Rational Expressions and Equations

1. (a) Factor the numerator:

$\frac{6x^2 - 12x}{3x} = \frac{6x(x-2)}{3x} = 2(x-2) = 2x - 4, \quad x \neq 0$

(b) Factor numerator and denominator:

$\frac{x^2 - 9}{x^2 - x - 6} = \frac{(x-3)(x+3)}{(x-3)(x+2)} = \frac{x+3}{x+2}, \quad x \neq 3,\ x \neq -2$

(Note: $x \neq 3$ because it makes the original denominator zero, even though it canceled.)

2. (a) Multiply and factor:

$\frac{2x}{x+3} \cdot \frac{x^2-9}{4x^2} = \frac{2x \cdot (x-3)(x+3)}{(x+3) \cdot 4x^2}$

Cancel $(x+3)$ and one factor of $x$:

$= \frac{2(x-3)}{4x} = \frac{x-3}{2x}, \quad x \neq 0,\ x \neq -3$

(b) Rewrite as multiplication by the reciprocal:

$\frac{x^2-4}{x+1} \div \frac{x-2}{x^2-1} = \frac{x^2-4}{x+1} \cdot \frac{x^2-1}{x-2}$

Factor: $x^2 - 4 = (x-2)(x+2)$ and $x^2 - 1 = (x-1)(x+1)$:

$= \frac{(x-2)(x+2)}{x+1} \cdot \frac{(x-1)(x+1)}{x-2}$

Cancel $(x-2)$ and $(x+1)$:

$= (x+2)(x-1) = x^2 + x - 2, \quad x \neq 2,\ x \neq \pm 1$

3. The LCD is $(x-2)(x+1)$:

$\frac{3}{x-2} + \frac{5}{x+1} = \frac{3(x+1)}{(x-2)(x+1)} + \frac{5(x-2)}{(x-2)(x+1)}$

$= \frac{3x + 3 + 5x - 10}{(x-2)(x+1)} = \frac{8x - 7}{(x-2)(x+1)}$

This does not factor further.

$\frac{8x-7}{(x-2)(x+1)}, \quad x \neq 2,\ x \neq -1$

4. Restrictions: $x \neq 3$ (denominator cannot be zero).

Multiply both sides by $(x-3)$:

$x = 9 + 2(x-3)$ $x = 9 + 2x - 6$ $x = 3 + 2x$ $-x = 3$ $x = -3$

Check restriction: $x = -3 \neq 3$, so it is allowed.

Check in original:

$\frac{-3}{-3-3} = \frac{-3}{-6} = \frac{1}{2}$

$\frac{9}{-3-3} + 2 = \frac{9}{-6} + 2 = -\frac{3}{2} + 2 = \frac{1}{2} \checkmark$

Solution: $x = -3$ (no extraneous solutions).

5. Restrictions: $x \neq 0$ and $x \neq -4$.

Multiply both sides by $x(x+4)$:

$2(x+4) + 3x = x(x+4)$ $2x + 8 + 3x = x^2 + 4x$ $5x + 8 = x^2 + 4x$ $0 = x^2 - x - 8$

Use the quadratic formula ($a=1$, $b=-1$, $c=-8$):

$x = \frac{1 \pm \sqrt{1 + 32}}{2} = \frac{1 \pm \sqrt{33}}{2}$

$x = \frac{1 + \sqrt{33}}{2} \approx 3.37 \quad \text{or} \quad x = \frac{1 - \sqrt{33}}{2} \approx -2.37$

Neither solution equals $0$ or $-4$, so both are valid — no extraneous solutions.

$x = \frac{1 \pm \sqrt{33}}{2}$

6. Pipe 1 fills $\frac{1}{6}$ of the tank per hour. Pipe 2 fills $\frac{1}{4}$ per hour. Together, in $t$ hours:

$\frac{t}{6} + \frac{t}{4} = 1$

Multiply through by the LCD $= 12$:

$2t + 3t = 12$ $5t = 12$ $t = \frac{12}{5} = 2.4 \text{ hours}$

Both pipes together fill the tank in $\dfrac{12}{5}$ hours (2 hours 24 minutes).

Check: $\frac{2.4}{6} + \frac{2.4}{4} = 0.4 + 0.6 = 1.0$ tank ✓