Exercise 48 Page 46

Before we can solve this equation, we need to isolate the absolute value expression using the Properties of Equality. An absolute value measures an expression's distance from a midpoint on a number line. |8-3h|= 3h-7 This equation means that the distance is 3h-7, either in the positive direction or the negative direction. |8-3h|= 3h-7 ⇒ l8-3h= 3h-7 8-3h= - (3h-7) To find the solutions to the absolute value equation, we need to solve both of these cases for h.

|8-3h|=3h-7

lc 8-3h ≥ 0:8-3h = (3h-7) & (I) 8-3h < 0:8-3h = - (3h-7) & (II)

lc8-3h=3h-7 & (I) 8-3h= -(3h-7) & (II)

Distr

(II): Distribute - 1

l8-3h=3h-7 8-3h= -3h+7

▼

Solve for h

(I), (II): LHS-8=RHS-8

l-3h=3h-15 -3h= -3h-1

SubEqn

(I): LHS-3h=RHS-3h

l- 6h=- 15 -3h= -3h-1

AddEqn

(II): LHS+3h=RHS+3h

l-6h= -15 0 ≠ -1

DivEqn

(I): .LHS /(- 6).=.RHS /(- 6).

lh= -15- 6 0 ≠ -1

DivNegNeg

(I): - a/- b=a/b

lh= 156 0 ≠ -1

ReduceFrac

(I): a/b=.a /3./.b /3.

lh= 52 0 ≠ -1

Solving the second equation resulted in a contradiction. Thus, there is only one possible solution to the equation. We will now check it.

Checking Our Answers

When solving an absolute value equation, it is important to check for extraneous solutions. We can check our answer by substituting it back into the original equation.

|8-3h|=3h-7

Substitute

h= 5/2

|8-3( 5/2)|? =3( 5/2)-7

MoveLeftFacToNum

a*b/c= a* b/c

|8-15/2|? =15/2-7

NumberToFrac

a = 2* a/2

|16/2-15/2|? =15/2-14/2

SubTerms

Subtract terms

|1/2|? =1/2

AbsPos

|1/2|=1/2

1/2=1/2

We can see that 52 satisfies our equation.