gierhame said:

So my work for the second one is:

__3__ + __4x__

x+1 x+3

__x__ + __6__

x+1 x+3

I will guess that you mean the following:

. . . . .\(\displaystyle \frac{\left(\frac{3}{x\, +\, 1}\, +\, \frac{4x}{x\, +\, 3}\right)}{\left(\frac{x}{x\, +\, 1}\, +\, \frac{6}{x\, +\, 3}\right)}\)

I

*think* you then converted everything to a common denominator:

. . . . .\(\displaystyle \frac{\left(\frac{3(x\, +\, 3)}{(x\, +\, 1)(x\, +\, 3)}\, +\, \frac{4x(x\, +\, 1)}{(x\, +\, 3)(x\, +\, 1)}\right)}{\left(\frac{x(x\, +\, 3)}{(x\, +\, 1)(x\, +\, 3)}\, +\, \frac{6(x\, +\, 1)}{(x\, +\, 3)(x\, +\, 1)}\right)}\)

Your work gets a bit confusing at this point. I

*think* you switched to just showing your work for one of the numerators...?

To continue clearly:

We multiply through the various parentheses, and simplify the polynomials:

. . . . .\(\displaystyle \frac{\left(\frac{3x\, +\, 9\, +\, 4x^2\, +\, 4x}{(x\, +\, 1)(x\, +\, 3)}\right)}{\left(\frac{x^2\, +\, 3x\, +\, 6x\, +\, 6}{(x\, +\, 1)(x\, +\, 3)}\right)}\)

. . . . .\(\displaystyle \frac{\left(\frac{4x^2\, +\, 7x\, +\, 9}{(x\, +\, 1)(x\, +\, 3)}\right)}{\left(\frac{x^2\, +\, 9x\, +\, 6}{(x\, +\, 1)(x\, +\, 3)}\right)}\)

To simplify when dividing by a fraction, one "flips" and converts to multiplication, which then leads, in this case, to cancellation:

. . . . .\(\displaystyle \left(\frac{4x^2\, +\, 7x\, +\, 9}{(x\, +\, 1)(x\, +\, 3)}\right)\, \left(\frac{(x\, +\, 1)(x\, +\, 3)}{x^2\, +\, 9x\, +\, 6}\right)\)

. . . . .\(\displaystyle \left(\frac{4x^2\, +\, 7x\, +\, 9}{1}\right)\, \left(\frac{1}{x^2\, +\, 9x\, +\, 6}\right)\)

. . . . .\(\displaystyle \frac{4x^2\, +\, 7x\, +\, 9}{x^2\, +\, 9x\, +\, 6}\)

You have factored one of the quadratics. Now factor the other one, and cancel the common factor to arrive at the fully-simplified form.

Eliz.