Question

If $|x+1|=|3x-2|$, what are the possible values for $x$?

• A. $\frac{1}{4}$ and $\frac{3}{4}$
• B. $\frac{1}{4}$ and $\frac{3}{2}$
• C. $\frac{2}{3}$ and $\frac{3}{2}$
• D. $\frac{2}{3}$ and $\frac{4}{3}$
• E. $\frac{3}{4}$ and $\frac{4}{3}$

Answer 1

The correct option is B $\frac{1}{4}$ and $\frac{3}{2}$

This is a complex absolute value problem, so you first must decide on an approach. The equation $|x+1|=|3x-2|$ has one variable $\left(x\right)$ and several constants ($1,3$ and $-2$). Thus, you should take an algebraic approach.

In theory, with two absolute value expressions, you would set up four cases. However, those four cases collapse to just two cases: (1) the two expressions inside the absolute value symbols are given the same sign, and (2) the two expressions are given the opposite sign.

Case (1): Same Sign

$x+1=3x-2$

$3=2x$

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

Case (2): Opposite sign

$x+1=-(3x-2)=-3x+2$

$4x=1$

$x=\frac{1}{4}$

Testing each solution in the original equation, you verify that both solutions are valid:

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

$\left|\frac{5}{2}\right|=|\frac{9}{2}-2|$

$\frac{5}{2}=\frac{5}{2}$

$|\frac{1}{4}+1|=|3\left(\frac{1}{4}\right)-2|$

$\left|\frac{5}{4}\right|=|\frac{3}{4}-2|=\left|\frac{-5}{4}\right|$

$\frac{5}{4}=\frac{5}{4}$