Before we can solve this equation, we need to isolate the absolute value expression using the Properties of Equality.

4 ∣ 2 x + 7 ∣ = 16

$LHS / 4 = RHS / 4$

∣ 2 x + 7 ∣ = 4

Solving for $x$

An absolute value measures an expression's distance from a midpoint on a number line.

∣ 2 x + 7 ∣ = 4

This equation means that the distance is

4,

either in the

positive

direction or the

negative

direction.

∣ 2 x + 7 ∣ = 4 \Rightarrow 2 x + 7 = - 4 2 x + 7 = - 4

To find the solutions to the absolute value equation, we need to solve both of these cases for

x

.

∣ 2 x + 7 ∣ = 4

$2 x + 7 \geq 0 : 2 x + 7 = 4 2 x + 7 < 0 : 2 x + 7 = - 4 (I) (II)$

2 x + 7 = 4 2 x + 7 = - 4 (I) (II)

$(I), (II):$ $LHS - 7 = RHS - 7$

2 x = - 3 2 x = - 11

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

x_{1} = - \frac{3}{2} x_{2} = - - \frac{1 1}{2}

Both

- \frac{3}{2}

and

- \frac{1 1}{2}

are solutions to the absolute value equation.

Checking Our Answers

When solving an absolute value equation, it is important to check for extraneous solutions. We can check our answers by substituting them back into the original equation. Let's start with

x = - \frac{3}{2} .

4 ∣ 2 x + 7 ∣ = 16

Substitute

$x = - \frac{3}{2}$

4 ∣ ∣ ∣ ∣ ∣ 2 (- \frac{3}{2}) + 7 ∣ ∣ ∣ ∣ ∣ = ? 16

▼

Simplify

MoveNegFracToNum

Put minus sign in numerator

4 ∣ ∣ ∣ ∣ ∣ 2 (\frac{- 3}{2}) + 7 ∣ ∣ ∣ ∣ ∣ = ? 16

DenomMultFracToNumber

$2 \cdot \frac{a}{2} = a$

4 ∣ - 3 + 7 ∣ = ? 16

AddTerms

Add terms

4 ∣ 4 ∣ = ? 16

AbsPos

$∣ 4 ∣ = 4$

4 (4) = ? 16

Multiply

16 = 16

We end with a true statement, so

x = - \frac{3}{2}

is not an extraneous solution. We will check

x = - \frac{1 1}{2}

in the same way.

4 ∣ x + 7 ∣ = 16

Substitute

$x = - \frac{1 1}{2}$

4 ∣ ∣ ∣ ∣ ∣ 2 (- \frac{1 1}{2}) + 7 ∣ ∣ ∣ ∣ ∣ = ? 16

▼

Simplify

MoveNegFracToNum

Put minus sign in numerator

4 ∣ ∣ ∣ ∣ ∣ 2 (\frac{- 1 1}{2}) + 7 ∣ ∣ ∣ ∣ ∣ = ? 16

DenomMultFracToNumber

$2 \cdot \frac{a}{2} = a$

4 ∣ - 11 + 7 ∣ = ? 16

AddTerms

Add terms

4 ∣ - 4 ∣ = ? 16

AbsNeg

$∣ - 4 ∣ = 4$

4 (4) = ? 16

Multiply

16 = 16

Again we end up with a true statement. Therefore,

x = - \frac{3}{2}

and

x = - \frac{1 1}{2}

satisfy the original equation and neither of them is an extraneous solution.

Rewriting Our Answers

Although

x = - \frac{3}{2}

and

x = - \frac{1 1}{2}

are perfectly valid solutions, we can also rewrite them as mixed numbers. Let's start with

x = - \frac{3}{2} .

x = - \frac{3}{2}

Rewrite

Rewrite $3$ as $2 + 1$

x = - \frac{2 + 1}{2}

WriteSumFrac

Write as a sum of fractions

x = - (\frac{2}{2} + \frac{1}{2})

CalcQuot

Calculate quotient

x = - (1 + \frac{1}{2})

Rewrite

Rewrite $1 + \frac{1}{2}$ as $1 \frac{1}{2}$

x = - 1 \frac{1}{2}

Using a similar process for

x = - \frac{1 1}{2},

we will find that

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

Solving for x

Checking Our Answers

Rewriting Our Answers

Solving for $x$

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