How do you solve $∣ ∣ ∣ x - \frac{1}{3} ∣ ∣ ∣ - 2 = 1 ?$ $abs(x-1/3)-2=1?$ ?

Question

2882 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-22T08:14:36

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

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

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

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

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

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

GeoGebra Classic app

Answer 2 · 2018-04-22T08:19:49

$x_{1} = \frac{10}{3}$ $x_1 = 10/3$
$x_{2} = - \frac{8}{3}$ $x_2 =-8/3$

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

In this case:

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

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

$\pm (x - \frac{1}{3}) = 3$ $+-(x-1/3) = 3$

Solve for x:

$x_{1} = \frac{10}{3}$ $x_1 = 10/3$
$x_{2} = - \frac{8}{3}$ $x_2 =-8/3$

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

GeoGebra Classic app

How do you solve abs(x-1/3)-2=1?∣∣∣x−13∣∣∣−2=1?abs(x-1/3)-2=1??