Exercise 20 Page 46

An absolute value measures an expression's distance from a midpoint on a number line. |2z-3|= 4z-1 This equation means that the distance is 4z-1, either in the positive direction or the negative direction. |2z-3|= 4z-1 ⇒ l2z-3= 4z-1 2z-3= - (4z-1) To find the solutions to the absolute value equation, we need to solve both of these cases for z.

|2z-3|=4z-1

lc 2z-3 ≥ 0:2z-3 = (4z-1) & (I) 2z-3 < 0:2z-3 = - (4z-1) & (II)

lc2z-3=4z-1 & (I) 2z-3= -(4z-1) & (II)

Distr

(II): Distribute -1

l2z-3=4z-1 2z-3= -4z+1

▼

Solve for z

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

l2z=4z+2 2z= -4z+4

SubEqn

(I): LHS-4z=RHS-4z

l- 2z=2 2z= -4z+4

AddEqn

(II): LHS+4z=RHS+4z

l-2z=2 6z=4

DivEqn

(I): .LHS /(-2).=.RHS /(-2).

lz=- 1 6z=4

DivEqn

(II): .LHS /6.=.RHS /6.

lz=- 1 z= 46

ReduceFrac

(II): a/b=.a /2./.b /2.

lz = -1 z= 23

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 -1.

|2z-3|=4z-1

Substitute

z= -1

|2( -1)-3|? =4( -1)-1

MultPosNeg

a(- b)=- a * b

|-2-3|? =-4-1

SubTerms

Subtract terms

|-5|? =-5

AbsNeg

|-5|=5

5 ≠ -5

We will check 23 in the same way.

|2z-3|=4z-1

Substitute

z= 2/3

|2( 2/3)-3|? =4( 2/3)-1

MoveLeftFacToNum

a*b/c= a* b/c

|4/3-3|? =8/3-1

NumberToFrac

a = 3* a/3

|4/3-9/3|? =8/3-3/3

SubFrac

Subtract fractions

|-5/3|? =5/3

AbsNeg

|-5/3|=5/3

5/3=5/3

We see that 23 satisfies the original equation. However, -1 is an extraneous solution.