Solve the equation |(2x − 3)/7| = (1/2)·|(x + 3)/4|.

Step-by-step solution

1. 1

   Move the positive constants outside the bars. Since $|A/c| = |A|/c$ for $c > 0$,

   $\frac{|2x-3|}{7} = \frac12\cdot\frac{|x+3|}{4} = \frac{|x+3|}{8}$

   The denominators $7$, $4$ and the factor $\tfrac12$ are all positive, so no sign bookkeeping is needed for them.

2. 2

   Clear the denominators. Multiplying both sides by $56$ (the least common multiple of $7$ and $8$):

   $8|2x-3| = 7|x+3|$

   The equation is now free of fractions, with only integers outside the two absolute values.

3. 3

   Square both sides safely. Both sides are non-negative, so squaring is a reversible step here and no extraneous roots can be introduced by it:

   $64(2x-3)^2 = 49(x+3)^2$

4. 4

   Expand and collect into a quadratic.

   $64\left(4x^2-12x+9\right) = 49\left(x^2+6x+9\right)$

   $256x^2 - 768x + 576 = 49x^2 + 294x + 441$

   $207x^2 - 1062x + 135 = 0 \quad \xrightarrow{\ \div 9\ } \quad 23x^2 - 118x + 15 = 0$

5. 5

   Solve the quadratic. With $a = 23$, $b = -118$, $c = 15$:

   $\Delta = 118^2 - 4(23)(15) = 13924 - 1380 = 12544 = 112^2$

   $x = \frac{118 \pm 112}{46} \quad \Longrightarrow \quad x = \frac{230}{46} = 5 \quad \text{or} \quad x = \frac{6}{46} = \frac{3}{23}$

   The perfect-square discriminant is the sign that the original problem was built to have rational answers.

6. 6

   Verify both roots in the original equation. For $x = 5$: $\left|\tfrac{7}{7}\right| = 1$ and $\tfrac12\left|\tfrac{8}{4}\right| = 1$. For $x = \tfrac{3}{23}$: $\left|\tfrac{-63/23}{7}\right| = \tfrac{9}{23}$ and $\tfrac12\left|\tfrac{72/23}{4}\right| = \tfrac{9}{23}$. Both check exactly, so the solution set is $\left\{5,\ \tfrac{3}{23}\right\}$.