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Pearson Algebra 2 Common Core, 2011

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Pearson Algebra 2 Common Core, 2011 View details

6. Absolute Value Equations and Inequalities

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Exercise 54 Page 46

Absolute values can be interpreted as the distance from a midpoint.

|x|=3

Practice makes perfect

Absolute values can be interpreted as the distance away from a midpoint. For one-variable absolute value equations, this distance can be represented by two points on a number line, such as the one given in the exercise.

Because our equation needs a distance from a midpoint, we should begin by finding the halfway point between the two given values. We can do this by calculating their mean.

Mean=- 3+ 3/2= 0 Now we need to find the distance between this midpoint and each of the given points.

We see that both -3 and 3 are 3 units away from 0. Written as an equation, we can show that the difference between a number x and the midpoint is equal to the distance we found above. |x- midpoint|=distance |x- 0|=3 Let's simplify the equation. |x-0|=3 ⇒ |x|=3

Subchapter links

Lesson Check p.45

Practice and Problem-Solving Exercises p.46-48

Standardized Test Prep p.48

Mixed Review p.48