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 : 2 ∣ 4 w - 1 ∣ = 3 ∣ 4 w + 2 ∣ ∣ 2 (4 w - 1) ∣ = ∣ 3 (4 w + 2) ∣ 2 (4 w - 1) = - 3 (4 w + 2)

Let's solve for

w

in each of these equations simultaneously.

2 ∣ 4 w - 1 ∣ = 3 ∣ 4 w + 2 ∣

$2 (4 w - 1) \geq 0 : 2 (4 w - 1) = 3 (4 w + 2) 2 (4 w - 1) < 0 : 2 (4 w - 1) = - 3 (4 w + 2) (I) (II)$

2 (4 w - 1) = 3 (4 w + 2) 2 (4 w - 1) = - 3 (4 w + 2) (I) (II)

$(I), (II):$ Distribute $2$

8 w - 2 = 3 (4 w + 2) 8 w - 2 = - 3 (4 w + 2)

$(I), (II):$ Distribute $3$

8 w - 2 = 12 w + 6 8 w - 2 = - (12 w + 6)

Distr

$(II):$ Distribute $- 1$

8 w - 2 = 12 w + 6 8 w - 2 = - 12 w - 6

▼

(I), (II):

Solve for

w

$(I), (II):$ $LHS + 2 = RHS + 2$

8 w = 12 w + 8 8 w = - 12 w - 4

SubEqn

$(I):$ $LHS - 12 w = RHS - 12 w$

- 4 w = 8 8 w = - 12 w - 4

DivEqn

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

w = - 2 8 w = - 12 w - 4

AddEqn

$(II):$ $LHS + 12 w = RHS + 12 w$

w = - 2 20 w = - 4

DivEqn

$(II):$ $LHS / 20 = RHS / 20$

w = - 2 w = - \frac{4}{2 0}

ReduceFrac

$(II):$ $\frac{a}{b} = \frac{a / 4}{b / 4}$

w_{1} = - 2 w_{2} = - \frac{1}{5}

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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