Exercise 6 Page 29

Before we can solve this equation, we need to isolate the absolute value expression using the Properties of Equality.

Solving for x

An absolute value measures an expression's distance from a midpoint on a number line. |5x-1|= 4This equation means that the distance is 4, either in the positive direction or the negative direction. |5x-1|= 4 ⇒ l5x-1= 4 5x-1= - 4 To find the solutions to the absolute value equation, we need to solve both of these cases for x.

| 5x-1|=4

lc 5x-1 ≥ 0:5x-1 = 4 & (I) 5x-1 < 0:5x-1 = - 4 & (II)

lc5x-1=4 & (I) 5x-1=- 4 & (II)

(I), (II): LHS+1=RHS+1

l5x=5 5x=- 3

(I), (II): .LHS /5.=.RHS /5.

lx_1=1 x_2=- 35

Both 1 and - 35 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 x= 1.

- 2|5x-1|-3=- 11

Substitute

x= 1

- 2|5( 1)-1|-3? =- 11

▼

Simplify

OneMult

1* a=a

- 2|5-1|-3? =- 11

SubTerm

Subtract term

- 2|4|-3? =- 11

AbsPos

|4|=4

- 2(4)-3? =- 11

Multiply

Multiply

- 8-3? =- 11

SubTerm

Subtract term

- 11=- 11

We end with a true statement, so x=1 is not an extraneous solution. We will check x= - 35 in the same way.

- 2|5x-1|-3=- 11

Substitute

x= -3/5

- 2|5( -3/5)-1|-3? =- 11

▼

Simplify

MoveNegFracToNum

Put minus sign in numerator

- 2|5(-3/5)-1|-3? =- 11

DenomMultFracToNumber

5 * a/5= a

- 2|-3-1|-3? =- 11

SubTerm

Subtract term

- 2|-4|-3? =- 11

AbsNeg

|-4|=4

- 2(4)-3? =- 11

Multiply

Multiply

- 8-3? =- 11

SubTerm

Subtract term

- 11=- 11

Again we end up with a true statement. Therefore, x=1 and x=- 35 satisfy the original equation and neither of them is an extraneous solution.