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 ∣ = ∣ 2 x + 5 ∣ ∣ 3 (x - 4) ∣ = ∣ 2 x + 5 ∣ 3 (x - 4) = - (2 x + 5)

Let's solve for

x

in each of these equations simultaneously.

3 ∣ x - 4 ∣ = ∣ 2 x + 5 ∣

$3 (x - 4) \geq 0 : 3 (x - 4) = (2 x + 5) 3 (x - 4) < 0 : 3 (x - 4) = - (2 x + 5) (I) (II)$

x + 8 = 2 x + 1 x + 8 = - (2 x + 1) (I) (II)

▼

Solve for

x

Distr

$(II):$ Distribute $- 1$

x + 8 = 2 x + 1 x + 8 = - 2 x - 1

$(I), (II):$ $LHS - 8 = RHS - 8$

x = 2 x - 7 x = - 2 x - 9

SubEqn

$(I):$ $LHS - 2 x = RHS - 2 x$

- x = - 7 x = - 2 x - 9

DivEqn

$(I):$ $LHS / (- 1) = RHS / (- 1)$

x = 7 x = - 2 x - 9

AddEqn

$(II):$ $LHS + 2 x = RHS + 2 x$

x = 7 3 x = - 9

DivEqn

$(II):$ $LHS / 3 = RHS / 3$

x_{1} = 7 x_{2} = - 3

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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