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Pearson Algebra 1 Common Core, 2011

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Pearson Algebra 1 Common Core, 2011 View details

7. Absolute Value Equations and Inequalities

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Exercise 53 Page 211

How many cases do you have after you remove the absolute value?

y=3.4 or y=-0.6

Practice makes perfect

Before we can solve this equation, we need to isolate the absolute value expression using the Properties of Equality.

3|5y-7|-6=24

LHS+6=RHS+6

3|5y-7|-6+6=24+6

Add terms

3|5y-7|=30

.LHS /3.=.RHS /3.

3|5y-7|/3=30/3

Calculate quotient

|5y-7|=10

An absolute value measures an expression's distance from a midpoint on a number line.

|5y-7|= 10 This equation means that the distance is 10, either in the positive direction or the negative direction. |5y-7|= 10 ⇒ l5y-7= 10 5y-7= -10 To find the solutions to the absolute value equation, we need to solve both of these cases for y.

| 5y-7|=10

lc 5y-7 ≥ 0:5y-7 = 10 & (I) 5y-7 < 0:5y-7 = - 10 & (II)

lc5y-7=10 & (I) 5y-7=-10 & (II)

(I), (II): LHS+7=RHS+7

lc5y=17 & (I) 5y=-3 & (II)

(I), (II): .LHS /5.=.RHS /5.

ly_1=3.4 y_2=-0.6

Both y=3.4 and y=-0.6 are solutions to the absolute value equation.

Subchapter links

Lesson Check p.210

Practice and Problem-Solving Exercises p.211-213

Standardized Test Prep p.213

Mixed Review p.213