The problem asks us to solve the equation $\frac{2q-7}{3} = \frac{q+6}{3q-4}$ for the variable $q$.

1. Problem Description

The problem asks us to solve the equation $\frac{2q-7}{3} = \frac{q+6}{3q-4}$ for the variable $q$.

2. Solution Steps

First, we cross-multiply to eliminate the fractions:

$(2q-7)(3q-4) = 3(q+6)$

Expanding both sides of the equation, we get:

$6q^2 - 8q - 21q + 28 = 3q + 18$

$6q^2 - 29q + 28 = 3q + 18$

Next, we move all terms to one side of the equation to obtain a quadratic equation:

$6q^2 - 29q - 3q + 28 - 18 = 0$

$6q^2 - 32q + 10 = 0$

We can simplify the quadratic equation by dividing all terms by 2:

$3q^2 - 16q + 5 = 0$

Now, we can solve the quadratic equation for $q$. We can factor the quadratic:

$(3q-1)(q-5) = 0$

This gives us two possible solutions for $q$:

$3q - 1 = 0 \implies 3q = 1 \implies q = \frac{1}{3}$

$q - 5 = 0 \implies q = 5$

Therefore, the solutions for $q$ are $\frac{1}{3}$ and $5$.

3. Final Answer

The solutions for $q$ are $q = \frac{1}{3}$ and $q = 5$.