Math Help - Equation w/ diff. denominators. Help!

1. Equation w/ diff. denominators. Help!

I have this equation that I have trouble comprehending the answer to it. It's been a while since I took Algebra II and this is practice in Pre Cal book. I want to be able to find out about the Lowest Common Denominator. Thanks for your help.
5 / 2x+3 + 4 / 2x-3 = 14x+3 / 4x^2-9

2. Originally Posted by lantern4christ
I have this equation that I have trouble comprehending the answer to it. It's been a while since I took Algebra II and this is practice in Pre Cal book. I want to be able to find out about the Lowest Common Denominator. Thanks for your help.
5 / 2x+3 + 4 / 2x-3 = 14x+3 / 4x^2-9
Hint:

$\frac{a}{b} + \frac{c}{d} = \frac{ad}{bd} + \frac{bc}{bd} = \frac{ad + bc}{bd}$.

In other words, multiply the top and bottom of the first fraction by the bottom of the second, and multiply the top and bottom of the second fraction by the bottom of the first.

$\frac{5}{2x + 3} + \frac{4}{2x - 3} = \frac{5}{2x + 3}\cdot\frac{2x - 3}{2x - 3} + \frac{4}{2x - 3}\cdot\frac{2x + 3}{2x + 3}$

$= \frac{5(2x - 3)}{(2x + 3)(2x - 3)} + \frac{4(2x + 3)}{(2x + 3)(2x - 3)}$

$= \frac{10x - 15}{4x^2 - 9} + \frac{8x + 12}{4x^2 - 9}$

$= \frac{18x - 3}{4x^2 - 9}$.

To solve your original equation, you would have...

$\frac{18x - 3}{4x^2 - 9} = \frac{14x + 3}{4x^2 - 9}$

$18x - 3 = 14x + 3$

$4x = 6$

$x = \frac{3}{2}$.

3. Originally Posted by lantern4christ
I have this equation that I have trouble comprehending the answer to it. It's been a while since I took Algebra II and this is practice in Pre Cal book. I want to be able to find out about the Lowest Common Denominator. Thanks for your help.
5 / 2x+3 + 4 / 2x-3 = 14x+3 / 4x^2-9
Please use parentheses! What you really mean is 5/(2x+3)+ 4/(2x-3)= 3/(4x^2- 9).

You can get rid of the fractions completely by multiplying each fraction by the lest common denominator. What is that least common denominator? The denominators are 2x+ 3, 2x- 3, and $4x^2- 9= ( ?)( ?)$.