# Math Help - Variable Factions help

1. ## Variable Factions help

$\frac{2}{15x} + \frac{7}{18}$

Can someone please go through the details on how to solve this? The correct answer is $\frac{12+35x}{90x}$

Edit: Title should read variable Fractions obviously my mistake.

2. Originally Posted by cmf0106
$\frac{2}{15x} + \frac{7}{18}$

Can someone please go through the details on how to solve this? The correct answer is $\frac{12+35x}{90x}$

Edit: Title should read variable Fractions obviously my mistake.
$15x=3\cdot 5 \cdot x$
$18=2 \cdot 3^2$

$LCD=90x$

$\frac{90x(\frac{2}{15x})}{90x}+\frac{90x(\frac{7}{ 18})}{90x}$

$\frac{12}{90x}+\frac{35x}{90x}=\frac{12+35x}{90x}$

3. I think that clarifies that question a few more on the same subject though.

1.in this example for canceling variables we have
$\frac{2+x}{3} = \frac{2}{x} + \frac{x}{x} = \frac{2}{x} +1$

Note that in this example the x in the denominator is applied to the 2 and x in the numerator. Shown here $\frac{2+x}{x}$, the x in the denominator goes to both numerators thus $\frac{2}{x} + \frac{x}{x}$

2.However in this example
$\frac{(2 + 3x)(x-1)}{2 + 3x}$ For the solution the author does not apply the whole denominator to both of the numerators and this is confusing me greatly. Working this problem out we have $\frac{(2 + 3x)(x-1)}{2 + 3x}=\frac{2+3x}{2+3x} * \frac{x-1}{1}= x -1$

why is the 2 + 3x not applied to both parts of this numerator resulting in $\frac{2+3x}{2+3x} * \frac{x-1}{2+3x}$?

4. Originally Posted by cmf0106
I think that clarifies that question a few more on the same subject though.

1.in this example for canceling variables we have
$\frac{2+x}{\color{red}3} = \frac{2}{x} + \frac{x}{x} = \frac{2}{x} +1$

Note that in this example the x in the denominator is applied to the 2 and x in the numerator. Shown here $\frac{2+x}{x}$, the x in the denominator goes to both numerators thus $\frac{2}{x} + \frac{x}{x}$

I assume you intended to put an x in the denominator instead of a 3.

2.However in this example
$\frac{(2 + 3x)(x-1)}{2 + 3x}$ For the solution the author does not apply the whole denominator to both of the numerators and this is confusing me greatly. Working this problem out we have $\frac{(2 + 3x)(x-1)}{2 + 3x}=\frac{2+3x}{2+3x} * \frac{x-1}{1}= x -1$

why is the 2 + 3x not applied to both parts of this numerator resulting in $\frac{2+3x}{2+3x} * \frac{x-1}{2+3x}$?

The first expression is made up of addends (terms that are added together). You must have a common denominator to combine them.

Your second expression is a product. You do not need a common denominator. You simply multiply numerators and denominators and factor out any common elements in the product or along the way.

$\frac{(2 + 3x)(x-1)}{2 + 3x}=\frac{2+3x}{2+3x} * \frac{x-1}{1}=\not[\frac{2+3x}{2+3x}] * \frac{x-1}{1}= \frac{1}{1} * \frac{x-1}{1}=x -1$

$\frac{2+3x}{2+3x}$ reduces to $\frac{1}{1}$

5. Originally Posted by masters
The first expression is made up of addends (terms that are added together). You must have a common denominator to combine them.

Your second expression is a product. You do not need a common denominator. You simply multiply numerators and denominators and factor out any common elements in the product or along the way.

$\frac{(2 + 3x)(x-1)}{2 + 3x}=\frac{2+3x}{2+3x} * \frac{x-1}{1}=\not[\frac{2+3x}{2+3x}] * \frac{x-1}{1}= \frac{1}{1} * \frac{x-1}{1}=x -1$

$\frac{2+3x}{2+3x}$ reduces to $\frac{1}{1}$
Look at a numerical example:

$\frac{(2)(3)}{6} \neq \frac{2}{6} * \frac{3}{6} \neq \frac{6}{36} \neq \frac{1}{6}$

$\frac{(2)(3)}{6}=\frac{2}{6} * \frac{3}{1} = \frac{6}{6} = 1$
$\frac{ab}{c}=\frac{a}{c}*\frac{b}{1}$
$\frac{ab}{c}=\frac{a}{c} * \frac{b}{c} = \frac{ab}{c^2} \neq \frac{ab}{c}$