Exercise 19 Page 227

An absolute value measures an expression's distance from a midpoint on a number line. |4k-2|= 11 This equation means that the distance is 11, either in the positive direction or the negative direction. |4k-2|= 11 ⇒ l4k-2= 11 4k-2= - 11 To find the solutions to the absolute value equation, we need to solve both of these cases for n.

| 4k-2|=11

lc 4k-2 ≥ 0:4k-2 = 11 & (I) 4k-2 < 0:4k-2 = - 11 & (II)

lc4k-2=11 & (I) 4k-2=- 11 & (II)

(I), (II): LHS+2=RHS+2

l4k=13 4k=- 9

(I), (II): .LHS /4.=.RHS /4.

lk_1= 134 k_2= - 94

MoveNegNumToFrac

(II): Put minus sign in front of fraction

lk_1= 134 k_2=- 94

Both 134 and - 94 are solutions to the absolute value equation.

Checking Our Answers

When solving an absolute value equation, it is important to check for extraneous solutions. We can check our answers by substituting them back into the original equation. Let's start with k_1= 134.

|4k-2|=11

Substitute

k= 13/4

|4( 13/4)-2|? =11

Multiply

Multiply

|13-2|? =11

SubTerm

Subtract term

|11|? =11

AbsPos

|11|=11

11=11

We will check k_2=- 94 in the same way.

|4k-2|=11

Substitute

k= -9/4

|4( -9/4)-2|? =11

MultPosNeg

a(- b)=- a * b

|- 9-2|? =11

SubTerm

Subtract term

|- 11|? =11

AbsNeg

|-11|=11

11=11

We see that both 134 and - 94 satisfy the original equation, so they are both proper solutions.