Question

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

Answer 1

`\boxed{\sf \frac{3x - 11}{ {x}^(2) + 2x - 3}}`

Explanation:

`\sf \implies (5)/(x + 3) - (2)/(x - 1) \\ \\ \sf Put \: each \: term \: in \: (5)/(x + 3) - (2)/(x - 1) \: over \: the \: common \\ \sf denominator \:(x + 3)(x - 1) : \\ \sf \implies (5(x - 1))/((x - 1)(x + 3)) - (2(x + 3))/((x - 1)(x + 3)) \\ \\ \sf (5(x - 1))/((x - 1)(x + 3)) - (2(x + 3))/((x - 1)(x + 3)) = (5(x - 1) - 2(x + 3))/((x - 1)(x + 3)) : \\ \sf \implies (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

last option.

Explanation:

See attached.

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