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 : ∣ 4 n - 15 ∣ = ∣ n ∣ ∣ 4 n - 15 ∣ = ∣ n ∣ 4 n - 15 = - n

Let's solve for

n

in each of these equations simultaneously.

∣ 4 n - 15 ∣ = ∣ n ∣

$4 n - 15 \geq 0 : 4 n - 15 = n 4 n - 15 < 0 : 4 n - 15 = - n (I) (II)$

4 n - 15 = n 4 n - 15 = - n (I) (II)

▼

Solve for

n

$(I), (II):$ $LHS + 15 = RHS + 15$

4 n = n + 15 4 n = - n + 15

SubEqn

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

3 n = 15 4 n = - n + 15

AddEqn

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

3 n = 15 5 n = 15

DivEqn

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

n = 5 5 n = 15

DivEqn

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

n_{1} = 5 n_{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.

{{ 'ml-heading-exercise' | message }} {{ exercise.name }} {{ 'ml-label-page' | message }} {{ exercise.page }}

{{ 'ml-solutions-missing-heading' | message }}