Exercise 10 Page 31

When the absolute value of an expression is equal to an expression, either the expressions are equal or the opposites of the expressions are equal. Let's look at an example equation. |ax+b|=cx For this equation, there are two possible cases to consider. ax+b=cx or ax+b=- cx To solve the given absolute value equation, we will write two equations similar to those above when we remove the absolute value.

|x+6|=2x

lc x+6 ≥ 0:x+6 = 2x & (I) x+6 < 0:x+6 = - 2x & (II)

lcx+6=2x & (I) x+6=- 2x & (II)

▼

Solve for x

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

l6=x 6=- 3x

DivEqn

(II):.LHS /(- 3).=.RHS /(- 3).

l6=x - 2=x

RearrangeEqn

(II): Rearrange equation

lx= 6 x= - 2

After solving an absolute value equation, it is necessary to check for extraneous solutions. To do this, we substitute the found solutions into the given equation and determine if they make true statements. Let's start with x= 6.

|x+6|=2x

Substitute

x= 6

|{\color{#0000FF}{6}}+6|\stackrel{?}=2( {\color{#0000FF}{6}})

▼

Simplify

AddTerms

Add terms

|12|\stackrel{?}=2(6)

Multiply

Multiply

|12|\stackrel{?}=12

AbsPos

|a|=a

12=12 ✓

We end with a true statement, so x=6 is not an extraneous solution. Now let's check x= -2.

|x+6|=2x

Substitute

x= - 2

|{\color{#009600}{\text{-} 2}}+6|\stackrel{?}=2({\color{#009600}{\text{-} 2}})

▼

Solve for x

AddTerms

Add terms

|4|\stackrel{?}=2(\text{-}2)

Multiply

Multiply

|4|\stackrel{?}=\text{-} 4

AbsPos

|a|=a

4≠- 4 *

This time we end with a false statement, so x=- 2 is an extraneous solution.