# simplifying (5x/4x+3) - (6/4x^2+3x) + (2/x), {(3/x+1)+(....

#### gierhame

##### New member
I can't figure out how to simplify these two problems. I have tried multiple times but I can seem to find the right answer.

The first problem is
(5x/4x+3) - (6/4x^2+3x) + (2/x)
I know you have to come up with a common denominator and I keep getting
(8x^2+11x-6)/(4x^2+3x)
I don't know what I am doing wrong.

The second is
{(3/x+1)+(4x/x+3)} / {(x/x+1) + (6/x+3)}
The answer I came up with was 13/3x+6
Once again I can't figure it out.

#### o_O

##### Full Member
Re: simplifying

Show us your work. It would be much easier for us to pinpoint your mistake.

#### Subhotosh Khan

##### Super Moderator
Staff member
Re: simplifying

gierhame said:
I can't figure out how to simplify these two problems. I have tried multiple times but I can seem to find the right answer.

The first problem is
(5x/4x+3) - (6/4x^2+3x) + (2/x)
I know you have to come up with a common denominator and I keep getting
(8x^2+11x-6)/(4x^2+3x)
I don't know what I am doing wrong.

The second is
{(3/x+1)+(4x/x+3)} / {(x/x+1) + (6/x+3)}
The answer I came up with was 13/3x+6
Once again I can't figure it out.

$$\displaystyle \frac{5x}{4x+3}\, + \, \frac{6}{4x^2+3x}\, + \, \frac{2}{x}$$

$$\displaystyle =\, \frac{5x}{4x+3}\, + \, \frac{6}{x\cdot (4x+3)}\, + \, \frac{2}{x}$$

$$\displaystyle =\, \frac{5x\cdot x\, + \, 6 \, + \, 2\cdot (4x+3)}{x\cdot (4x+3)}$$

$$\displaystyle =\, \frac{5x^2\, + \, 8x \, + \, 12}{x\cdot (4x+3)}$$

#### Loren

##### Senior Member
Re: simplifying

(5x/4x+3) - (6/4x^2+3x) + (2/x) means $$\displaystyle (\frac{5x}{4}x+3) - (\frac{6}{4}x^2 +3x)+\frac{2}{x}$$.

If that is not what you mean you need to clarify by using grouping symbols correctly.

#### gierhame

##### New member
So my work for the second one is:

3 + 4x
x+1 x+3

x + 6
x+1 x+3

=x+3)(3) + x+3)(4x)
(x+1)(x+3) (x+1)(x-1)

(x+3)(x) + (x+1)((6)
(x+1)(x+3) (x+1)(x+3)

=4x[sup:35wn23nw]2[/sup:35wn23nw]-9x+9 times (x+1)(x+3)

(x+1)(x+3) x[sup:35wn23nw]2[/sup:35wn23nw]+9x+6

=(4x-3)(x+3)
x[sup:35wn23nw]2[/sup:35wn23nw]+9x+6

#### stapel

##### Super Moderator
Staff member
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.

#### gierhame

##### New member
can the top expression be factored, because I can't find a way to make it work?

#### stapel

##### Super Moderator
Staff member
gierhame said:
can the top expression be factored, because I can't find a way to make it work?