3-5Solving Inequalities with Variables on Both Sides

EQ: How can you get the variables on one side of the inequality?

Some inequalities have variable terms on both sides of the inequality symbol. You can solve these inequalities like you solved equations with variables on both sides.

Use the properties of inequality to “collect” all the variable terms on one side and all the constant terms on the other side.

Example 1

Solve the inequality and graph the solutions.

y ≤ 4y + 18–y –y

0 ≤ 3y + 18–18 – 18

–18 ≤ 3y

–6 ≤ y (or y –6)

To collect the variable terms on one side, subtract y from both sides.

Since 18 is added to 3y, subtract 18 from both sides to undo the addition.

Since y is multiplied by 3, divide both sides by 3 to undo the multiplication.

–10 –8 –6 –4 –2 0 2 4 6 8 10

4m – 3 < 2m + 6To collect the variable terms on one

side, subtract 2m from both sides.–2m – 2m

2m – 3 < + 6

Since 3 is subtracted from 2m, add 3 to both sides to undo the subtraction

+ 3 + 3

2m < 9 Since m is multiplied by 2, divide both sides by 2 to undo the multiplication.

4 5 6

Your Turn...

Solve the inequality and graph the solutions.

Solve the inequality and graph the solutions.

A Little Harder…

5t + 1 < –2t – 6

5t + 1 < –2t – 6+2t +2t

7t + 1 < –6– 1 < –1

7t < –77t < –77 7t < –1

–5 –4 –3 –2 –1 0 1 2 3 4 5

To collect the variable terms on one side, add 2t to both sides.

Since 1 is added to 7t, subtract 1 from both sides to undo the addition.

Since t is multiplied by 7, divide both sides by 7 to undo the multiplication.

Real Life Application

A-Plus Advertising charges a fee of $24 plus $0.10 per flyer to print and deliver flyers. Print and More charges $0.25 per flyer. For how many flyers is the cost at A-Plus Advertising less than the cost of Print and More?

Let f represent the number of flyers printed.

24 + 0.10 • f < 0.25 • f

plus

$0.10per

flyer

is lessthan

# of flyers.

A-Plus

Advertising

fee of $24

Print and

More’s cost

per flyer

# of flyers

times times

Check It Out! Example 2 Continued

24 + 0.10f < 0.25f

–0.10f –0.10f

24 < 0.15f

160 < f

To collect the variable terms, subtract 0.10f from both sides.

Since f is multiplied by 0.15, divide both sides by 0.15 to undo the multiplication.

More than 160 flyers must be delivered to make A-Plus Advertising the lower cost company.

You may need to simplify one or both sides of an inequality before solving it. Look for like terms to combine and places to use the Distributive Property.

Distributive Property!

Solve the inequality and graph the solutions.

Distribute 2 on the left side of the inequality.

To collect the variable terms, subtract 2k from both sides.

2(k – 3) > 3 + 3k

2k + 2(–3) > 3 + 3k

2k – 6 > 3 + 3k–2k – 2k

–6 > 3 + k–3 –3

–9 > k

Since 3 is added to k, subtract 3 from both sides to undo the addition.

–12 –9 –6 –3 0 3

k < -9

Even Harder…

Solve the inequality and graph the solutions.

5(2 – r) ≥ 3(r – 2)

5(2 – r) ≥ 3(r – 2)

5(2) – 5(r) ≥ 3(r) + 3(–2)

10 – 5r ≥ 3r – 6+6 +6

16 − 5r ≥ 3r+ 5r +5r

16 ≥ 8r

Distribute 5 on the left side of the inequality and distribute 3 on the right side of the inequality.

Since 6 is subtracted from 3r, add 6 to both sides to undo the subtraction.

Since 5r is subtracted from 16 add 5r to both sides to undo the subtraction.

Check It Out! Example 3a Continued

–6 –2 0 2–4 4

16 ≥ 8rSince r is multiplied by 8, divide

both sides by 8 to undo the multiplication.

2 ≥ r

What About Decimals?

Solve the inequality and graph the solutions.

0.5x – 0.3 + 1.9x < 0.3x + 6

2.4x – 0.3 < 0.3x + 6+ 0.3 + 0.3 2.4x < 0.3x + 6.3

–0.3x –0.3x

2.1x < 6.3

Since 0.3 is subtracted from 2.4x, add 0.3 to both sides.

Since 0.3x is added to 6.3, subtract 0.3x from both sides.

x < 3

Since x is multiplied by 2.1, divide both sides by 2.1.

Simplify.2.4x – 0.3 < 0.3x + 6

There are special cases of inequalities called identities and contradictions.

Example 4A: Identities and Contradictions

Solve the inequality.

2x – 7 ≤ 5 + 2x

2x – 7 ≤ 5 + 2x–2x –2x

–7 ≤ 5Subtract 2x from both sides.

True statement.

The inequality 2x − 7 ≤ 5 + 2x is an identity. All values of x make the inequality true. Therefore, all real numbers are solutions.

4(y – 1) ≥ 4y + 2

4(y – 1) ≥ 4y + 2

4(y) + 4(–1) ≥ 4y + 2

4y – 4 ≥ 4y + 2

Distribute 4 on the left side.

Check It Out! Example 4a

Solve the inequality.

–4y –4y

–4 ≥ 2

Subtract 4y from both sides.

False statement.

No values of y make the inequality true. There are no solutions.

Solve the inequality.

x – 2 < x + 1

x – 2 < x + 1 –x –x

–2 < 1Subtract x from both sides.True statement.

All values of x make the inequality true. All real numbers are solutions.

Check It Out! Example 4b

Homework

Workbook page 23Pick any 8 problems

Lesson Quiz: Part I

Solve each inequality and graph the solutions.

1. t < 5t + 24 t > –6

2. 5x – 9 ≤ 4.1x – 81 x ≤ –80

b < 133. 4b + 4(1 – b) > b – 9

Lesson Quiz: Part II

4. Rick bought a photo printer and supplies for $186.90, which will allow him to print photos for $0.29 each. A photo store charges $0.55 to print each photo. How many photos must Rick print before his total cost is less than getting prints made at the photo store?

Rick must print more than 718 photos.

Lesson Quiz: Part III

Solve each inequality.

5. 2y – 2 ≥ 2(y + 7)

contradiction, no solution

6. 2(–6r – 5) < –3(4r + 2)

identity, all real numbers