Q.Solve the following system of inequalities $\dfrac{2x+1}{7x-1} > 5$, $\dfrac{x+7}{x-8} > 2$.

Concept: Linear Inequality Solutions — solving rational inequalities by clearing denominators carefully, considering sign changes.

Step 1: Solve $\dfrac{2x+1}{7x-1} > 5$.

Bring all terms to one side:

$\dfrac{2x+1}{7x-1} - 5 > 0 \implies \dfrac{2x+1 - 5(7x-1)}{7x-1} > 0 \implies \dfrac{-33x + 6}{7x-1} > 0$.

Multiply numerator and denominator by $-1$ (reverses inequality):

$\dfrac{33x - 6}{7x-1} < 0$.

Critical points: $x = \frac{6}{33} = \frac{2}{11}$ and $x = \frac{1}{7}$.

Sign analysis gives: $\frac{1}{7} < x < \frac{2}{11}$.

Step 2: Solve $\dfrac{x+7}{x-8} > 2$.

$\dfrac{x+7}{x-8} - 2 > 0 \implies \dfrac{x+7 - 2(x-8)}{x-8} > 0 \implies \dfrac{-x + 23}{x-8} > 0$. …

Concept and Intuition

When you see a rational inequality like $\frac{2x+1}{7x-1} > 5$, your first instinct might be to multiply both sides by the denominator. That's dangerous — because the denominator $7x-1$ could be positive or negative depending on $x$, and multiplying by a negative number flips the inequality sign. Instead, the cleanest approach is to bring everything to one side, combine into a single fraction, and then study where that fraction is positive or negative.

The same logic applies to the second inequality. Once we solve each separately, the system asks for values of $x$ that satisfy both — so we take the intersection of the two solution sets.

Let's work through it step by step.

Solving $\dfrac{2x+1}{7x-1} > 5$

1. Bring $5$ to the left side:

2. Combine into a single fraction. Write $5$ as $\frac{5(7x-1)}{7x-1}$:

3. Simplify the numerator:

So we have:

4. Factor where possible. The numerator: $-33x + 6 = -3(11x - 2)$. So:

Since $-3$ is a negative constant, we can multiply both sides by $-1$ (which flips the inequality) to simplify:

5. Now we have a simple rational inequality: $\frac{11x - 2}{7x-1} < 0$. This fraction is negative when the numerator and denominator have opposite signs.

Find the critical points (where numerator or denominator equals zero):

• Numerator zero: $11x - 2 = 0 \implies x = \frac{2}{11}$
• Denominator zero: $7x - 1 = 0 \implies x = \frac{1}{7}$

Note that $x = \frac{1}{7}$ is excluded from the domain (division by zero).

6. Arrange the critical points on the number line: $\frac{2}{11} \approx 0.1818$ and $\frac{1}{7} \approx 0.1429$. So $\frac{2}{11} > \frac{1}{7}$. The order is:

7. Test the sign of $\frac{11x - 2}{7x-1}$ in each interval:

We need the fraction $< 0$, so the solution is: …