Question

add.
\\(\frac{8}{x - 7} + \frac{x - 4}{x - 3}\\)
simplify your answer as much as

Explanation:

Step1: Find common denominator

The common denominator of $(x-7)$ and $(x-3)$ is $(x-7)(x-3)$. Rewrite each fraction with this denominator:
$$\frac{8(x-3)}{(x-7)(x-3)} + \frac{(x-4)(x-7)}{(x-7)(x-3)}$$

Step2: Expand numerators

Calculate the expanded form of each numerator:
$$8(x-3) = 8x - 24$$
$$(x-4)(x-7) = x^2 -7x -4x +28 = x^2 -11x +28$$

Step3: Combine numerators

Add the expanded numerators over the common denominator:
$$\frac{(8x - 24) + (x^2 -11x +28)}{(x-7)(x-3)}$$

Step4: Simplify numerator

Combine like terms in the numerator:
$$x^2 + (8x -11x) + (-24 +28) = x^2 -3x +4$$

Step5: Final simplified form

Write the simplified fraction, noting the denominator cannot be factored to cancel with the numerator:
$$\frac{x^2 -3x +4}{(x-7)(x-3)}$$
(or expanded denominator: $\frac{x^2 -3x +4}{x^2 -10x +21}$)

Answer:

$\frac{x^2 - 3x + 4}{(x-7)(x-3)}$ (or equivalently $\frac{x^2 - 3x + 4}{x^2 - 10x + 21}$)