GRE Quantitative Comparison: Determining range of values satisfying equation involving absolute values

Question

Consider the equation \begin{equation} |2a-1|+|3b+2|=0 \end{equation}

Which of the following is true:

• $a>b$
• $a<b$
• $a=b$
• The relationship cannot be determined.

How can one solve for the range of possible values of $a$ and $b$? Thank you very much!

Answer 1

Use:

$$|x|\ge0$$

$$|x|+|y|=0\iff |x|=0 \text{ and } |y|=0$$ $$|2a-1|=0\implies a=\frac12$$

$$|3b+2|=0\implies b=-\frac23 $$

$$a\gt b$$

Answer 2

solve 2a-1 =0 and 3b+2 = 0 since we need to solve RHS=0; you will get a=1/2 and b=-2/3. Therefore, a>b.