Holt Algebra 1 3-5 Solving Inequalities with Variables on Both Sides Warm Up Solve each equation. 1. 2x = 7x + 15 2. 5. Solve and graph 5(2 – b) > 5 2.

Holt Algebra 1

3-5 Solving Inequalities with Variables on Both Sides

Warm UpSolve each equation. 1. 2x = 7x + 15

2.

5. Solve and graph 5(2 – b) > 52.

3. 2(3z + 1) = –2(z + 3)

4. 3(p – 1) = 3p + 2

x = –3

b < –3

–5 –3 –2 –1–4 0–6

3y – 21 = 4 – 2y y = 5

z = –1

no solution

Holt Algebra 1


Solve inequalities that contain variable terms on both sides.

Objective

Holt Algebra 1


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.

Holt Algebra 1


Directions: Solve the inequality and graph the solutions.

Holt Algebra 1


Example 1

y ≤ 4y + 18

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

Holt Algebra 1


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

Example 2

Holt Algebra 1


Example 3

4x ≥ 7x + 6

4x ≥ 7x + 6–7x –7x

–3x ≥ 6

x ≤ –2

To collect the variable terms on one side, subtract 7x from both sides.

Since x is multiplied by –3, divide both sides by –3 to undo the multiplication. Change ≥ to ≤.

–10 –8 –6 –4 –2 0 2 4 6 8 10

Holt Algebra 1


Example 4

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.

Holt Algebra 1


Example 5: Business Application

The Home Cleaning Company charges $312 to power-wash the siding of a house plus $12 for each window. Power Clean charges $36 per window, and the price includes power-washing the siding. How many windows must a house have to make the total cost from The Home Cleaning Company less expensive than Power Clean?

Let w be the number of windows.

Holt Algebra 1


Example 5 Continued

312 + 12 • w < 36 • w

312 + 12w < 36w– 12w –12w

312 < 24w

13 < w

The Home Cleaning Company is less expensive for houses with more than 13 windows.

To collect the variable terms, subtract 12w from both sides.

Since w is multiplied by 24, divide both sides by 24 to undo the multiplication.

HomeCleaningCompany

siding charge

plus$12 per window

# of windows

is lessthan

PowerClean

cost per window

# ofwindows.times

times

Holt Algebra 1


Example 6

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

Holt Algebra 1


Example 6 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.

Holt Algebra 1


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.

Holt Algebra 1


Example 7

2(k – 3) > 6 + 3k – 3

2(k – 3) > 3 + 3k Distribute 2 on the left side of the inequality.

2k + 2(–3) > 3 + 3k

2k – 6 > 3 + 3k–2k – 2k

–6 > 3 + k

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

–3 –3

–9 > k

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

Holt Algebra 1


Example 7 Continued

–9 > k

–12 –9 –6 –3 0 3

Holt Algebra 1


Example 8

0.9y ≥ 0.4y – 0.5

0.9y ≥ 0.4y – 0.5–0.4y –0.4y

0.5y ≥ – 0.5

0.5y ≥ –0.5 0.5 0.5

y ≥ –1

To collect the variable terms, subtract 0.4y from both sides.

Since y is multiplied by 0.5, divide both sides by 0.5 to undo the multiplication.

–5 –4 –3 –2 –1 0 1 2 3 4 5

Holt Algebra 1


Example 9

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.

Holt Algebra 1


Example 9 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

Holt Algebra 1


Example 10

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

Holt Algebra 1


Example 10 Continued

x < 3

–5 –4 –3 –2 –1 0 1 2 3 4 5

Holt Algebra 1


There are special cases of inequalities called identities and contradictions.

Holt Algebra 1


Holt Algebra 1


Example 11

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.

Holt Algebra 1


2(3y – 2) – 4 ≥ 3(2y + 7)

2(3y – 2) – 4 ≥ 3(2y + 7)

2(3y) – 2(2) – 4 ≥ 3(2y) + 3(7)

6y – 4 – 4 ≥ 6y + 21

6y – 8 ≥ 6y + 21

Distribute 2 on the left side and 3 on the right side.

Example 12

–6y –6y

–8 ≥ 21

Subtract 6y from both sides.

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

Holt Algebra 1


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.

Example 13

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.

Holt Algebra 1


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.

Example 14

Holt Algebra 1


Lesson Summary: 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

Holt Algebra 1


Lesson Summary: 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.

Holt Algebra 1


Lesson Summary: 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

Holt Algebra 1 3-5 Solving Inequalities with Variables on Both Sides Warm Up Solve each equation. 1. 2x = 7x + 15 2. 5. Solve and graph 5(2 – b) > 5 2.

Documents

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sides directions

holt algebra

variable terms

solution slide

objective slide

constant terms

home cleaning company