# How do you solve the equation abs(2(1/3+1/2x))=1?

Nov 26, 2017

$x = \frac{1}{3} \text{ or } x = - \frac{5}{3}$

#### Explanation:

$\text{distribute the factor}$

rArr|(2(1/3+1/2x)|=|2/3+x|

$\text{the value inside the absolute value function can be}$
$\textcolor{b l u e}{\text{positive or negative}}$

$\textcolor{b l u e}{\text{first solution}}$

$\frac{2}{3} + x = 1 \Rightarrow x = 1 - \frac{2}{3} = \frac{1}{3}$

$\textcolor{b l u e}{\text{second solution}}$

$- \left(\frac{2}{3} + x\right) = 1$

$\Rightarrow - \frac{2}{3} - x = 1$

$\Rightarrow - x = 1 + \frac{2}{3} = \frac{5}{3} \Rightarrow x = - \frac{5}{3}$

$\textcolor{b l u e}{\text{As a check}}$

x=1/3"

$\Rightarrow | \frac{2}{3} + \frac{1}{3} | = | 1 | = 1$

$x = - \frac{5}{3}$

$\Rightarrow | \frac{2}{3} - \frac{5}{3} | = | - 1 | = 1$

Nov 26, 2017

Use the piecewise definition of the absolute value function to separate the equation into two equations and then solve each equation.

#### Explanation:

The piecewise definition of the absolute value function is:

|f(x)| = {(f(x); f(x) >= 0),(-f(x); f(x) < 0):}

In this case $f \left(x\right) = 2 \left(\frac{1}{3} + \frac{1}{2} x\right)$

Substitute into the definition:

|2(1/3+1/2x)| = {(2(1/3+1/2x); 2(1/3+1/2x) >= 0),(-2(1/3+1/2x); 2(1/3+1/2x) < 0):}

Simplify the domain restrictions:

|2(1/3+1/2x)| = {(2(1/3+1/2x); x >= -2/3),(-2(1/3+1/2x); x < -2/3):}

Separate the given equation into two equations with its respective domain restriction:

2(1/3+1/2x) = 1; x >=-2/3 and -2(1/3+1/2x) = 1; x < -2/3

Multiply the second equation by -1:

2(1/3+1/2x) = 1; x >=-2/3 and 2(1/3+1/2x) = -1; x < -2/3

Distribute the two in both equations:

2/3+x = 1; x >=-2/3 and 2/3+x = -1; x < -2/3

Subtract $\frac{2}{3}$ from both sides of both equations:

x = 1/3; x >=-2/3 and x = -5/3; x < -2/3

The domain restrictions can be dropped, because neither equation violates them:

$x = \frac{1}{3}$ and $x = - \frac{5}{3}$