Question

simplify and reduce to the lowest terms. (leave answer in factored form.) (\frac{3x}{x^2 - 6x + 9} + \frac{7}{x - 3} =) (\boxed{}) excluded value(s) (x
eq) (\boxed{})

Explanation:

Step1: Factor the denominator

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

Step2: Find common denominator

The common denominator is $(x-3)^2$. Rewrite the second fraction:
$\frac{7}{x-3} = \frac{7(x-3)}{(x-3)^2}$

Step3: Combine the fractions

$\frac{3x}{(x-3)^2} + \frac{7(x-3)}{(x-3)^2} = \frac{3x + 7(x-3)}{(x-3)^2}$

Step4: Simplify the numerator

$3x + 7x - 21 = 10x - 21$

Step5: Identify excluded values

Denominator cannot be zero, so $x-3
eq 0$

Answer:

Simplified form: $\frac{10x - 21}{(x-3)^2}$
Excluded Value(s) $x
eq 3$