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 x - 4 ∣ = ∣ 3 x - 5 ∣ ∣ 3 x - 4 ∣ = ∣ 3 x - 5 ∣ 3 x - 4 = - (3 x - 5)

Let's solve for

x

in each of these equations simultaneously.

∣ 3 x - 4 ∣ = ∣ 3 x - 5 ∣

$3 x - 4 \geq 0 : 3 x - 4 = (3 x - 5) 3 x - 4 < 0 : 3 x - 4 = - (3 x - 5) (I) (II)$

3 x - 4 = 3 x - 5 3 x - 4 = - (3 x - 5) (I) (II)

▼

Solve for

x

Distr

$(II):$ Distribute $- 1$

3 x - 4 = 3 x - 5 3 x - 4 = - 3 x + 5

SubEqn

$(I):$ $LHS - 3 x = RHS - 3 x$

- 4 \neq = - 5 3 x - 4 = - 3 x + 5

AddEqn

$(II):$ $LHS + 3 x = RHS + 3 x$

- 4 \neq = - 5 6 x - 4 = 5

AddEqn

$(II):$ $LHS + 4 = RHS + 4$

- 4 \neq = - 5 6 x = 9

DivEqn

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

- 4 \neq = - 5 x = \frac{9}{6}

ReduceFrac

$(II):$ $\frac{a}{b} = \frac{a / 3}{b / 3}$

- 4 \neq = - 5 x = \frac{3}{2}

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

x = \frac{3}{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 a true statement is made.

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