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 : ∣ f - 6 ∣ = ∣ f + 8 ∣ ∣ f - 6 ∣ = ∣ f + 8 ∣ f - 6 = - (f + 8)

Let's solve for

f

in each of these equations simultaneously.

∣ f - 6 ∣ = ∣ f + 8 ∣

$f - 6 \geq 0 : f - 6 = (f + 8) f - 6 < 0 : f - 6 = - (f + 8) (I) (II)$

f - 6 = f + 8 f - 6 = - (f + 8) (I) (II)

▼

Solve for

f

Distr

$(II):$ Distribute $- 1$

f - 6 = f + 8 f - 6 = - f - 8

SubEqn

$(I):$ $LHS - f = RHS - f$

- 6 \neq = 8 f - 6 = - f - 8

AddEqn

$(II):$ $LHS + f = RHS + f$

- 6 \neq = 8 2 f - 6 = - 8

AddEqn

$(II):$ $LHS + 6 = RHS + 6$

- 6 \neq = 8 2 f = - 2

DivEqn

$(II):$ $LHS / 2 = RHS / 2$

- 6 \neq = 8 f = - 1

While solving the first equation, we encountered a contradiction, so the first equation has no solution. The solution to the second equation is

f = - 1 .

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.