Exercise 88 Page 177

Begin by isolating the absolute value term.

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We will begin by substituting the equation for y into the given equation. Then, we will isolate the absolute value term.

y<6

Substitute

y= 2|6-3x|+4

2|6-3x|+4<6

SubIneq

LHS-4

2|6-3x|<2

DivIneq

.LHS /2.<.RHS /2.

|6-3x|<1

Now that we isolated the absolute value term, we should consider two cases, positive case and negative case. |6-3x|< 1 ⇒ lPositive Case: 6-3x<1 Negative Case: 6-3x>-1 Let's solve both cases for x.

l6-3x<1 6-3x>-1

SubIneq

(I): LHS-6

l-3x<-5 6-3x>-1

DivNegIneq

(I): Divide by (-3) and flip inequality sign

lx>5/3 6-3x>-1

SubIneq

(II): LHS-6>RHS-6

lx>5/3 -3x>-7

DivNegIneq

(II): Divide by (-3) and flip inequality sign

lx>5/3 x< 7/3

As a result of our operations, we can see that the set of values greater than 53 and less than 73 describes x. Therefore, the correct option is A.

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Exercises p.174-177

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Exercise 88 Page 177

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