Solve the inequality: (x-2)/(4x+3) (3x+4)/(5-6x)

The given inequality is:

Subtract $\frac{3x+4}{5-6x}$ from both sides to get:

The common denominator is $(4x+3)(5-6x)$.

First, expand $(x-2)(5-6x)$:

Next, expand $(3x+4)(4x+3)$:

Now, subtract the second expanded expression from the first:

Combine like terms:

Factor out $-2$ from the numerator:

The inequality becomes:

Consider the quadratic expression $9x^2 + 4x + 11$. Its discriminant is $\Delta = b^2 - 4ac = (4)^2 - 4(9)(11) = 16 - 396 = -380$.
Since the discriminant is negative ($\Delta < 0$) and the leading coefficient ($a=9$) is positive, the quadratic $9x^2 + 4x + 11$ is always positive for all real values of $x$.
Therefore, the numerator $-2(9x^2 + 4x + 11)$ is always negative (a negative number multiplied by a positive number).

For the entire fraction to be greater than or equal to zero, and knowing the numerator is always negative, the denominator must also be negative.
Also, the denominator cannot be zero, so we must have $(4x+3)(5-6x) < 0$.

Set each factor in the denominator to zero to find the critical points:
$4x+3 = 0 \implies 4x = -3 \implies x = -\frac{3}{4}$
$5-6x = 0 \implies 5 = 6x \implies x = \frac{5}{6}$
These critical points divide the number line into three intervals: $(-\infty, -\frac{3}{4})$, $(-\frac{3}{4}, \frac{5}{6})$, and $(\frac{5}{6}, \infty)$.

• For $x < -\frac{3}{4}$ (e.g., $x=-1$): $(4(-1)+3)(5-6(-1)) = (-1)(5+6) = (-1)(11) = -11$. This is negative, so the inequality holds.
• For $-\frac{3}{4} < x < \frac{5}{6}$ (e.g., $x=0$): $(4(0)+3)(5-6(0)) = (3)(5) = 15$. This is positive, so the inequality does not hold.
• For $x > \frac{5}{6}$ (e.g., $x=1$): $(4(1)+3)(5-6(1)) = (7)(5-6) = (7)(-1) = -7$. This is negative, so the inequality holds.

The solution is where $(4x+3)(5-6x) < 0$, which is $x < -\frac{3}{4}$ or $x > \frac{5}{6}$.

The final answer is $x < -\frac{3}{4} \text{ or } x > \frac{5}{6}$.