# How do you solve abs(x-1/3)-2=1??

Apr 22, 2018

$x = \frac{10}{3}$ or $- \frac{8}{3}$

#### Explanation:

As $| x - \frac{1}{3} | - 2 = 1$, we have

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

hence either $x - \frac{1}{3} = 3$ i.e. $x = 3 + \frac{1}{3} = \frac{10}{3}$

or $x - \frac{1}{3} = - 3$ i.e. $x = - 3 + \frac{1}{3} = - \frac{8}{3}$

Graphically this solution can be plotted as the intersection points of the functions on the LHS and the RHS as shown below :- Apr 22, 2018

${x}_{1} = \frac{10}{3}$
${x}_{2} = - \frac{8}{3}$

#### Explanation:

Any algebraic problem with absolute values you always have to isolate the absolute value on one side of the equation.

In this case:

$| x - \frac{1}{3} | = 3$

remember that for absolute values you will have 2 answers because both positive and negative values have a positive absolute value:

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

Solve for x:

${x}_{1} = \frac{10}{3}$
${x}_{2} = - \frac{8}{3}$

Graphically this solution can be plotted as the intersection points of the functions on the LHS and the RHS as shown below :- 