Big Ideas Math Integrated I, 2016
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Big Ideas Math Integrated I, 2016 View details
Chapter Review
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Exercise 16 Page 45

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

w=3 and w=-1/5

Practice makes perfect

Before we begin, let's isolate the absolute value on the left-hand side.

-2|5w-7|+9=-7
-2|5w-7|+9-9=-7-9
-2|5w-7|=-16
|5w-7|=8

An absolute value measures an expression's distance from a midpoint on a number line. |5w-7|= 8

This equation means that the distance is 8, either in the positive direction or the negative direction. |5w-7|= 8 ⇒ l5w-7= 8 5w-7= -8 To find the solutions to the absolute value equation, we need to solve both of these cases for w.

| 5w-7|=8

lc 5w-7 ≥ 0:5w-7 = 8 & (I) 5w-7 < 0:5w-7 = - 8 & (II)

lc5w-7=8 & (I) 5w-7=-8 & (II)

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

l5w=15 5w=-1

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

lw=3 w=-1/5

Both 3 and - 15 are solutions to the given equation. By substituting w=3 and w=- 15 into the equation and evaluating, we can check if the solutions are correct. First, let's check w=3.

|5w-7|=8
|5( 3)-7|? =8
|15-7|? =8
|8|? =8
8=8 ✓

Because the left-hand and right-hand sides are equal, we know that the solution is correct. Now we will check the second solution by substituting - 15 for w.

|5w-7|=8
|5( -1/5)-7|? =8
|5( - 1/5)-7|? =8
|-1-7|? =8
|-8|? =8
8=8 ✓

Both sides of the equation are again equal. Therefore, both solutions are correct.