Core Connections Integrated II, 2015
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Core Connections Integrated II, 2015 View details
2. Section 8.2
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Exercise 45 Page 447

Practice makes perfect
a An absolute value measures an expression's distance from a midpoint on a number line.
|x|= 3

This equation means that the distance is 3, either in the positive direction or the negative direction. |x|= 3 ⇒ lx= 3 x= - 3 Both 3 and -3 are solutions to the absolute value equation. Let's graph these solutions on a number line.

b An absolute value measures an expression's distance from a midpoint on a number line.
|x|= 0

This equation means that the distance is 0, either in the positive direction or the negative direction. However, notice that ± 0 ⇔ 0, which means we only have one solution. |x|= 0 ⇒ x=0 Let's graph this on a number line.

c Before we can remove the absolute value equation, we have to isolate it.
|x|+2=6
|x|=4

Now we can remove the absolute value, which gives us a positive and a negative solution. |x|= 4 ⇒ lx= 4 x= - 4 Both -4 and 4 are solutions to the absolute value equation. Let's graph these solutions on a number line.

d Like in Part C, we have to isolate the absolute value before we can remove it.
3|x|=15
|x|=5

Now we can remove the absolute value, which gives us a positive and a negative solution. |x|= 5 ⇒ lx= 5 x= -5 Both -5 and 5 are solutions to the absolute value equation. Let's graph these solutions on a number line.

e Like in previous parts, if we remove the absolute value we get two equations.
|x-5|= 2 ⇒ lx-5= 2 x-5= - 2To find the solutions to the absolute value equation, we need to solve both of these cases for x.
| x-5|=2

lc x-5 ≥ 0:x-5 = 2 & (I) x-5 < 0:x-5 = - 2 & (II)

lcx-5=2 & (I) x-5=- 2 & (II)

(I), (II): LHS+5=RHS+5

lx_1=7 x_2=3
Both 7 and 3 are solutions to the absolute value equation. Let's graph these solutions on a number line.
f Like in previous parts, if we remove the absolute value we get two equations.
|x+3|= 7 ⇒ lx+3= 7 x+3= - 7To find the solutions to the absolute value equation, we need to solve both of these cases for n.
| x+3|=7

lc x+3 ≥ 0:x+3 = 7 & (I) x+3 < 0:x+3 = - 7 & (II)

lcx+3=7 & (I) x+3=- 7 & (II)

(I), (II): LHS-3=RHS-3

lx_1=4 x_2=- 10
Both 4 and -10 are solutions to the absolute value equation. Let's graph these solutions on a number line.