# Thread: Solving equations/fractions/common denominator problem

1. ## Solving equations/fractions/common denominator problem

For a sample word problem I'm looking at, after you determine the equation, you have to solve it. This is the first step... finding the common denominator

[IMG]file:///C:/Users/
I'm okay with it past that step, the answer then is 30(x+1) - 30x = x^2 + x which I have no clue how to get...

After that, it's just 30x + 30 - 30x = x^2 + x, then 30 = x^2 + x, 0 = x^2 + x -30, 0 = (x+6)(x-5), x+6=0 and x-5=0, x=-6, x=5 which is okay but the first part just confuses me so much, I have no clue how any of that works - I know obviously what a common denominator is but I have no clue what happened to any number on that fraction to make it turn into what it became... unfortunately since this is the first step on several problems (including problems that end up being in the ax^2+bx+c quadratic form), and I understand the parts after it (simple solving or finding the roots from the quadratic equation) but this I have no idea how to do...

2. so the problem you're having is determining the lowest common denominator? Or is it with multiplying everything out?

Assuming the former, basically you have 3 fractions there, each with a different denominator. The easiest way to get a common denominator is to go through each fraction and multiply it by the denominators of the other two fractions. for instance for
$\frac{0.5}{x}$ you multiple it by $\frac{x+1}{x+1}$ and $\frac{60}{60}$

For the second fraction, you multiple it by $\frac{x}{x}$ and $\frac{60}{60}$

so, for the left hand side, you end up with
$\frac{(60)(0.5)(x+1)}{(60)(x)(x+1)} - \frac{(60)(0.5)x}{(60)(x)(x+1)}$

from here you should probably be able to see how the left side of the image you have above was created. The same process applies to the right side of the equation. I hope this was of help.