holt mcdougal algebra 1 2-5 solving inequalities with variables on both sides 2-5 solving...

Holt McDougal Algebra 1

2-5 Solving Inequalities with Variables on Both Sides2-5 Solving Inequalities with Variables

on Both Sides

Holt Algebra 1

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Lesson QuizLesson Quiz



2-5 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



Solve inequalities that contain variable terms on both sides.

Objective



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 1A: Solving Inequalities with Variables on Both Sides

Solve the inequality and graph the solutions.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



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 1B: Solving Inequalities with Variables on Both Sides

Solve the inequality and graph the solutions.




Check It Out! Example 1a

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




Check It Out! Example 1b

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.



Example 2: 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.



Example 2 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



Check It Out! Example 2

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.



Example 3A: Simplify Each Side Before Solving


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.



Example 3A Continued

–9 > k

–12 –9 –6 –3 0 3



Example 3B: Simplify Each Side Before Solving

Solve the inequality and graph the solution.

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





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





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



Check It Out! Example 3b Continued

x < 3

–5 –4 –3 –2 –1 0 1 2 3 4 5



Some inequalities are true no matter what value is substituted for the variable. For these inequalities, all real numbers are solutions.

Some inequalities are false no matter what value is substituted for the variable. These inequalities have no solutions.

If both sides of an inequality are fully simplified and the same variable term appears on both sides, then the inequality has all real numbers as solutions or it has no solutions. Look at the other terms in the inequality to decide which is the case.



Additional Example 4A: All Real Numbers as Solutions or No Solutions

Solve the inequality.

2x – 7 ≤ 5 + 2x

The same variable term (2x) appears on both sides. Look at the other terms.

For any number 2x, subtracting 7 will always result in a lower number than adding 5.

All values of x make the inequality true.

All real numbers are solutions.



2(3y – 2) – 4 ≥ 3(2y + 7)


Additional Example 4B: All Real Numbers as Solutions or No Solutions

Distribute 2 on the left side and 3 on the right side and combine like terms.

6y – 8 ≥ 6y + 21

The same variable term (6y) appears on both sides. Look at the other terms.

For any number 6y, subtracting 8 will never result in a higher number than adding 21.

No values of y make the inequality true.

There are no solutions.



4(y – 1) ≥ 4y + 2

4y – 4 ≥ 4y + 2 Distribute 4 on the left side.



The same variable term (4y) appears on both sides. Look at the other terms.

For any number 4y, subtracting 4 will never result in a higher number than adding 2.

No values of y make the inequality true.

There are no solutions.




x – 2 < x + 1


The same variable term (x) appears on both sides. Look at the other terms.

For any number x, subtracting 2 will always result in a lesser number than adding 1.

All values of x make the inequality true.

All real numbers are solutions.



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)

no solutions

6. 2(–6r – 5) < –3(4r + 2)

all real numbers

holt mcdougal algebra 1 2-5 solving inequalities with variables on both sides 2-5 solving...

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properties of inequality