Question

Im having trouble with this question... please help me.

Answer 1

Answer:

$\boxed{\sf \frac{3x - 11}{ {x}^{2} + 2x - 3}}$

Step-by-step explanation:

$\sf \implies \frac{5}{x + 3} - \frac{2}{x - 1} \\ \\ \sf Put \: each \: term \: in \: \frac{5}{x + 3} - \frac{2}{x - 1} \: over \: the \: common \\ \sf denominator \:(x + 3)(x - 1) : \\ \sf \implies \frac{5(x - 1)}{(x - 1)(x + 3)} - \frac{2(x + 3)}{(x - 1)(x + 3)} \\ \\ \sf \frac{5(x - 1)}{(x - 1)(x + 3)} - \frac{2(x + 3)}{(x - 1)(x + 3)} = \frac{5(x - 1) - 2(x + 3)}{(x - 1)(x + 3)} : \\ \sf \implies \frac{5(x - 1) - 2(x + 3)}{(x - 1)(x + 3)} \\ \\ \sf 5(x - 1) = 5x - 5 : \\ \sf \implies \frac{ \boxed{ \sf 5x - 5} - 2(x + 3)}{(x - 1)(x + 3)} \\ \\ \sf - 2(x + 3) = - 2x - 6 : \\ \sf \implies \frac{5x - 5 + \boxed{ \sf - 2x - 6}}{(x - 1)(x + 3)} \\ \\ \sf Grouping \: like \: terms, \: 5x - 5 - 2x - 6 = \\ \sf (5x - 2x) + ( - 5 - 6) : \\ \sf \implies \frac{ \boxed{ \sf (5x - 2x) + ( - 5 - 6)}}{(x - 1)(x + 3)} \\ \\ \sf 5x - 2x = 3x : \\ \sf \implies \frac{ \boxed{ \sf 3x} + ( - 5 - 6)}{(x - 1)(x + 3)} \\ \\ \sf - 5 - 6 = - 11 : \\ \sf \implies \frac{3x + \boxed{ - 11}}{(x - 1)(x + 3)}$

$\sf (x - 1)(x + 3) = x(x + 3) - 1(x + 3) : \\ \sf \implies \frac{3x - 11}{ \boxed{ \sf x(x + 3) - 1(x + 3)}} \\ \\ \sf x(x + 3) = {x}^{2} + 3x : \\ \sf \implies \frac{3x - 11}{ \boxed{ \sf {x}^{2} + 3x} - 1(x + 3)} \\ \\ \sf - 1(x + 3) = - x - 3 \\ \sf \implies \frac{3x - 11}{ {x}^{2} + 3x + \boxed{ \sf - x - 3} } \\ \\ \sf 3x - x = 2x : \\ \sf \implies \frac{3x - 11}{ {x}^{2} + 2x - 3}$

Answer 2

Answer:

last option.

Step-by-step explanation:

See attached.

I'm not sure why you crossed out the last option, but it seems to be correct.