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 k - 2 ∣ = 2 ∣ k + 2 ∣ ∣ 3 k - 2 ∣ = ∣ 2 (k + 2) ∣ 2 + x = - 2 (k + 2)

Let's solve for

k

in each of these equations simultaneously.

∣ 3 k - 2 ∣ = 2 ∣ k + 2 ∣

$3 k - 2 \geq 0 : 3 k - 2 = 2 (k + 2) 3 k - 2 < 0 : 3 k - 2 = - 2 (k + 2) (I) (II)$

3 k - 2 = 2 (k + 2) 3 k - 2 = - 2 (k + 2) (I) (II)

▼

Solve for

k

Distr

$(I):$ Distribute $2$

3 k - 2 = 2 k + 4 3 k - 2 = - 2 (k + 2) (I) (II)

Distr

$(II):$ Distribute $- 2$

3 k - 2 = 2 k + 4 3 k - 2 = - 2 k - 4 (I) (II)

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

3 k = 2 k + 6 3 k = - 2 k - 2

SubEqn

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

k = 6 3 k = - 2 k - 2

AddEqn

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

k = 6 5 k = - 2

DivEqn

$(II):$ $LHS / 5 = RHS / 5$

k_{1} = 6 k_{2} = - \frac{2}{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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