaces and 3 bracelets.

(2, 8)

(5, 3)

6-6 solving linear inequalities a linear inequality is similar to a linear equation, but the equal...

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