Complex Fractions with variable in denominator

**street1030** · February 15th 2011, 02:47 PM

2 / 5 / x + 4

Having trouble solving this problem and have no clue where to start. I'm trying to eliminate any fractions in the denominator.

Couln't figure out how to format this right so I just have it typed out. It's 2 divided by 5 over x + 4

**e^(i*pi)** · February 15th 2011, 02:51 PM

Use \dfrac for a fraction: $\dfrac{2}{5x+4}$

However, I see no imaginary part and hence no complex number here. That form is the simplest it's going to get since you can't eliminate x from the denominator.

What are you hoping to achieve?

EDIT:

Assuming you mean $\dfrac{2}{\left(\frac{5}{x+4}\right)}$ then recall that dividing by a fraction is the same as flipping and multiplying it. Thus $\dfrac{2}{\left(\frac{5}{x+4}\right)} = 2 \times \dfrac{x+4}{5}$ which has no x in the denominator.

**topsquark** · February 15th 2011, 02:51 PM

Originally Posted by street1030

$2 / \fraction{5}{x + 4}$

Having trouble solving this problem and have no clue where to start. I'm trying to eliminate any fractions in the denominator.

$\displaystyle \frac{2}{ \frac{5}{x + 4}}$ Is this it?

Like any other complex fraction you need to clear the fraction out of the denominator by multiplying the top and bottom of the fraction by the factor that clears the denominator.

$\displaystyle = \left ( \frac{2}{ \frac{5}{x + 4}} \right ) \cdot \frac{x + 4}{x + 4}$

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

You finish up from here.

-Dan

**Plato** · February 15th 2011, 02:53 PM

Originally Posted by street1030

2 / 5 / x + 4
Couln't figure out how to format this right so I just have it typed out. It's 2 divided by 5 over x + 4

$\dfrac{\frac{2}{5}}{x+4}=\dfrac{2}{5x+20}$

**street1030** · February 15th 2011, 02:58 PM

Sorry for the confusion guys. Forgot to preview post before posting. It's supposed to be $\dfrac{2}{5}$ all over $x + 4$

**e^(i*pi)** · February 15th 2011, 03:07 PM

That would be Plato's answer then and if you're wondering he/she has "dropped" the 5 into the denominator and then distributed it

**street1030** · February 15th 2011, 03:11 PM

Having trouble understanding how we went from $\displaystyle = \frac{2(x + 4)}{ \frac{5}{x + 4} \cdot (x + 4)}$

to Platos

$\dfrac{2}{5x+20}$

Did you cross multiply to cancel out the x + 4 in the denominator and the numerator. Then just distribute the 5 to the remaining x + 4?

**Plato** · February 15th 2011, 03:20 PM

Originally Posted by street1030

Did you cross multiply to cancel out the x + 4 in the denominator and the numerator. Then just distribute the 5 to the remaining x + 4?

$\dfrac{\frac{2}{5}}{x+4}=\dfrac{5\left(\frac{2}{5} \right)}{5(x+4)}=\dfrac{2}{5x+20}$

**street1030** · February 15th 2011, 03:24 PM

Alright thanks guys! Going to go test this out on some more problems.

**topsquark** · February 15th 2011, 03:32 PM

Originally Posted by street1030

Having trouble understanding how we went from $\displaystyle = \frac{2(x + 4)}{ \frac{5}{x + 4} \cdot (x + 4)}$

to Platos

$\dfrac{2}{5x+20}$

Did you cross multiply to cancel out the x + 4 in the denominator and the numerator. Then just distribute the 5 to the remaining x + 4?

They are not the same. I was using a different fraction than the one that Plato simplified for you.

-Dan

**street1030** · February 15th 2011, 03:40 PM

O yeah. I need to do the one you're doing topsquark. So the way you did it. you get? 2x + 8 / 5?

**topsquark** · February 15th 2011, 04:03 PM

Originally Posted by street1030

O yeah. I need to do the one you're doing topsquark. So the way you did it. you get? 2x + 8 / 5?

(2x + 8)/5. Yes. Please note my use of the parenthesis. What you wrote is 2x + (8/5).

-Dan

**street1030** · February 15th 2011, 04:12 PM

Thanks a bunch

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