Exercise 6 Page 45

Let's start by reviewing how the absolute value function works. The absolute value of a real number |x| is the distance from 0 on the number line. Then, |2|=2 and |-2|=2.

To take both possibilities into account when solving an absolute value equation, we isolate the expression inside the absolute value and then derive 2 equations. One using the negative value, and the other using the positive value. We can see an example below. Now, we can derive the positive and the negative case. Derived Equations 0.5cm [0.8em] x+4 = 5x 1cm x+4 =-5x If we are solving an absolute value equation and one of the derived equations solution is not a solution to the original equation, that is called an extraneous solution. Let's solve the derived equations shown above.

x+4 = 5x

SubEqn

LHS-x=RHS-x

x+4-x = 5x-x

SubTerms

Subtract terms

4= 4x

DivEqn

.LHS /4.=.RHS /4.

4/4 = 4x/4

CalcQuot

Calculate quotient

1=x

RearrangeEqn

Rearrange equation

x=1

Since the other equation just differ in the sign of the right-hand side, the solution is x=-1. Now we should check if these solutions make the original equation hold true. Let's first try using x=1

|x+4|= 5x

Substitute

x= 1

| 1+4| ? = 5( 1)

Multiply

Multiply

|1+4| ? = 5

AddTerms

Add terms

|5|? = 5

AbsPos

|5|=5

5=5

The other solution is x=-1. Let's check it.

|x+4|= 5x

Substitute

x= - 1

| -1+4| ? = 5( - 1)

Multiply

Multiply

|-1+4| ? = - 5

AddTerms

Add terms

|3| ? = - 5

AbsPos

|3|=3

3≠ - 5

Notice that this is not a solution to the original equation, even if it is to the derived equation. In this case, x=-1 is an extraneous solution.