Rational and Radical Equations

1. The denominators cannot be zero, so:

$x \neq 1, \quad x \neq -2$

Multiply both sides by the least common denominator $(x - 1)(x + 2)$:

$2(x + 2) + 3(x - 1) = (x - 1)(x + 2)$

Expand:

$2x + 4 + 3x - 3 = x^2 + x - 2$

$5x + 1 = x^2 + x - 2$

Move all terms to one side:

$x^2 - 4x - 3 = 0$

Use the quadratic formula:

$x = \frac{4 \pm \sqrt{(-4)^2 - 4(1)(-3)}}{2} = \frac{4 \pm \sqrt{28}}{2}$

$x = 2 \pm \sqrt{7}$

Neither value is $1$ or $-2$, so both are allowed by the restrictions.

Check: If $x$ satisfies $x^2 - 4x - 3 = 0$, then the equation obtained after multiplying by the nonzero LCD is true. Since neither solution makes the LCD zero, both solutions are valid.

The solutions are:

$\boxed{x = 2 + \sqrt{7} \text{ or } x = 2 - \sqrt{7}}$

2. The denominator cannot be zero, so $x \neq 2$.

Multiply both sides by $x - 2$:

$x + 4 = 3(x - 2) - 10$

Expand and solve:

$x + 4 = 3x - 6 - 10$

$x + 4 = 3x - 16$

$20 = 2x$

$x = 10$

Since $10 \neq 2$, the solution does not make the denominator zero.

Verification:

$\frac{10 + 4}{10 - 2} = \frac{14}{8} = \frac{7}{4}$

$3 - \frac{10}{10 - 2} = 3 - \frac{10}{8} = 3 - \frac{5}{4} = \frac{7}{4}$

The solution is:

$\boxed{x = 10}$

3. a) Pipe A fills $\frac{1}{x}$ of the tank per hour. Pipe B fills $\frac{1}{x + 3}$ of the tank per hour. Together, they fill $\frac{1}{2}$ of the tank per hour:

$\frac{1}{x} + \frac{1}{x + 3} = \frac{1}{2}$

b) The restrictions are $x \neq 0$ and $x \neq -3$. In context, $x$ must also be positive.

Multiply both sides by $2x(x + 3)$:

$2(x + 3) + 2x = x(x + 3)$

$2x + 6 + 2x = x^2 + 3x$

$4x + 6 = x^2 + 3x$

$0 = x^2 - x - 6$

Factor:

$0 = (x - 3)(x + 2)$

So:

$x = 3 \quad \text{or} \quad x = -2$

c) The value $x = -2$ is not valid because time cannot be negative. It also would make Pipe B's time $x + 3 = 1$ hour, but Pipe A's time would be $-2$ hours, which has no meaning in this context.

Therefore Pipe A alone takes $3$ hours, and Pipe B alone takes:

$x + 3 = 3 + 3 = 6 \text{ hours}$

Verification:

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

Together they fill the tank in $2$ hours. $\checkmark$

4. The square root is defined when $2x + 7 \geq 0$. Also, because the left side is nonnegative, the right side must satisfy $x + 1 \geq 0$. Any solution must have $x \geq -1$.

Square both sides:

$\sqrt{2x + 7} = x + 1$

$2x + 7 = (x + 1)^2$

$2x + 7 = x^2 + 2x + 1$

$7 = x^2 + 1$

$x^2 = 6$

$x = \pm \sqrt{6}$

Only $x = \sqrt{6}$ satisfies $x \geq -1$. Check it in the original equation:

$\sqrt{2\sqrt{6} + 7} \stackrel{?}{=} \sqrt{6} + 1$

Square the right side:

$(\sqrt{6} + 1)^2 = 6 + 2\sqrt{6} + 1 = 7 + 2\sqrt{6}$

This matches the expression under the radical, so:

$\sqrt{2\sqrt{6} + 7} = \sqrt{6} + 1$

The solution is:

$\boxed{x = \sqrt{6}}$

5. The square roots are defined when:

$x + 5 \geq 0 \quad \text{and} \quad x - 1 \geq 0$

So $x \geq 1$.

Move one radical to the other side:

$\sqrt{x + 5} = 2 + \sqrt{x - 1}$

Square both sides:

$x + 5 = (2 + \sqrt{x - 1})^2$

$x + 5 = 4 + 4\sqrt{x - 1} + x - 1$

$x + 5 = x + 3 + 4\sqrt{x - 1}$

Subtract $x + 3$ from both sides:

$2 = 4\sqrt{x - 1}$

$\frac{1}{2} = \sqrt{x - 1}$

Square again:

$\frac{1}{4} = x - 1$

$x = \frac{5}{4}$

Check in the original equation:

$\sqrt{\frac{5}{4} + 5} - \sqrt{\frac{5}{4} - 1} = \sqrt{\frac{25}{4}} - \sqrt{\frac{1}{4}}$

$= \frac{5}{2} - \frac{1}{2} = 2$

The solution is:

$\boxed{x = \frac{5}{4}}$