ALGEBRA 1

Lesson 6-5 Warm-Up

ALGEBRA 1

“Linear Inequalities” (6-5)

What is the solution of an inequality?

What is a linear inequality?

Solution of an Inequality: All of the coordinates on a graph that make the inequality true

Linear Inequality: the shaded region on a graph that contains all of the solutions of an inequality (the region is defined by a boundary line)

Tips: When the inequality uses the or sign, use a dashed boundary line to show points on the line are not solutions to the inequality. When the inequality uses the ≤ or ≥ sign, use a solid boundary line to show points on the line are solutions to the inequality. For an inequalities in the form of y or y ≤, shade below the boundary line. For an inequalities in the form of y or y ≥, shade above the boundary line. Always check to make sure you shaded the correct side of the boundary line by testing one of the solutions in the shaded area (the solution into should makes the inequality a true statement).

ALGEBRA 1

Graph y > –2x + 1.

First, graph the boundary line y = –2x + 1.

The coordinates of the points on the boundary line do not make the inequality true. So, use a dashed line.

Shade above the boundary line.

Check: The point (0, 2) is in the region of the graph of the inequality.See if (0, 2) satisfies the inequality.y > –2x + 12 > –2(0) + 12 > 1

Substitute (0, 2) for (x, y).

Linear InequalitiesLESSON 6-5

Additional Examples

ALGEBRA 1

Graph 4x – 3y 9.>–

Solve 4x – 3y 9 for y.>–4x – 3y 9 >–

43

<–y x – 3 Divide each side by –3. Reverse the inequality symbol.

>––3y –4x + 9 Subtract 4x from each side.

The coordinates of the points on the boundary line make the inequality true. So, use a solid line.

Graph y = x – 3.43

Since y x – 3, shade below the

boundary line.

43

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Linear InequalitiesLESSON 6-5

Additional Examples

ALGEBRA 1

“Linear Inequalities” (6-5)

What if the inequality is in the form of Ax + By = C?

If the inequality is in the form of Ax + By = C:

1.First, find the x and y intercepts (found by substituting 0 for x to find y and vice-versa).

2.Then, choose a test point on the boundary line to determine if the line is dashed (line doesn’t include solutions) or solid (line includes solutions).

3.Finally, choose a test point above and / or below the boundary line to determine if the solutions are above or below the line).

ALGEBRA 1

Suppose your budget allows you to spend no morethan $24 for decorations for a party. Streamers cost $2 aroll and tablecloths cost $6 each. Use intercepts to graphthe inequality that represents the situation. Find threepossible combinations of streamers and tablecloths youcan buy.

Words: cost of plus cost of is less than total budget streamers tablecloths or equal to

Define: Let s = the number of rolls of streamers.

Let t = the number of tablecloths.

Equation: 2 s + 6 t 24≤

Linear InequalitiesLESSON 6-5

Additional Examples

ALGEBRA 1

(continued)

Graph 2s + 6t 24 by graphing the intercepts (12, 0) and (0, 4).

The coordinates of the points on the boundary line make the inequality true. So, use a solid line.

Graph only in Quadrant I, since you cannot buy a negative amount of decorations.

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Test the point (0, 0). 2s + 6t 24 2(0) + 6(0) 24 Substitute (0, 0) for (s, t). 0 24 Since the inequality is true, (0, 0) is a solution.

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Linear InequalitiesLESSON 6-5

Additional Examples

ALGEBRA 1

(continued)

Shade the region containing (0, 0). The graph below shows all the possible solutions of the problem.

Since the boundary line is included in the graph, the intercepts are also solutions to the inequality.

The solution (9, 1) means that if you buy 9 rolls of streamers, you can buy 1 tablecloth. Three solutions are (9, 1), (6, 2), and (3, 3).

Linear InequalitiesLESSON 6-5

Additional Examples

ALGEBRA 1

1. Determine whether (4, 1) is a solution of 3x + 2y 10.

Graph each inequality.

2. x > –2 3. 5x – 2y > 10 4. 2x + 6y 0

yes>–

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Linear InequalitiesLESSON 6-5

Lesson Quiz