# 2.7: Solve Linear Inequalities

Learning Objectives

By the end of this section, you will be able to:

• Graph inequalities on the number line
• Solve inequalities using the Subtraction and Addition Properties of inequality
• Solve inequalities using the Division and Multiplication Properties of inequality
• Solve inequalities that require simplification
• Translate to an inequality and solve

Note

Before you get started, take this readiness quiz.

1. Translate from algebra to English: (15>x).
If you missed this problem, review Exercise 1.3.1.
2. Solve: (n−9=−42).
If you missed this problem, review Exercise 2.1.7.
3. Solve: (−5p=−23).
If you missed this problem, review Exercise 2.2.1.
4. Solve: (3a−12=7a−20).
If you missed this problem, review Exercise 2.3.22.

## Graph Inequalities on the Number Line

Do you remember what it means for a number to be a solution to an equation? A solution of an equation is a value of a variable that makes a true statement when substituted into the equation.

What about the solution of an inequality? What number would make the inequality (x > 3) true? Are you thinking, ‘x could be 4’? That’s correct, but x could be 5 too, or 20, or even 3.001. Any number greater than 3 is a solution to the inequality (x > 3).

We show the solutions to the inequality (x > 3) on the number line by shading in all the numbers to the right of 3, to show that all numbers greater than 3 are solutions. Because the number 3 itself is not a solution, we put an open parenthesis at 3. The graph of (x > 3) is shown in Figure (PageIndex{1}). Please note that the following convention is used: light blue arrows point in the positive direction and dark blue arrows point in the negative direction.

The graph of the inequality (x geq 3) is very much like the graph of (x > 3), but now we need to show that 3 is a solution, too. We do that by putting a bracket at (x = 3), as shown in Figure (PageIndex{2}).

Notice that the open parentheses symbol, (, shows that the endpoint of the inequality is not included. The open bracket symbol, [, shows that the endpoint is included.

Exercise (PageIndex{1})

Graph on the number line:

1. (xleq 1)
2. (x<5)
3. (x>−1)

1. (xleq 1) This means all numbers less than or equal to 1. We shade in all the numbers on the number line to the left of 1 and put a bracket at x=1 to show that it is included.

2. (x<5) This means all numbers less than 5, but not including 5. We shade in all the numbers on the number line to the left of 5 and put a parenthesis at x=5 to show it is not included.

3. (x>−1) This means all numbers greater than −1, but not including −1. We shade in all the numbers on the number line to the right of −1, then put a parenthesis at x=−1 to show it is not included.

Exercise (PageIndex{2})

Graph on the number line:

1. (xleq −1)
2. (x>2)
3. (x<3)

Exercise (PageIndex{3})

Graph on the number line:

1. (x>−2)
2. (x<−3)
3. (xgeq −1)

We can also represent inequalities using interval notation. As we saw above, the inequality (x>3) means all numbers greater than 3. There is no upper end to the solution to this inequality. In interval notation, we express (x>3) as ((3, infty)). The symbol (infty) is read as ‘infinity’. It is not an actual number. Figure (PageIndex{3}) shows both the number line and the interval notation.

The inequality (xleq 1) means all numbers less than or equal to 1. There is no lower end to those numbers. We write (xleq 1) in interval notation as ((-infty, 1]). The symbol (-infty) is read as ‘negative infinity’. Figure (PageIndex{4}) shows both the number line and interval notation.

INEQUALITIES, NUMBER LINES, AND INTERVAL NOTATION

Did you notice how the parenthesis or bracket in the interval notation matches the symbol at the endpoint of the arrow? These relationships are shown in Figure (PageIndex{5}).

Exercise (PageIndex{5})

Graph on the number line and write in interval notation:

1. (x>2)
2. (xleq −1.5)
3. (xgeq frac{3}{4})

Exercise (PageIndex{6})

Graph on the number line and write in interval notation:

1. (xleq −4)
2. (xgeq 0.5)
3. (x<-frac{2}{3})

## Solve Inequalities using the Subtraction and Addition Properties of Inequality

The Subtraction and Addition Properties of Equality state that if two quantities are equal, when we add or subtract the same amount from both quantities, the results will be equal.

PROPERTIES OF EQUALITY

[egin{array} { l l } { extbf { Subtraction Property of Equality } } & { extbf { Addition Property of Equality } } { ext { For any numbers } a , b , ext { and } c , } & { ext { For any numbers } a , b , ext { and } c } { ext { if } qquad quad a = b , } & { ext { if } qquad quad a = b } { ext { then } a - c = b - c . } & { ext { then } a + c = b + c } end{array}]

Similar properties hold true for inequalities.

 For example, we know that −4 is less than 2. If we subtract 5 from both quantities, is the left side still less than the right side? We get −9 on the left and −3 on the right. And we know −9 is less than −3. The inequality sign stayed the same.

Similarly we could show that the inequality also stays the same for addition.

PROPERTIES OF INEQUALITY

[egin{array} { l l } { extbf { Subtraction Property of Inequality } } & { extbf { Addition Property of Inequality } } { ext { For any numbers } a , b , ext { and } c , } & { ext { For any numbers } a , b , ext { and } c } { ext { if }qquad quad a < b } & { ext { if } qquad quad a < b } { ext { then } a - c < b - c . } & { ext { then } a + c < b + c } { ext { if } qquad quad a > b } & { ext { if } qquad quad a > b } { ext { then } a - c > b - c . } & { ext { then } a + c > b + c } end{array}]

We use these properties to solve inequalities, taking the same steps we used to solve equations. Solving the inequality (x+5>9), the steps would look like this:

[egin{array}{rrll} {} &{x + 5} &{ >} &{9} { ext{Subtract 5 from both sides to isolate }x.} &{x + 5 - 5} &{ >} &{9 - 5} {} &{x} &{ >} &{4} end{array}]

Any number greater than 4 is a solution to this inequality.

Exercise (PageIndex{7})

Solve the inequality (n - frac{1}{2} leq frac{5}{8}), graph the solution on the number line, and write the solution in interval notation.

 Add (frac{1}{2}) to both sides of the inequality. Simplify. Graph the solution on the number line. Write the solution in interval notation.

Exercise (PageIndex{8})

Solve the inequality, graph the solution on the number line, and write the solution in interval notation.

(p - frac{3}{4} geq frac{1}{6})

Exercise (PageIndex{9})

Solve the inequality, graph the solution on the number line, and write the solution in interval notation.

(r - frac{1}{3} leq frac{7}{12})

## Solve Inequalities using the Division and Multiplication Properties of Inequality

The Division and Multiplication Properties of Equality state that if two quantities are equal, when we divide or multiply both quantities by the same amount, the results will also be equal (provided we don’t divide by 0).

PROPERTIES OF EQUALITY

[egin{array}{ll} { extbf{Division Property of Equality}} &{ extbf{MUltiplication Property of Equality}} { ext{For any numbers a, b, c, and c} eq 0} &{ ext{For any numbers a, b, c}} { ext{if } qquad a = b} &{ ext{if} qquad quad a = b} { ext{then }quad frac{a}{c} = frac{b}{c}} &{ ext{then } quad ac = bc} end{array}]

Are there similar properties for inequalities? What happens to an inequality when we divide or multiply both sides by a constant?

Consider some numerical examples.

 Divide both sides by 5. Multiply both sides by 5. Simplify. Fill in the inequality signs.

The inequality signs stayed the same.

Does the inequality stay the same when we divide or multiply by a negative number?

 Divide both sides by -5. Multiply both sides by -5. Simplify. Fill in the inequality signs.

The inequality signs reversed their direction.

When we divide or multiply an inequality by a positive number, the inequality sign stays the same. When we divide or multiply an inequality by a negative number, the inequality sign reverses.

Here are the Division and Multiplication Properties of Inequality for easy reference.

DIVISION AND MULTIPLICATION PROPERTIES OF INEQUALITY

For any real numbers a,b,c

[egin{array}{ll} { ext{if } a < b ext{ and } c > 0, ext{ then}} &{frac{a}{c} < frac{b}{c} ext{ and } ac < bc} { ext{if } a > b ext{ and } c > 0, ext{ then}} &{frac{a}{c} > frac{b}{c} ext{ and } ac > bc} { ext{if } a < b ext{ and } c < 0, ext{ then}} &{frac{a}{c} > frac{b}{c} ext{ and } ac > bc} { ext{if } a > b ext{ and } c < 0, ext{ then}} &{frac{a}{c} < frac{b}{c} ext{ and } ac < bc} end{array}]

When we divide or multiply an inequality by a:

• positive number, the inequality stays the same.
• negative number, the inequality reverses.

Exercise (PageIndex{11})

Solve the inequality, graph the solution on the number line, and write the solution in interval notation.

(9c>72)

(c>8)

((8, infty))

Exercise (PageIndex{12})

Solve the inequality, graph the solution on the number line, and write the solution in interval notation.

(12dleq 60)

(dleq 5)

((-infty, 5])

Exercise (PageIndex{14})

Solve each inequality, graph the solution on the number line, and write the solution in interval notation.

(−8q<32)

(q>−4)

Exercise (PageIndex{15})

Solve each inequality, graph the solution on the number line, and write the solution in interval notation.

(−7rleq −70)

SOLVING INEQUALITIES

Sometimes when solving an inequality, the variable ends up on the right. We can rewrite the inequality in reverse to get the variable to the left.

[egin{array}{l} x > a ext{ has the same meaning as } a < x end{array}]

Think about it as “If Xavier is taller than Alex, then Alex is shorter than Xavier.”

Exercise (PageIndex{17})

Solve the inequality, graph the solution on the number line, and write the solution in interval notation.

(24 leq frac{3}{8}m)

Exercise (PageIndex{18})

Solve the inequality, graph the solution on the number line, and write the solution in interval notation.

(-24 < frac{4}{3}n)

Exercise (PageIndex{20})

Solve the inequality, graph the solution on the number line, and write the solution in interval notation.

(frac{k}{-12}leq 15)

Exercise (PageIndex{21})

Solve the inequality, graph the solution on the number line, and write the solution in interval notation.

(frac{u}{-4}geq -16)

​​​​​

## Solve Inequalities That Require Simplification

Most inequalities will take more than one step to solve. We follow the same steps we used in the general strategy for solving linear equations, but be sure to pay close attention during multiplication or division.

Exercise (PageIndex{23})

Solve the inequality (3qgeq 7q−23), graph the solution on the number line, and write the solution in interval notation.

Exercise (PageIndex{24})

Solve the inequality (6x<10x+19), graph the solution on the number line, and write the solution in interval notation.

Exercise (PageIndex{25})

Solve the inequality (8p+3(p−12)>7p−28) graph the solution on the number line, and write the solution in interval notation.

 Simplify each side as much as possible. 8p+3(p−12)>7p−28 Distribute. 8p+3p−36>7p−28 Combine like terms. 11p−36>7p−28 Subtract 7p from both sides to collect the variables on the left. 11p−36−7p>7p−28−7p Simplify. 4p−36>−28 Add 36 to both sides to collect the constants on the right. 4p−36+36>−28+36 Simplify. 4p>8 Divide both sides of the inequality by 4; the inequality stays the same. (frac{4p}{4}>84) Simplify. (p>2) Graph the solution on the number line. Write the solution in interval notation. ((2, infty))

Exercise (PageIndex{26})

Solve the inequality (9y+2(y+6)>5y−24), graph the solution on the number line, and write the solution in interval notation.

Exercise (PageIndex{27})

Solve the inequality (6u+8(u−1)>10u+32), graph the solution on the number line, and write the solution in interval notation.

Just like some equations are identities and some are contradictions, inequalities may be identities or contradictions, too. We recognize these forms when we are left with only constants as we solve the inequality. If the result is a true statement, we have an identity. If the result is a false statement, we have a contradiction.

Exercise (PageIndex{28})

Solve the inequality (8x−2(5−x)<4(x+9)+6x), graph the solution on the number line, and write the solution in interval notation.

 Simplify each side as much as possible. 8x−2(5−x)<4(x+9)+6x Distribute. 8x−10+2x<4x+36+6x Combine like terms. 10x−10<10x+36 Subtract 10x from both sides to collect the variables on the left. 10x−10−10x<10x+36−10x Simplify. −10<36 The xx’s are gone, and we have a true statement. The inequality is an identity.The solution is all real numbers. Graph the solution on the number line. Write the solution in interval notation. ((-infty, infty))

Exercise (PageIndex{29})

Solve the inequality (4b−3(3−b)>5(b−6)+2b), graph the solution on the number line, and write the solution in interval notation.

Exercise (PageIndex{30})

Solve the inequality (9h−7(2−h)<8(h+11)+8h), graph the solution on the number line, and write the solution in interval notation.

Exercise (PageIndex{32})

Solve the inequality (frac{1}{4}x - frac{1}{12}x > frac{1}{6}x + frac{7}{8}), graph the solution on the number line, and write the solution in interval notation.

Exercise (PageIndex{33})

Solve the inequality (frac{2}{5}z - frac{1}{3}z < frac{1}{15}z - frac{3}{5}), graph the solution on the number line, and write the solution in interval notation.

## Translate to an Inequality and Solve

To translate English sentences into inequalities, we need to recognize the phrases that indicate the inequality. Some words are easy, like ‘more than’ and ‘less than’. But others are not as obvious.

Think about the phrase ‘at least’ – what does it mean to be ‘at least 21 years old’? It means 21 or more. The phrase ‘at least’ is the same as ‘greater than or equal to’.

Table (PageIndex{4}) shows some common phrases that indicate inequalities.

>(geq)<(leq)
" data-valign="middle" class="lt-math-15134">is greater thanis greater than or equal tois less thanis less than or equal to
" data-valign="middle" class="lt-math-15134">is more thanis at leastis smaller thanis at most
" data-valign="middle" class="lt-math-15134">is larger thanis no less thanhas fewer thanis no more than
" data-valign="middle" class="lt-math-15134">exceedsis the minimumis lower thanis the maximum
Table (PageIndex{4})

Exercise (PageIndex{34})

Translate and solve. Then write the solution in interval notation and graph on the number line.

Twelve times c is no more than 96.

 Translate. Solve—divide both sides by 12. Simplify. Write in interval notation. Graph on the number line.

Exercise (PageIndex{35})

Translate and solve. Then write the solution in interval notation and graph on the number line.

Twenty times y is at most 100

Exercise (PageIndex{36})

Translate and solve. Then write the solution in interval notation and graph on the number line.

Nine times z is no less than 135

Exercise (PageIndex{37})

Translate and solve. Then write the solution in interval notation and graph on the number line.

Thirty less than x is at least 45.

 Translate. Solve—add 30 to both sides. Simplify. Write in interval notation. Graph on the number line.

Exercise (PageIndex{38})

Translate and solve. Then write the solution in interval notation and graph on the number line.

Nineteen less than p is no less than 47

Exercise (PageIndex{39})

Translate and solve. Then write the solution in interval notation and graph on the number line.

Four more than a is at most 15.

## Key Concepts

• Subtraction Property of Inequality
For any numbers a, b, and c,
if aif a>b then a−c>b−c.
For any numbers a, b, and c,
if aif a>b then a+c>b+c.
• Division and Multiplication Properties of Inequality
For any numbers a, b, and c,
if a0, then acbc.
if a>b and c>0, then ac>bc and ac>bc.
if abc and ac>bc.
if a>b and c<0, then ac
• When we divide or multiply an inequality by a:
• positive number, the inequality stays the same.
• negative number, the inequality reverses.

## Solving Two Step Linear Inequalities

To solve a two-step inequality, undo the addition or subtraction first, using inverse operations , and then undo the multiplication or division.

The inverse operation of addition is subtraction and vice versa.

Similarly, the inverse operation of multiplication is division and vice versa.

Note that, whenever you multiply or divide both sides of an inequality by a negative number, reverse the inequality.

First, we need to isolate the variable term on one side of the inequality. Here, on the left, 1 is added to the variable term, 2 x . The inverse operation of addition is subtraction. So, subtract 1 from both sides.

Now, we have the variable x multiplied by 2 . The inverse operation of multiplication is division. So, divide both sides by 2 .

That is, the inequality is true for all values of x which are less than 3 .

Therefore, the solutions to the inequality 2 x + 1 < 7 are all numbers less than 3 .

First we need to isolate the variable term &minus 3 x on the left. The inverse operation of subtraction is addition. So, add 8 to both sides.

To isolate the variable x , divide both sides by &minus 3 .

Note that, whenever you multiply or divide both sides of an inequality by a negative number, reverse the inequality.

Therefore, the solutions to the inequality &minus 3 x &minus 8 &ge &minus 2 are all numbers less than or equal to &minus 2 .

## Solving Linear Inequalities

Overview
• Section 2.7 in the textbook:
– Graphing inequalities on a number line &
interval notation
– Using the Addition Property of Inequality
– Using the Multiplication Property of Inequality
– Solving inequalities using both properties

Graphing Inequalities on a
Number Line & Interval Notation

Solution Set
• Solution set – all values that satisfy an
inequality
• Often a solution set is expressed using a
number line

Graphing Inequalities on a Number
Line
• Consider the inequality x > 1
–What are some values of x that make the
inequality true?

• <2, 3, 4, 5, …>
– Thus x can be any value greater than 1

• Which direction on the number line indicates
increasing values?
– Can x = 1 be a solution to the inequality?

• Since x = 1 is not in the solution set, we put a
parenthesis around 1 on the number line

• Now consider x ≥ 1
– Graphed ALMOST the same way EXCEPT

• Is x = 1 included in the solution set?
– Since x = 1 is part of the solution set, we put a bracket around 1
on the number line

• Given x < a or x > a:
– Parenthesis goes around a on the number line (not
inclusive)

• Given x ≤ a or x ≥ a:
– Bracket goes around a on the number line (inclusive)

Interval Notation
• Easy once the graph is obtained

• Represents the “endpoints” of the graph
– First “value” is what is shaded furthest to the
left on the graph
– Second “value” is what is shaded furthest to
the right on the graph
– A shaded arrow represents ∞

• Parentheses ALWAYS go around infinity

Graphing Inequalities on a Number
Line & Interval Notation (Example)
Ex 1: Graph on a number line AND write in
interval notation:

• Works the same way as the Addition
Property of EQuality

• When a number is being added or
subtracted to the variable:
– Add the OPPOSITE number to BOTH SIDES

• Consider 2 < 7
–What happens when we add 2 to both sides?
–What happens when we subtract 5?

(Example)
Ex 2: Solve, graph, AND write the solution
set in interval notation:

Multiplication Property of Inequality
• ALMOST the same as the Multiplication
Property of EQuality

• When the variable is being multiplied by a
number:
– Divide BOTH SIDES by the number
INCLUDING THE SIGN

• Consider 4 > 2
–What happens when we divide by 2?
–What happens when we divide by -2?

• Thus, when we DIVIDE an INEQUALITY
by a NEGATIVE number:
– Switch the direction of the inequality
– Failing to SWITCH the inequality when
DIVIDING by a NEGATIVE number is a
common mistake!

Multiplication Property of
Inequality (Example)
Ex 3: Solve, graph, AND write the solution
set in interval notation:

Solving Inequalities Using
Both Properties

Solving Inequalities Using Both
Properties (Example)
Ex 4: Solve, graph, AND write the solution
set in interval notation:

Ex 5: Solve, graph, AND write the solution
set in interval notation:

Summary
• After studying these slides, you should know
how to do the following:
– Graph an inequality on a number line
– Understand the Addition and Multiplication Properties
of Inequality
– Solve and graph inequalities

– See the list of suggested problems for 2.7

• Next lesson
– Solving Absolute Value Equations (Section E.1)

## Solving Linear Inequalities

An inequality is a sentence with , &le, or &ge as its verb. An example is 3x - 5 0 are true, then ac bc is true.
Similar statements hold for a &le b.

When both sides of an inequality are multiplied by a negative number, we must reverse the inequality sign.
First-degree inequalities with one variable, like those in Example 1 below, are linear inequalities.

EXAMPLE 1 Solve each of the following. Then graph the solution set.
a) 3x - 5
Solve 13 - 7x &ge 10x - 4
Subtracting 10x
13 - 17x &ge -4

Dividing by -17 and reversing the inequality sign
x &le 1

The solution set is , or (-&infin, 1]. The graph of the solution set is shown below.

### Compound Inequalities

When two inequalities are joined by the word and or the word or, a compound inequality is formed. A compound inequality like
-3 1. Then graph the solution set.
Solution We have

 2x - 5 &le -7 or 2x - 5 > 1. 2x &le -2 or 2x > 6 Adding 5 x &le -1 or x > 3. Dividing by 2
The solution set is or x > 3>. We can also write the solution using interval notation and the symbol for the union or inclusion of both sets: (-&infin -1] (3, &infin). The graph of the solution set is shown below.

To check, we graph y1 = 2x - 5, y2 = -7, and y3 = 1. Note that for or x > 3>, y1 &le y2 or y1 > y3.

### Inequalities with Absolute Value

Inequalities sometimes contain absolute-value notation. The following properties are used to solve them.
For a > 0 and an algebraic expression X:
|X| a is equivalent to X a.
Similar statements hold for |X| &le a and |X| &ge a.

## Mathematics PreCalculus Mathematics at Nebraska

In the previous section, we talked about linear equations, which are mathematical statements that indicate that two expressions are equal. But what if we would like to compare expressions that are different "sizes"?

###### In this section, you will.

represent solutions to inequalities using a variety of representations

### Subsection Linear Inequalities

The symbol (gt) is called an , and the statement (agt b) is called an . There are four inequality symbols:

Inequalities that include the symbols (gt) or (lt) are called those that include (ge) or (le) are called .

If an inequality has one or more variables in it, a of that inequality is any set of values that can replace the variables to produce a true statement. For example, (3) is a solution for (2x>2) since (2cdot3>2) is a true statement. However, (3) is not the only solution! In fact, any number greater than (1) is a solution. While linear equations have exactly one, zero, or infinitely many solutions, linear inequalities can have much more complex solution sets.

Solving linear inequalities is very similar to solving linear equalities. The main difference is in multiplying and dividing: if we multiply or divide both sides of an inequality by a negative number, the direction of the inequality must be reversed. For example, if we multiply both sides of the inequality

Because of this property, the rules for solving linear equations must be revised slightly for solving linear inequalities.

###### To Solve a Linear Inequality

We may add the same number to both sides of an inequality or subtract the same number from both sides of an inequality without changing its solutions.

We may multiply or divide both sides of an inequality by a positive number.

If we multiply or divide both sides of an inequality by a negative number, we must reverse the direction of the inequality symbol.

###### Example 53

Solve the inequality (4 - 3x ge -17 ext<.>)

Use the rules above to isolate (x) on one side of the inequality.

Notice that we reversed the direction of the inequality when we divided by (-3 ext<.>) Any number less than or equal to (7) is a solution of the inequality.

A involves two inequality symbols. To solve a compound inequality, we use the same steps as before, applying the operations on all three "sides" of the inequality symbols.

###### Example 54

Solve (4 le 3x + 10 le 16 ext<.>)

We isolate (x) by performing the same operations on all three sides of the inequality.

The solutions are all numbers between (-2) and (2 ext<,>) inclusive.

###### Caution 55

Notice in the previous example that in a compound inequality, both of the inequality symbols are in the same "direction". You typically would not see something like (4lt x geq 8) or (-3geq x leq 2 ext<.>)

### Subsection Interval Notation

A common way to represent solutions of an inequality is with . An interval is a set that consists of all the real numbers between two numbers (a) and (b ext<.>)

The set (-2 le x le 2) includes its endpoints (-2) and (2 ext<,>) so we call it a , and we denote it by ([-2, 2]) (see Figure56a). The square brackets tell us that the endpoints are included in the interval. An interval that does not include its endpoints, such as (-2 lt x lt 2 ext<,>) is called an , and we denote it with round brackets, ((-2, 2)) (see Figure56b).

###### Caution 57

Do not confuse the open interval ((-2, 2)) with the point ((-2, 2) ext) The notation is the same, so you must decide from the context whether an interval or a point is being discussed.

We can also discuss , such as (xlt 3) and (xge -1 ext<,>) shown in Figure58. We denote the interval (xlt 3) by ((-infty, 3) ext<,>) and the interval (xge -1) by ([-1, infty) ext<.>) The symbol (infty ext<,>) for infinity, does not represent a specific real number but rather indicates that the interval continues forever along the real line. We always use round brackets next to (pminfty) in infinite intervals.

Finally, we can combine two or more intervals into a larger set. For example, the set consisting of (xlt -1) or (xgt 2 ext<,>) shown in Figure59, is the of two intervals and is denoted by ((-infty,-2) cup (2,infty) ext<.>)

Many solutions of inequalities are intervals or unions of intervals.

###### Example 60

Write each of the solution sets with interval notation and graph the solution set on a number line.

## Solving Systems of Linear Inequalities

Solutions to a system of linear inequalities are the ordered pairs that solve all the inequalities in the system. Therefore, to solve these systems, graph the solution sets of the inequalities on the same set of axes and determine where they intersect. This intersection, or overlap, defines the region of common ordered pair solutions.

Example 1: Graph the solution set: < − 2 x + y > − 4 3 x − 6 y ≥ 6 .

Solution: To facilitate the graphing process, we first solve for y.

For the first inequality, we use a dashed boundary defined by y = 2 x − 4 and shade all points above the line. For the second inequality, we use a solid boundary defined by y = 1 2 x − 1 and shade all points below. The intersection is darkened.

Now we present the solution with only the intersection shaded.

Example 2: Graph the solution set: < − 2 x + 3 y > 6 4 x − 6 y > 12 .

Solution: Begin by solving both inequalities for y.

Use a dashed line for each boundary. For the first inequality, shade all points above the boundary. For the second inequality, shade all points below the boundary.

As you can see, there is no intersection of these two shaded regions. Therefore, there are no simultaneous solutions.

Example 3: Graph the solution set: < y ≥ − 4 y < x + 3 y ≤ − 3 x + 3 .

Solution: The intersection of all the shaded regions forms the triangular region as pictured darkened below:

After graphing all three inequalities on the same set of axes, we determine that the intersection lies in the triangular region pictured.

The graphic suggests that (−1, 1) is a common point. As a check, substitute that point into the inequalities and verify that it solves all three conditions.

### Key Takeaway

• To solve systems of linear inequalities, graph the solution sets of each inequality on the same set of axes and determine where they intersect.

### Topic Exercises

Part A: Solving Systems of Linear Inequalities

Determine whether the given point is a solution to the given system of linear equations.

## DISTANCE-RATE-TIME PROBLEMS

### OBJECTIVES

Upon completing this section you should be able to:

1. Identify distance-rate-time problems.
2. Apply the distance formula to solve problems in this group.

Another formula often found in verbal problems is

d = rt (distance = rate X time),

which is the distance formula for constant rate. Given the rate at which an object is moving and the time that it moves at this rate, we can find the distance the object moves.

Example 1 A car travels at the rate of 55 mph for four hours. How far does it travel?

Using the formula d = rt, substitute r = 55 and t = 4, obtaining

Thus, the distance traveled is 220 miles.

 How far would the car travel in one hour? In two hours? In three hours?

If the distance and rate are both given, we can find the time.

Example 2 How long will it take a plane whose ground speed is 530 mph to travel 2,120 miles?

In this case, let d = 2,120 and r = 530 in the formula

It will therefore take four hours to travel the distance.

 Always check to see that the units agree. That is, if the speed is given in miles per hour then the distance must also be in miles.

We can also solve for the rate if we are given the distance and time.

Example 3 A person walked 21 miles in hours. What was the person's average rate?

We substitute d = 21 and t = 3.5 in the formula

The person's average rate was six miles per hour.

One type of distance problem involves two objects leaving from the same point and traveling in the same direction.

 You could also leave the time in fraction form. Then

Example 4 A bank robber leaves town heading north at an average speed of 120 kilometers per hour. The sheriff leaves two hours later in a plane that travels at 200 kilometers per hour. How long will it take the sheriff to catch the robber?

In this table x represents the time the sheriff takes to catch the robber. Notice that the distance 120(x + 2) and 200x come from the formula d = rt.

 Use a table such as this one when two or more moving objects are involved. It allows for easy comparisons.

The equation comes from the fact that the sheriff and the robber will have traveled the same distance when the sheriff catches the robber. Thus,

The sheriff will catch the robber in three hours.

 Notice that since the sheriff left two hours later, the robber will have been on the run for x + 2 hours when he is caught. When the sheriff catches the robber, they will each have traveled the same distance.

Check: The bank robber has traveled for five hours at 120 km/ hr for a distance of 5(120) = 600 km.

The sheriff has traveled for three hours at 200 km/hr for a distance of 3(200) = 600 km.

Another type of distance problem involves two objects leaving from the same point and traveling in opposite directions.

Example 5 Pamela and Sue start at the same point and walk in opposite directions. The rate at which they are moving away from one another is 11 mph. At the end of three hours Pamela stops and Sue continues to walk for another hour. At the end of that time Pamela has walked twice as far as Sue. How far apart are they?

We use the following table:

 The problem tells us the time and lets us represent the distance. To represent the value for r we divide the distance by the time.

The sum of their rates is 11 mph.

Thus, they are 36 miles apart.

Check: Pamela's rate is Sue's rate is Their total rate is 11 mph. Also, Pamela's distance (24) is twice Sue's distance (12).

 Again, check to make sure the answers agree with the original problem.

Still another type of distance problem involves two objects that leave from two different points and travel toward each other.

Example 6 Juan and Steven started 36 miles apart and walked toward each other, meeting in three hours. If Juan walked two miles per hour faster than Steven, find the rate of each.

 If we let x represent the rate at which Steven walked, then Juan's rate would be represented by x + 2.

First set up the following table:

 We represent the distance as the product of the rate and time (d = rt).

The total distance they have traveled is 36 miles. Thus,

Juan's rate = 7 mph
Steven's rate = 5 mph.

Check: Juan's rate (7) is two miles per hour faster than Steven's (5). Also 3(7) + 3(5) = 21 + 15 = 36.

Within the class of problems using d = rt is a subclass of problems concerned with parallel and opposing forces.

 Parallel forces travel in the same direction, and opposing forces travel in opposite directions. The normal speed of the plane will be reduced by the speed of the wind.

Example 7 A plane, whose speed in still air is 550 mph, flies against a headwind of 50 mph. How long will it take to travel 1,500 miles?

We use the formula d = rt. The distance is 1,500 miles and the rate of the plane against the wind will be its still air speed (550 mph) reduced by the headwind (50 mph).

Thus, it will take three hours to travel the distance.
Check: 3(550 - 50) = 3(500) = 1,500

Example 8 Mike can row his boat from the hunting lodge upstream to the park in five hours. He can row back from the park to the lodge downstream in three hours. If Mike can row x kilometers per hour in still water, and if the stream is flowing at the rate of two kilometers per hour, how far is it from the lodge to the park?

In working the problem we assume that the rate of the stream will increase or decrease the rate of the boat by two kilometers per hour. Since x represents Mike's speed in still water, we obtain

Setting the distance upstream equal to the distance downstream, we obtain

Notice that x is not the solution to the problem but is Mike's rate of rowing in still water. However, the question asked is "What is the distance from the lodge to the park?" To answer this use either the distance upstream or downstream since they are the same. Using the upstream distance, we have

 When Mike is rowing upstream, he is opposing the direction of the current. When he rows downstream, he is parallel to the stream's current.

 See if you get the same answer using the distance downstream.

## 2.7: Solve Linear Inequalities

1. Solve the following inequality and give the solution in both inequality and interval notation.

[4left( ight) - 1 > 5 - 7left( <4 - z> ight)]

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We know that the process of solving a linear inequality is pretty much the same process as solving a linear equation. We do basic algebraic manipulations to get all the (z)’s on one side of the inequality and the numbers on the other side. Just remember that what you do to one side of the inequality you have to do to the other side as well. So, let’s go through the solution process for this linear inequality.

First, we should clear out the parenthesis on both sides and do any simplification that we can. Doing this gives,

[egin4z + 8 - 1 > 5 - 28 + 7z 4z + 7 > - 23 + 7zend] Show Step 2

We can now subtract 7(z) from both sides and subtract 7 to both sides to get,

Note that we could just have easily subtracted 4(z) from both sides and added 23 to both sides. Each will get the same result in the end.

For the final step we need to divide both sides by -3. Recall however that because we are dividing by a negative number we need to switch the direction of the inequality to get,

So, the inequality form of the solution is ( equire box[2pt,border:1px solid black]<>) and the interval notation form of the solution is ( equire box[2pt,border:1px solid black] < ight)>>) .

## 2.7: Solve Linear Inequalities

6. Solve the following inequality and give the solution in both inequality and interval notation.

Show All Steps Hide All Steps

Just like with single inequalities solving these follow pretty much the same process as solving a linear equation. The only difference between this and a single inequality is that we now have three parts of the inequality and so we just need to remember that what we do to one part we need to do to all parts.

Also, recall that the main goal is to get the variable all by itself in the middle and all the numbers on the two outer parts of the inequality.

So, let’s start by subtracting (frac<1><6>) from all the parts. This gives,

Finally, all we need to do is multiply all three parts by -2 to get,

Don’t forget that because we were multiplying everything by a negative number we needed to switch the direction of the inequalities.

So, the inequality form of the solution is ( equire box[2pt,border:1px solid black]<< - frac<<23>> <3>le x < - frac<<11>><3>>>) (we flipped the inequality around to get the smaller number on the left as that is a more “standard” form). The interval notation form of the solution is ( equire box[2pt,border:1px solid black]<><3>, - frac<<11>><3>> ight)>>) .

For the interval notation form remember that the smaller number is always on the left (hence the reason for flipping the inequality form above!) and be careful with parenthesis and square brackets. We use parenthesis if we don’t include the number and square brackets if we do include the number.