Question

I need help asap!

And show work if possible so I can figure out how to solve similar questions.

Answer 1

A.)
`(x)/(7)`

B.)
`(x+1)/(x-3)`

C.)
`(3(x+12))/((x-3)(x+8))` =
`(3x+36)/(x^2+5x-24)`

D.)
`(2(x-7))/((x+4)(x+6))` =
`(2x-14)/(x^2+10x+24)`

Explanation:

A.)
`(x^2+4x+3)/(x^2-x-2)* (x^2-2x)/(7x+21)\\`, Firstly you have to factor the denominator (bottom terms on lower part of the fraction.)

`(x^2+4x+3)/((x-2)(x+1)) * (x^2-2x)/(7(x+3))` + also make sure that the denominator
`\\eq 0` if it's asking for domain, now we do the same for the numerator.

`((x+1)(x+3))/((x-2)(x+1)) * (x(x-2))/(7(x+3))`, now we cross-multiply. Get rid of the common factors you see.
`(x+1)/(x+1)*(x)/(7)`, then simplify further
`1*(x)/(7)` =
`(x)/(7)`

B.) Remember to KCF, keep the first fraction, turn division sign into multiplication, flip second fraction

`(x^2-x-20)/(x^2-25) / (x^2+x-12)/(x^2+6x+5)`,
`(x^2-x-20)/(x^2-25) * (x^2+6x+5)/(x^2+x-12)`, now do the steps i did in step a.

`((x+4)(x-5))/((x-5)(x+5)) * ((x+5)(x+1))/((x+4)(x-3))`, simplify by cross-multiplying equals to
`1* ((x+1))/((x-3)) = (x+1)/(x-3)`

C.)
`(12)/(x^2+5x-24) +(3)/(x-3)`, multiply to get same factors for denominator on both sides then add and then simplify.
`(12)/((x+8)(x-3)) +(3)/(x-3)`, for 2nd fraction multiply by x+8,

`(12)/((x+8)(x-3)) +(3(x+8))/((x-3)(x+8))` =
`(12)/((x+8)(x-3)) +(3x+24)/((x-3)(x+8)) = (3(x+12))/((x-3)(x+8))`

D.) Same as c, just multiply to get same denominator but for both

`(3(x+6))/((x+4)(x+6)) - (x+4)/((x+4)(x+6)) = (3x+18)/((x+4)(x+6)) -(x+4)/((x+4)(x+6))`

finally =
`(2(x-7))/((x+4)(x+6))`