1. The problem is to solve the inequality $$\frac{12q}{x} > 43$$ for $x$. 2. We want to isolate $x$ on one side. Since $x$ is in the denominator, we multiply both sides by $x$, but we must consider the sign of $x$ because multiplying or dividing by a negative number reverses the inequality. 3. Case 1: $x > 0$ Multiply both sides by $x$ (positive, so inequality direction stays the same): $$12q > 43x$$ 4. Divide both sides by 43: $$\frac{12q}{\cancel{43}} > x \cancel{\frac{1}{43}}$$ So, $$x < \frac{12q}{43}$$ 5. Case 2: $x < 0$ Multiply both sides by $x$ (negative, so inequality direction reverses): $$12q < 43x$$ 6. Divide both sides by 43: $$\frac{12q}{\cancel{43}} < x \cancel{\frac{1}{43}}$$ So, $$x > \frac{12q}{43}$$ 7. Since the problem states $x > 43$, we combine this with the above results: - For $x > 0$, $x < \frac{12q}{43}$ and $x > 43$ must both hold. - For $x < 0$, $x > \frac{12q}{43}$ and $x < 0$ must both hold. 8. The solution depends on the value of $q$ and the relation between $\frac{12q}{43}$ and 43. Summary: - If $x > 0$, then $43 < x < \frac{12q}{43}$. - If $x < 0$, then $\frac{12q}{43} < x < 0$. This is the solution for $x$ given the inequality and the condition $x > 43$.