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6.1 Solving Linear Inequalities in One VariablePowerPoint Presentation

6.1 Solving Linear Inequalities in One Variable

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6.1 Solving Linear Inequalities in One Variable

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A linear inequalityin onevariable is an inequality which can be put into the form

ax + b > c

where a, b, and c are real numbers.

Note that the “>” can be replaced by , <, or .

Examples: Linear inequalities in one variable.

2x+ 3> 4

can be written –4x+ (–2)<–5.

2x – 2 < 6x – 5

6x+ 1 3(x – 5)

can be written 6x+1–15.

0

0

-4

-4

-3

-3

-2

-2

-1

-1

1

1

2

2

3

3

4

4

The solution set for an inequality can be expressed in two ways.

Example: Express the solution set of x <3 in two ways.

Open circle indicate that the number is not included in the solution set.

{x | x<3}

1. Set-builder notation:

°

2. Graph on the real line:

Example:Express the solution set of x 4 in two ways.

Closed circle indicate that the number is included in the solution set.

{x | x4}

1. Set-builder notation:

•

2. Graph on the real line:

A solution of an inequality in one variable is a number which, when substituted for the variable, results in a true inequality.

Examples: Are any of the values of x given below solutions of 2x > 5?

?

?

x = 2

2(2) > 5

4 > 5

False

2 is not a solution.

?

?

x = 2.6

2(2.6) > 5

5.2 > 5

True

2.6 is a solution.

?

?

x = 3

2(3) > 5

6 > 5

True

3 is a solution.

?

?

x = 1.5

2(1.5) > 5

3 > 5

False

1.5 is not a solution.

The solution set of an inequality is the set of all solutions.

The graph of a linear inequality in one variable is the graph on the real number line of all solutions of the inequality.

Verbal Phrase

Inequality

Graph

All real numbers less than 2

x < 2

All real numbers greater than -1

x > -1

All real numbers less than or

equal to 4

x< 4

All real numbers greater than or equal to -3

x> -3

+ 4

+ 4

Addition and Subtraction Properties

- If a > b and c is a real number, then a > b, a + c > b + c, and a – c > b – c have the same solution set.

- If a < b and c is a real number, then a < b, a + c < b + c, and a – c < b – c have the same solution set.

Example: Solve x – 4 > 7.

Add 4 to each side of the inequality.

x– 4 > 7

x > 11

Set-builder notation.

{x | x > 11}

– 2x

– 2x

Example: Solve 3x2x + 5.

Subtract 2x from each side.

3x2x + 5

x 5

{x | x5}

Set-builder notation.

Multiplication and Division Properties

- If c > 0 the inequalities a > b,ac > bc, and>have the same solution set.

- If c < 0 the inequalities a > b,ac < bc, and<have the same solution set.

Example: Solve 4x 12.

4x 12

Divide by 4.

x 3

4 is greater than 0, so the inequality sign remains the same.

Example: Solve .

(–3)

(–3)

Multiply by –3.

–3 is less than 0, so the inequality sign changes.

x >

Example: Solve x + 5 < 9x + 1.

–8x + 5 < 1

Subtract 9x from both sides.

–8x < –4

Subtract 5 from both sides.

Divide both sides by –8 and simplify.

Inequality sign changes because of division by a negative number.

Solution set in set-builder notation.

Example: Solve .

Subtract from both sides.

-20

-10

0

10

20

30

40

50

60

Add 2 to both sides.

Multiply both sides by 5.

•

Solution set as a graph.

Example: A cell phone company offers its customers a rate of $89 per month for 350 minutes, or a rate of $40 per month plus $0.50 for each minute used.

How many minutes per month can a customer who chooses the second plan use before the charges exceed those of the first plan?

Let x = the number of minutes used.

Solve the inequality 0.50x + 40 89.

0.50x 49

Subtract 40.

x 24.5

Divide by 0.5.

The customer can use up to 24.5 minutes per month before the cost of the second plan exceeds the cost of the first plan.

A compound inequality is formed by joining two inequalities with “and” or “or.”

0

-4

-3

-2

-1

1

2

3

4

Example: Solve x + 2 < 5 and 2x – 6 > – 8.

Solve the first inequality.

Solve the second inequality.

x +2 < 5

2x – 6 > – 8

x < 3

2x > – 2

Subtract 2.

Add 6.

{x | x < 3}

x > – 1

Divide by 2.

Solution set

{x | x > –1}

Solution set

The solution set of the “and” compound inequality is the intersection of the two solution sets.

°

º

0

-4

-3

-2

-1

1

2

3

4

When solving compound inequalities, it is possible to work with both inequalities at once.

This inequality means 11 < 6x + 5 and 6x + 5 < 29.

Example: Solve 11 < 6x + 5 < 29.

6 < 6x < 24

Subtract 5 from each of the three parts.

1 < x < 4

Divide 6 into each of the three parts.

Solution set.

Example: Solve .

0

-4

-3

-2

-1

1

2

3

4

Subtract 6 from each part.

Multiply each part by –2.

Multiplication by a negative number changes the inequality sign for each part.

•

•

Solution set.

0

-4

-3

-2

-1

1

2

3

4

Example: Solve x + 5 > 6 or 2x < –4.

Solve the first inequality.

Solve the second inequality.

x + 5 > 6

2x < –4

x > 1

x < –2

{ x | x > 1}

{ x | x < –2}

Solution set

Solution set

Since the inequalities are joined by “or” the solution set is the union of the solution sets.

°

°

There are 4 types of absolute value inequalities and equivalent inequalities

- |x| < a
- |x| <a
- |x| > a
- |x| >a

Translating Absolute Value Inequalities

- The inequality |ax + b| < c is equivalent to -c < ax + b < c
- The inequality |ax + b| > c is equivalent to ax + b < -c or
ax + b > c

0

-4

-3

-2

-1

1

2

3

4

º º

Example Solve |x - 4| < 3

- -3 < x - 4 < 3
- 1 < x < 7
- The solution set is {x| 1 < x < 7} and the interval is (1, 7)

0

-4

-3

-2

-1

1

2

3

4

Example Solve |4x - 1| < 9

- -9 ≤ 4x - 1 ≤ 9
- -8 ≤ 4x≤ 10
- -2 ≤x≤ 5/2
- The interval solution is [-2, 5/2]

0

-4

-3

-2

-1

1

2

3

4

º º

Solve |x| > a

- Solve |x + 1| > 2
- x + 1 < -2 or x + 1 > 2
- x < -3 or x > 1
- The solution interval is (-∞, -3) U (1, ∞)

0

-4

-3

-2

-1

1

2

3

4

• •

Solve |x| ≥a

- Solve |2x - 8| ≥ 4
- 2x – 8 ≤ -4 or 2x - 8 ≥ 4
- 2x≤ 4 or 2x≥ 12
- x≤ 2 or x≥ 6
- The solution interval is (-∞, 2] U [ 6, ∞)

- A linear inequality in two variables can be written in any one of these forms:
- Ax + By < C
- Ax + By > C
- Ax + By ≤ C
- Ax + By ≥ C

- An ordered pair (x, y) is a solution of the linear inequality if the inequality is TRUE when x and y are substituted into the inequality.

- Which ordered pair is a solution of
5x - 2y ≤ 6?

- (0, -3)
- (5, 5)
- (1, -2)
- (3, 3)

- The graph of a linear inequality is the set of all points in a coordinate plane that represent solutions of the inequality.
- We represent the boundary line of the inequality by drawing the function represented in the inequality.

- The boundary line will be a:
- Solid line when ≤ and ≥ are used.
- Dashed line when < and > are used.

- Our graph will be shaded on one side of the boundary line to show where the solutions of the inequality are located.

Here are some steps to help graph linear inequalities:

- Graph the boundary line for the inequality. Remember:
- ≤ and ≥ will use a solid curve.
- < and > will use a dashed curve.

- Test a point NOT on the boundary line to determine which side of the line includes the solutions. (The origin is always an easy point to test, but make sure your line does not pass through the origin)
- If your test point is a solution (makes a TRUE statement), shade THAT side of the boundary line.
- If your test points is NOT a solution (makes a FALSE statement), shade the opposite side of the boundary line.

y

5

x

-5

-5

5

- Graph the inequality x ≤ 4 in a coordinate plane
- Decide whether to
use a solid or

dashed line.

- Use (0, 0) as a
test point.

- Shade where the
solutions will be.

y

5

x

-5

-5

5

- Graph y>x + 2 in a coordinate plane.
- Sketch the boundary line of the graph.
- Solid or dashed
line?

- Use (0, 0) as a
test point.

- Shade where the
solutions are.

y

5

x

-5

-5

5

- Graph y> -½x - 2 in a coordinate plane.
- Sketch the boundary line of the graph.
- Solid or dashed
line?

- Use (0, 0) as a
test point.

- Shade where the
solutions are.

y

5

x

-5

-5

5

- Graph 3x - 4y > 12 in a coordinate plane.
- Sketch the boundary line of the graph.
- Find the x- and
y-intercepts and

plot them.

- Find the x- and
- Solid or dashed
line?

- Use (0, 0) as a
test point.

- Shade where the
solutions are.

y

5

x

-5

-5

5

- Graph y < 2/5x in a coordinate plane.
- Sketch the boundary line of the graph.
- Find the x- and y-intercept and plot them.
- Both are the origin!

- Use the line’s slope
to graph another point.

- Solid or dashed
line?

- Use a test point
OTHER than the

origin.

- Shade where the
solutions are.