Question

1. what is the difference of \\(\frac{4x + 6}{x^2 + 7x + 10} - \frac{2x + 3}{x^2 + 6x + 8}\\)?\

\\(\frac{2x^2 - 9x - 9}{(x + 5)(x + 2)(x + 4)}, x \
eq -5, x \
eq -2, x \
eq -4\\)\
\\(\frac{2x^2 + 9x + 9}{(x + 5)(x + 2)(x + 4)}, x \
eq -5, x \
eq -2, x \
eq -4\\)\
\\(\frac{2x^2 - 9x + 9}{(x + 5)(x + 2)(x + 4)}, x \
eq -5, x \
eq -2, x \
eq -4\\)\
\\(\frac{2x^2 + 9x - 9}{(x + 5)(x + 2)(x + 4)}, x \
eq -5, x \
eq -2, x \
eq -4\\)

Explanation:

Step1: Factor denominators

First denominator: $x^2 +7x +10=(x+5)(x+2)$
Second denominator: $x^2 +6x +8=(x+2)(x+4)$
LCD: $(x+5)(x+2)(x+4)$

Step2: Rewrite fractions with LCD

First fraction: $\frac{(4x+6)(x+4)}{(x+5)(x+2)(x+4)}=\frac{4x^2+16x+6x+24}{LCD}=\frac{4x^2+22x+24}{LCD}$
Second fraction: $\frac{(2x+3)(x+5)}{(x+5)(x+2)(x+4)}=\frac{2x^2+10x+3x+15}{LCD}=\frac{2x^2+13x+15}{LCD}$

Step3: Subtract the fractions

$\frac{(4x^2+22x+24)-(2x^2+13x+15)}{LCD}=\frac{4x^2+22x+24-2x^2-13x-15}{LCD}=\frac{2x^2+9x+9}{LCD}$

Wait correction: Wait no, wait original problem is $\frac{4x+6}{x²+7x+10} - \frac{2x+3}{x²+6x+8}$
Wait 4x+6=2(2x+3), so first fraction: $\frac{2(2x+3)}{(x+5)(x+2)}$
Second fraction: $\frac{2x+3}{(x+2)(x+4)}$
Factor out (2x+3): $(2x+3)[\frac{2}{(x+5)(x+2)} - \frac{1}{(x+2)(x+4)}]$
Take LCD inside bracket: $\frac{2(x+4)-1(x+5)}{(x+5)(x+2)(x+4)}=\frac{2x+8 -x -5}{denom}=\frac{x+3}{denom}$
Multiply back: $(2x+3)(x+3)/denom=(2x²+6x+3x+9)/denom=(2x²+9x+9)/denom$

Wait but the options: Let me check again
Wait no wait original problem: $\frac{4x+6}{x²+7x+10} - \frac{2x+3}{x²+6x+8}$
Wait 4x+6=2(2x+3), x²+7x+10=(x+5)(x+2), x²+6x+8=(x+2)(x+4)
So rewrite: $\frac{2(2x+3)}{(x+5)(x+2)} - \frac{2x+3}{(x+2)(x+4)}$
Common denominator is (x+5)(x+2)(x+4)
First term: $\frac{2(2x+3)(x+4)}{(x+5)(x+2)(x+4)}=\frac{2(2x²+8x+3x+12)}{denom}=\frac{2(2x²+11x+12)}{denom}=\frac{4x²+22x+24}{denom}$
Second term: $\frac{(2x+3)(x+5)}{(x+5)(x+2)(x+4)}=\frac{2x²+10x+3x+15}{denom}=\frac{2x²+13x+15}{denom}$
Subtract: (4x²+22x+24)-(2x²+13x+15)=2x²+9x+9
So the result is $\frac{2x²+9x+9}{(x+5)(x+2)(x+4)}$ with restrictions x≠-5,-2,-4
Which is the second option: $\frac{2x²+9x+9}{(x+5)(x+2)(x+4)}, x≠-5,x≠-2,x≠-4$

Step1: Factor denominators

$x^2+7x+10=(x+5)(x+2)$, $x^2+6x+8=(x+2)(x+4)$

Step2: Rewrite with LCD

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

Step3: Expand numerators

$2(2x²+11x+12)-(2x²+13x+15)=4x²+22x+24-2x²-13x-15$

Step4: Simplify numerator

$2x²+9x+9$

Step5: Final result

$\frac{2x²+9x+9}{(x+5)(x+2)(x+4)}$ with restrictions.

Answer:

$\frac{2x^2 - 9x - 9}{(x+5)(x+2)(x+4)}, x
eq -5, x
eq -2, x
eq -4$