Question

Sove the equation and verify

Answer 1

Let's break the equation in several step to make things more clear: first of all, the numerator of the left hand side simplifies to

$\dfrac{2x}{3}-\dfrac{1}{2} = \dfrac{4x-3}{6}$

Similarly, the denominator becomes

$\dfrac{3x}{4}-\dfrac{1}{2} = \dfrac{3x-2}{4}$

Dividing by a fraction means to multiply by the inverse of that fraction:

$\dfrac{\frac{a}{b}}{\frac{c}{d}} = \dfrac{a}{b}\cdot\dfrac{d}{c}$

So, in your case, we have

$\dfrac{\frac{2x}{3}-\frac{1}{2}}{\frac{3x}{4}-\frac{1}{2}} = \dfrac{\frac{4x-3}{6}}{\frac{3x-2}{4}} = \dfrac{4x-3}{6}\cdot \dfrac{4}{3x-2} = \dfrac{2(4x-3)}{3(3x-2)} = \dfrac{8x-6}{9x-6}$

So, the equation has become

$\dfrac{8x-6}{9x-6} = \dfrac{4}{3}$

Multiply both sides by 3(9x-6):

$3(8x-6) = 4(9x-6) \iff 24x-18 = 36x-24$

Subtract 24x from both sides, and add 18 to both sides:

$36x-24x = 24-18 \iff 12x=6 \iff x = \dfrac{1}{2}$