Solve the inequality: (21 + 13x) / (2 + 3x²) >= 10

The given inequality is:

Subtract 10 from both sides:

Find a common denominator, which is $2 + 3x$:

Distribute the $-10$ in the numerator:

Combine like terms in the numerator:

The critical points are the values of $x$ that make the numerator zero or the denominator zero.
For the numerator:

For the denominator:

Note that $x = -\frac{2}{3}$ is an excluded value because the denominator cannot be zero.

The critical points $x = -\frac{2}{3}$ and $x = \frac{1}{17}$ divide the number line into three intervals:

1. $x < -\frac{2}{3}$
2. $-\frac{2}{3} < x < \frac{1}{17}$
3. $x > \frac{1}{17}$

We will test a value from each interval in the inequality $\frac{1 - 17x}{2 + 3x} \geq 0$.

• Interval 1: $x < -\frac{2}{3}$ (e.g., let $x = -1$) Numerator: $1 - 17(-1) = 1 + 17 = 18$ (positive) Denominator: $2 + 3(-1) = 2 - 3 = -1$ (negative) Fraction: $\frac{\text{positive}}{\text{negative}} = \text{negative}$. So, $\frac{1 - 17x}{2 + 3x} < 0$.

• Interval 2: $-\frac{2}{3} < x < \frac{1}{17}$ (e.g., let $x = 0$) Numerator: $1 - 17(0) = 1$ (positive) Denominator: $2 + 3(0) = 2$ (positive) Fraction: $\frac{\text{positive}}{\text{positive}} = \text{positive}$. So, $\frac{1 - 17x}{2 + 3x} > 0$.

• Interval 3: $x > \frac{1}{17}$ (e.g., let $x = 1$) Numerator: $1 - 17(1) = 1 - 17 = -16$ (negative) Denominator: $2 + 3(1) = 2 + 3 = 5$ (positive) Fraction: $\frac{\text{negative}}{\text{positive}} = \text{negative}$. So, $\frac{1 - 17x}{2 + 3x} < 0$.

We are looking for where $\frac{1 - 17x}{2 + 3x} \geq 0$. This occurs in Interval 2.
Since the inequality includes "equal to", $x = \frac{1}{17}$ is part of the solution (because it makes the numerator zero).
However, $x = -\frac{2}{3}$ is not part of the solution (because it makes the denominator zero).
Therefore, the solution is $-\frac{2}{3} < x \leq \frac{1}{17}$.