Question

1. $1-\frac{1}{u-5}=\frac{5u^2 + 28u - 12}{u^2 - 3u - 10}$

Explanation:

Step1: Factor the denominator

Factor $u^2 - 3u - 10$ into $(u-5)(u+2)$

Step2: Eliminate denominators

Multiply all terms by $(u-5)(u+2)$:
$$(1)(u-5)(u+2) - \frac{1}{u-5}(u-5)(u+2) = \frac{5u^2+28u-12}{(u-5)(u+2)}(u-5)(u+2)$$
Simplify to get:
$$(u-5)(u+2) - (u+2) = 5u^2+28u-12$$

Step3: Expand left-hand side

Expand and combine like terms:
$$u^2-3u-10 - u - 2 = u^2-4u-12$$

Step4: Set up quadratic equation

Equate to right-hand side and rearrange:
$$u^2-4u-12 = 5u^2+28u-12$$
$$0 = 4u^2+32u$$

Step5: Factor and solve

Factor the quadratic:
$$4u(u+8)=0$$
Solve for $u$:
$4u=0 \implies u=0$; $u+8=0 \implies u=-8$

Step6: Check for extraneous solutions

Verify $u=0$ and $u=-8$ do not make original denominators zero (denominators are zero at $u=5, -2$, so both are valid).

Answer:

$u=0$ and $u=-8$