Page 554 - Algebra 1

Simplifying Before Solving Inequalities
Example
2
Solve the inequality. Justify each step.
a. -15≤3(2x-1)≤39 SOLUTION
-15 ≤ 3(2x - 1) ≤ 39 -15 ≤ 6x - 3 ≤ 39 +__3 +__3 +__3 -12 ≤ 6x ≤ 42
_1 · -12 ≤ _1 · 6x ≤ _1 · 42 666
-2 ≤ x ≤ 7
Distributive Property
Addition Property of Inequality Simplify.
Multiplication Property of Inequality
Simplify.
b. -12 ≥ -6b - 18 OR -2(4 - b) ≥ 10 SOLUTION
-12 ≥ -6b - 18 OR -2(4 - b) ≥ 10 -12 ≥ -6b - 18 OR -8 + 2b ≥ 10 +_ 1_8 +_ 1_8 +__8 +__8
Distributive Property
Addition Property of Inequality Simplify.
Division Property of Inequality Simplify.
6 ≥ -6b OR 2b ≥ 18 6≤-6b OR_2b≥18
___ -6-6 22
-1 ≤ b OR b ≥ 9 Application: Zoology
Example
3
Zoologists randomly choose 5 zebras out of a herd of 20. Four zebras weigh 540 pounds, 550 pounds, 520 pounds, and 530 pounds, respectively. What could the weight of the fifth zebra be if the average weight of all 5 zebras is to be between 500 and 600 pounds?
SOLUTION Set up a compound inequality representing the situation and solve.
Maximum Weight
Hint
To find the mean (average) of a set of data, divide the sum of the data by the number of data in the set.
Minimum Weight Mean Weight
greater than or equal 540 + 550 + 520 + 530 + x ___
500≤ 540+550+520+530+x ≤600 5
less than or equal to500____5 to600
Online Connection www.SaxonMathResources.com
500·5≤ 2140+x ·5≤600·5 5
2500≤ 2140+x≤ 3000 -_ 2_14_0 -_ 2_14_0 -_ 2_14_0 360 ≤ x ≤ 860
The fifth zebra’s weight could be between 360 lb and 860 lb, inclusive.
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