ALGEBRA READINESS

LESSON 9-2Warm Up Lesson 9-2 Warm-Up

ALGEBRA READINESS

LESSON 9-2Warm Up Lesson 9-2 Warm-Up

ALGEBRA READINESS

“Solving Inequalities by Adding or Subtracting” (9-2)

(3-1)What are “equivalent inequalities”?

What is the “Addition Property of Inequality”?

Equivalent Inequalities: two inequalities with the same solutions

Example: x < 3 has the same solution as x + 4 < 7. As you can see in the picture below, if you take 4 blocks away from both sides of x + 4 7, you get x < 3.

Rule: Addition Property of Inequality: If you add the same number to both sides of an inequality, the inequality is equivalent to the original inequality.

Note: This property is also true for ≥ and ≤.

if a > b, then a + c > b + c if a < b, then a + c < b + c

Example: 3 > 1, so 3 + 1 > 1 + 1 Example: -5 < 4, so -5 + 2 < 4 + 2

4 > 2 -3 < 6

Notice the left side is stays 2 bigger than the right side in the example on the left and 9 less than the right side in the example on the right if you add the same amount to both sides of the inequality.

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How do you check to make sure an equality it written or graphed correctly?

To check an inequality, replace the variable with one of the solutions of the inequality

Example: Solve and graph x – 3 < 5.

x – 3 + 3 < 5 + 3 Add 3 to both sides.

x < 8 Simplify

Check (Step 1): Both sides of the inequality should be equal when you substitute the variable with the solution.

8 – 3 = 5 Substitute or “Plug In” 8 for x.

5 = 5 (True Statement) Simplify

Check (Step 2): Check to make sure the inequality symbol in pointing the correct way by substituting the variable with an solution.

7 – 3 < 5 7 is one of the solutions for x < 8.

4 < 5 (True Statement) Simplify

“Solving Inequalities by Adding or Subtracting” (9-2)

ALGEBRA READINESS

Solve p –3 < –5.

p – 3 < – 5

Add 3 to both sides to isolate the variable + 3 + 3

Step 1: Check whether your answer is a solution to the related equation.

Check:

p – 3 = – 5 Write the related equation.

– 5 = – 5

Substitute –2 for p.(–2) – 3 – 5

Solving Inequalities by Adding or SubtractingLESSON 9-2

Additional Examples

p < – 2 Simplify.+ 0

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(continued)

Substitute a number less than –2 for p.(–3) – 3 < – 5

Step 2: Check the inequality symbol by substituting into the inequality.

p – 3 < – 5

– 6 < – 5

Solving Inequalities by Adding or SubtractingLESSON 9-2

Additional Examples

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Solve 8 d – 2. Graph and check your solution.>

+ 2 + 2 Add 2 to each side.

10 d , or d 10 Simplify (Switch the sign if you switch the signaround).

> <

Check: 8 = d – 2 Check the computation.

8 (10) – 2 Substitute 10 for d.

8 = 88 ≥ d – 2 Check the direction of the inequality.

8 ≥ (9) – 2 Substitute 9 for d.

8 ≥ 7

Solving Inequalities by Addition and SubtractingLESSON 9-2

Additional Examples

8 d – 2 >

+ 0

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“Solving Inequalities by Adding and Subtracting” (9-2)

(3-1)What is the “Subtraction Property of Inequality”?

Rule: Subtraction Property of Inequality: If you subtract the same number to both sides of an inequality, the inequality is equivalent to the original inequality.

if a > b, then a - c > b - c if a < b, then a - c < b - c

Example: 3 > 1, so 3 - 1 > 1 - 1 Example: -5 < 4, so -5 - 2 < 4 - 2

Note: This property is also true for ≥ and ≤.

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The Drama Club can spend no more than $120 on

costumes. They spent $79. How much more can they spend?

≤

Words amount plus amount left is no more $120 spent to spend than

Let c = the amount left to spend on costumes.

Inequality 79 + c 120

Subtract 79 from both sides to isolate the variable.– 79 – 79

79 + c ≤ 120

The Drama Club can spend at most $41 more on costumes.

c ≤ 41 Simplify.

Solving Inequalities by Adding or SubtractingLESSON 9-2

Additional Examples

0 +

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Solve each inequality.

1. q – 5 8 2. r + 10 < 4

3. 6 + x 21

4. A lamp can use light bulbs of up to 75 watts. The lamp is using a 45-watt bulb. At most, how many watts are available for brighter light?

>–

<–

r < – 6

30 watts

x 15<–

q 13>–

Solving Inequalities by Adding or SubtractingLESSON 9-2

Lesson Quiz