When solving an equation involving absolute value expressions, we should consider what would happen if we removed the absolute value symbols. Let's look at an example equation.

∣ a x + b ∣ = ∣ c x + d ∣

Although we can make

4

statements about this equation, there are actually only two possible cases to consider.

Statement	Result
Both absolute values are positive.	$a x + b = c x + d$
Both absolute values are negative.	$- (a x + b) = - (c x + d)$
Only the left-hand side is negative.	$- (a x + b) = c x + d$
Only the right-hand side is negative.	$a x + b = - (c x + d)$

Because of the Properties of Equality, when the absolute values of two expressions are equal, either

the

expressions

are

equal

or

the

opposites

of

the

expressions

are

equal .

Now let's consider these two cases for the given equation.

Given Equation : First Equation : Second Equation : ∣ 3 h + 1 ∣ = 7 h ∣ 3 h + 1 ∣ = ∣ 7 h ∣ 3 h + 1 = - 7 h

Let's solve for

h

in each of these equations simultaneously.

∣ 3 h + 1 ∣ = 7 h

$3 h + 1 \geq 0 : 3 h + 1 = 7 h 3 h + 1 < 0 : 3 h + 1 = - 7 h (I) (II)$

3 h + 1 = 7 h 3 h + 1 = - 7 h (I) (II)

▼

Solve for

h

SubEqn

$LHS - 3 h = RHS - 3 h$

1 = 4 h 1 = - 10 h

DivEqn

$(I):$ $LHS / 4 = RHS / 4$

\frac{1}{4} = h 1 = - 10 h

DivEqn

$(II):$ $LHS / (- 10) = RHS / (- 10)$

\frac{1}{4} = h - \frac{1}{1 0} = h

RearrangeEqn

Rearrange equation

h = \frac{1}{4} h = - \frac{1}{1 0}

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 a true statement is made.

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