Chapter3.6

Evaluating Algebraic Expressions

3-6 Solving Inequalities by Adding or Subtracting

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California StandardsCalifornia Standards

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1. 5 7 2. –3 –4

3. 2.5 –2.7 4. –8 –7

Solve.5. 4 + y = 16 6. m – 7 = 14

7. –3 = 8 + w 8. 7 = t + 10

< >

> <

12 21

–11 –3



AF4.0 Students solve simple linear equations and inequalities over the rational numbers.

California Standards



When you add or subtract the same number on both sides of an inequality, the resulting statement will still be true.

–2 < 5+7 +7

5 < 12

You can find solution sets of inequalities the same way you find solutions of equations, by isolating the variable.



Solve and graph the inequality.

Additional Example 1A: Solving Inequalities by Adding or Subtracting

x + 3 > –5x + 3 > –5

–3 –3

x > –8

Since 3 is added x, subtract 3 from both sides.

–9 8 –7 6 5 4 3 2 1 0 1 2 3 4 5



Additional Example 1A Continued

Check

x + 3 > –5

–4 + 3 > –5?

–1 > –5?

Substitute –4 for x.

According to the graph –4 should be a solution and –9 should not be a solution.

So –4 is a solution.

x + 3 > –5

–9 + 3 > –5?

–6 > –5?


So –9 is not a solution.



m – 4 ≥ –2

m – 4 ≥ –2+ 4 + 4

m ≥ 2

Since 4 is subtracted from m, add 4 to both sides.

–1 0 1 2 3 4 5 6 7 8 9 10 11 12


Additional Example 1B: Solving Inequalities by Adding or Subtracting



r + 3 ≤ –3 r + 3 ≤ –3

– 3 –3

r ≤ –6

Since 3 is added to r, subtract 3 from both sides.

Additional Example 1C: Solving Inequalities by Adding or Subtracting

–9 8 –7 6 5 4 3 2 1 0 1 2 3 4 5




5 > n + 1

Since 1¼ is added to n, subtract 1¼ from both sides.

Additional Example 1D: Solving Inequalities by Adding or Subtracting


34

14

5 > n + 1 34

14

– 1 – 1 14

14

4 > n 12

–7 6 –5 4 3 2 1 0 1 2 3 4 5 6 7




Check It Out! Example 1A

x + 4 > –2x + 4 > –2

–4 –4

x > –6

Since 4 is added x, subtract 4 from both sides.

–9 8 –7 6 5 4 3 2 1 0 1 2 3 4 5



Check It Out! Example 1A Continued

Check

x + 4 > –2

2 + 4 > –2?

6 > –2?

Substitute 2 for x.

According to the graph 2 should be a solution and –8 should not be a solution.

So 2 is a solution.

x + 4 > –2

–8 + 4 > –2?

–4 > –2?


So –8 is not a solution.



w – 8 ≥ –3

w – 8 ≥ –3+ 8 + 8

w ≥ 5

Since 8 is subtracted from w, add 8 to both sides.

–1 0 1 2 3 4 5 6 7 8 9 10 11 12


Check It Out! Example 1B



c + 6 ≤ –1 c + 6 ≤ –1

– 6 – 6

c ≤ –7

Since 6 is added to c, subtract 6 from both sides.

Check It Out! Example 1C

–9 8 –7 6 5 4 3 2 1 0 1 2 3 4 5




3 > n + 1

Since 1 is added to n, subtract

1 from both sides.

Check It Out! Example 1D


23

13

3 > n + 1 23

13

– 1 – 1 13

13

2 > n 13

–7 6 –5 4 3 2 1 0 1 2 3 4 5 6 7

13

13



While training for a race, Ann’s goal is to run at least 3.5 miles each day. She has already run 1.8 miles today. Write and solve an inequality to find out how many more miles she must run today.

Additional Example 2: Sports Application

Let m = the number of additional miles.

1.8 + m ≥ 3.5–1.8 –1.8

m ≥ 1.7

Ann should run at least 1.7 more miles.

Since 1.8 is added to m, subtract 1.8 from both sides.

1.8 miles plus additional miles is at least 3.5 miles. 1.8 + m ≥ 3.5



2 is greater than 1.7. Substitute 2 for m.

Check

Additional Example 2 Continued

1.8 + m ≥ 3.5

1.8 + 2 ≥ 3.5 ?

3.8 ≥ 3.5 ?

x

1 is less than 1.7. Substitute 1 for m.

1.8 + m ≥ 3.5

1.8 + 1 ≥ 3.5 ?

2.8 ≥ 3.5 ?



Tim’s company produces recycled paper. They produce 60.5 lb of paper each day. They have already produced at least 20.2 lb today. Write and solve an inequality to find out how many more pounds Tim’s company must produce.

Check It Out! Example 2

Let p = the number of additional pounds of paper.

20.2 + p ≥ 60.5–20.2 –20.2

p ≥ 40.3Tim’s company should produce at least 40.3 lb more of paper.

Since 20.2 is added to p, subtract 20.2 from both sides.

20.2 lbs plus additional pounds is at least 60.5 lb.

20.2 + p ≥ 60.5



41 is greater than 40.3. Substitute 41 for p.

Check

Check It Out! Example 2 Continued

20.2 + p ≥ 60.5

20.2 + 41 ≥ 60.5 ?

61.2 ≥ 60.5 ?

x

40 is less than 40.3. Substitute 40 for p.

20.2 + p ≥ 60.5

20.2 + 40 ≥ 60.5 ?

60.2 ≥ 60.5 ?



Lesson Quiz: Part I

Solve and graph each inequality.

1. g – 7 < –3

2. 5 + s ≥ 4

3. –5.1 ≤ x – 5.1

4. 3 + y > 4

g < 4

1 0 1 2 3 4 5

4 3 2 1 0 1 2•

s ≥ –1

4 3 2 1 0 1 2•

x ≥ 0

15

y <

1/5 2/5 3/5 4/5 1 1 1/5

45



Lesson Quiz: Part II

5. Tasha is folding letters for a fundraiser. She

knows there are at least 300 letters, and she

has already folded 125 of them. Write and

solve an inequality to show how many more

letters she must fold.

125 + x ≥ 300; x ≥ 175

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Chapter3.6