Question

consider the expression \\(\frac{3x + 4}{x + 5} - \frac{2x - 1}{x + 3}\\)

part a
write the expression as one fraction in simplified form.
enter the correct answer in the box.

part b
what are the excluded values?
\\(\bigcirc\\) \\(x = -3, 5\\)
\\(\bigcirc\\) \\(x = -3, -5\\)
\\(\bigcirc\\) \\(x = 3, 5\\)
\\(\bigcirc\\) \\(x = 3, -5\\)

Explanation:

Step1: Find common denominator

The common denominator is $(x+5)(x+3)$. Rewrite each fraction:
$\frac{(3x+4)(x+3)}{(x+5)(x+3)} - \frac{(2x-1)(x+5)}{(x+5)(x+3)}$

Step2: Expand numerators

Calculate each numerator product:
$(3x+4)(x+3) = 3x^2 + 9x + 4x + 12 = 3x^2 + 13x + 12$
$(2x-1)(x+5) = 2x^2 + 10x - x - 5 = 2x^2 + 9x - 5$

Step3: Subtract numerators

Subtract the second expanded numerator from the first:
$(3x^2 + 13x + 12) - (2x^2 + 9x - 5) = 3x^2 +13x +12 -2x^2 -9x +5$
$= x^2 + 4x + 17$

Step4: Simplify the fraction

Combine into a single fraction:
$\frac{x^2 + 4x + 17}{(x+5)(x+3)}$

Step5: Find excluded values

Set denominators equal to 0:
$x+5=0 \implies x=-5$
$x+3=0 \implies x=-3$

Answer:

Part A:

$\frac{x^2 + 4x + 17}{(x+5)(x+3)}$

Part B:

$\boldsymbol{x = -3, -5}$