# Expressing as a single fraction

I'm trying to improve my maths so I can sit the GRE this time next year.

I'm stuck on a really silly question:

Express as a single fraction:

$$\frac{\frac{3x}{2y-7y}}{4x}$$

I'm trying to find a common denominator to cancel out the bottom line. Am I on the right track?

• Yes you are on the right track Oct 30, 2016 at 21:34
• 3x / (2y) - 7y/(4x) Oct 30, 2016 at 21:35
• I'm thinking I should be multiplying the 2y and the 7y by 4x, despite it being an expression rather than an equation. Oct 30, 2016 at 21:37
• I heard good things about New Zealands mathematics, this question kinda amazes me. Oct 30, 2016 at 21:37
• @NewZealan'sfinest What would the common denominator be here ? Oct 30, 2016 at 21:38

Yes, you are on the right track, but in order to do that you will need to multiply up by the denominators, or their multiples, to get the lowest common denominator, I get $4xy$.

Multiply the left fraction by $\frac{2x}{2x}$ and the right fraction by $\frac{y}{y}$

$\frac{3x}{2y}(\frac{2x}{2x})-\frac{7y}{4x}(\frac{y}{y})=\frac{6x^2}{4xy}-\frac{7y^2}{4xy}=\frac{6x^2-7y^2}{4xy}$

As you have stated in the comments that the actual question is: $\frac{3x}{2y-\frac{7y}{4x}}$, if I have understood you correctly, I will answer that now:

$\frac{3x}{2y(\frac{4x}{4x})-\frac{7y}{4x}}=\frac{3x}{\frac{2y-5y}{4x}}=\frac{3x(4x)}{-5y}=\frac{-12x^2}{5y}$

• Thanks so much for this explanation, it's super helpful. I wish I could give you five up votes. Oct 30, 2016 at 21:53
• I'm glad it's helpful, these questions have given me some difficulty in the pat as well
– JAP
Oct 30, 2016 at 21:57

$$\frac{12x^2 - 14y^2 }{8xy}$$

And again, dividing all by $2$,

$$\frac{6x^2 - 7y^2}{4xy}$$

• your eponym would be so proud. Oct 30, 2016 at 21:49