2x − 3 < 7x + 4

The inequality shown in the image is:

Subtract $\frac{2}{3x+1}$ from both sides to set the inequality to zero:

The common denominator is $(5x-2)(3x+1)$.

Expand the numerator:

Simplify the numerator:

Critical points are the values of $x$ that make the numerator or the denominator equal to zero.

• Numerator: $-x+7 = 0 \implies x = 7$.
• Denominator: $5x-2 = 0 \implies 5x = 2 \implies x = \frac{2}{5}$.
• Denominator: $3x+1 = 0 \implies 3x = -1 \implies x = -\frac{1}{3}$. Arrange the critical points in ascending order: $-\frac{1}{3}$, $\frac{2}{5}$, 7.

These critical points divide the number line into four intervals: $(-\infty, -\frac{1}{3})$, $(-\frac{1}{3}, \frac{2}{5})$, $(\frac{2}{5}, 7)$, and $(7, \infty)$. We test a value from each interval in the simplified expression $f(x) = \frac{-x+7}{(5x-2)(3x+1)}$.

• Interval 1: $x < -\frac{1}{3}$ (e.g., $x = -1$) $f(-1) = \frac{-(-1)+7}{(5(-1)-2)(3(-1)+1)} = \frac{8}{(-7)(-2)} = \frac{8}{14} > 0$.
• Interval 2: $-\frac{1}{3} < x < \frac{2}{5}$ (e.g., $x = 0$) $f(0) = \frac{-0+7}{(5(0)-2)(3(0)+1)} = \frac{7}{(-2)(1)} = -\frac{7}{2} < 0$.
• Interval 3: $\frac{2}{5} < x < 7$ (e.g., $x = 1$) $f(1) = \frac{-1+7}{(5(1)-2)(3(1)+1)} = \frac{6}{(3)(4)} = \frac{6}{12} > 0$.
• Interval 4: $x > 7$ (e.g., $x = 8$) $f(8) = \frac{-8+7}{(5(8)-2)(3(8)+1)} = \frac{-1}{(38)(25)} < 0$.

We are looking for where $f(x) \le 0$. This means the expression is negative or equal to zero.
The expression is negative in the intervals $(-\frac{1}{3}, \frac{2}{5})$ and $(7, \infty)$.
The expression is zero when $x=7$.
The values $x = -\frac{1}{3}$ and $x = \frac{2}{5}$ make the denominator zero, so they are excluded from the solution (represented by open intervals).
Combining these, the solution set is $x \in \left(-\frac{1}{3}, \frac{2}{5}\right) \cup [7, \infty)$.

Draw a number line. Place open circles at $x = -\frac{1}{3}$ and $x = \frac{2}{5}$ to indicate exclusion. Place a closed circle at $x = 7$ to indicate inclusion. Shade the regions that satisfy the inequality.

The solution is $\boxed{x \in \left(-\frac{1}{3}, \frac{2}{5}\right) \cup [7, \infty)}$.