The inequality frac{x-1}{3x+4}

Question

(A) Given the inequality: $\frac{x-1}{3x+4} - \frac{x-3}{3x-2} < 0$ Taking the common denominator: $\frac{(x-1)(3x-2) - (x-3)(3x+4)}{(3x+4)(3x-2)} < 0

Vedclass · Accepted Answer

$\left(-\frac{4}{3}, \frac{2}{3}\right)$

Vedclass · Answer

(A) Given the inequality: $\frac{x-1}{3x+4} - \frac{x-3}{3x-2} Taking the common denominator: $\frac{(x-1)(3x-2) - (x-3)(3x+4)}{(3x+4)(3x-2)} Expanding the numerators: $\frac{(3x^2 - 2x - 3x + 2) - (3x^2 + 4x - 9x - 12)}{(3x+4)(3x-2)} Simplifying the numerator: $\frac{(3x^2 - 5x + 2) - (3x^2 - 5x - 12)}{(3x+4)(3x-2)} $\frac{3x^2 - 5x + 2 - 3x^2 + 5x + 12}{(3x+4)(3x-2)} $\frac{14}{(3x+4)(3x-2)} Since the numerator $14$ is positive,the expression is negative only when the denominator is negative: $(3x+4)(3x-2) The roots of the denominator are $x = -\frac{4}{3}$ and $x = \frac{2}{3}$. Using the wavy curve method (sign scheme),the expression $(3x+4)(3x-2)$ is negative between the roots. Thus,$x \in \left(-\frac{4}{3}, \frac{2}{3}\right)$.

The inequality $\frac{x-1}{3x+4} < \frac{x-3}{3x-2}$ holds for all $x$ in the interval:

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