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 : ∣ 6 a - 5 ∣ = 4 a ∣ 6 a - 5 ∣ = ∣ 4 a ∣ 6 a - 5 = - 4 a

Let's solve for

a

in each of these equations simultaneously.

∣ 6 a - 5 ∣ = 4 a

$6 a - 5 \geq 0 : 6 a - 5 = 4 a 6 a - 5 < 0 : 6 a - 5 = - 4 a (I) (II)$

6 a - 5 = 4 a 6 a - 5 = - 4 a (I) (II)

▼

Solve for

a

SubEqn

$LHS - 6 a = RHS - 6 a$

- 5 = - 2 a - 5 = - 10 a

DivEqn

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

\frac{5}{2} = a - 5 = - 10 a

DivEqn

$(II):$ $LHS / (- 10) = RHS / (- 10)$

\frac{5}{2} = a \frac{5}{1 0} = a

RearrangeEqn

Rearrange equation

a = \frac{5}{2} a = \frac{5}{1 0}

ReduceFrac

$(II):$ $\frac{a}{b} = \frac{a / 5}{b / 5}$

a = \frac{5}{2} a = \frac{1}{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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