How to do rational expressions

Question

Never was much of a math student but I am brushing up on my arithmetic and algebra for college.

I am using sample questions from Accuplacer and then using video lectures and practice on Khan Academy. It is going well but sometimes the examples in the lecture are very trivial and I am currently doing very poorly with rational expressions in the form:

$$ \frac{1}{x+3} + \frac{1}{x} = 2x +\frac{3}{x(x+3)}$$

or

$$ \frac{u}{x} + \frac{5u}{x} - \frac{u}{5x} = \frac{29u}{5x}$$

The answers that I supplied are from a multiple choice list, so I need to be able to be able to get those answers on my own.

I have looked up the answers to these problems and I do not understand how it works. I am looking for any killer examples that would really help me grasp it.

This video lecture is where I am at when it comes to rational expressions. I can do some work when it is a 1 coefficient and when it is > 1. I lose confidence in my ability to teach myself the above two problems. Thank you.

Answer 1

In an equation you may do anything (except multiply or divide with 0) as long as you do it to both sides: so in this case the denominators can be canceled by multiplying every term with factors that will cancel the denominators: for example in the first expression you multiply EVERY TERM with $x(x+3)$. This will cancel factors in the with terms that have it in the denominator and you will be left with fraction-less expressions:

$$\frac{1}{x}\times x(x+3)+3\times x(x+3)+\frac{1}{x}\times x(x+3)=2x\times x(x+3)+\frac{3}{x(x+3)}\times x(x+3) $$ $$(x+3)+3\times x(x+3)+(x+3)=2x\times x(x+3)+3$$ $$...$$

Since we multiplied with x(x+3) its just important to note that it may never equal 0. So x may not equal 0 or -3 in the end. If it does label that answer as not applicable.

To determine what to multiply with factorize the denominators first and then multiply each term with the lowest common multiples of the denominators.

Answer 2

I'll take your two problems themselves. The key to do what you want to do is multiplying the numerator of fraction 1 by the denominator of fraction 2, and the numerator of fraction 2 by the denominator of fraction 1. Your first problem. $$\dfrac{1}{x+3} + \dfrac{1}{x}$$ Now how do we do this? Well firstly you have to make one common denominator, and no better way to do that then multiplying the two denominators. That means our denominator will be $x(x+3)$. But notice what we are doing here to get this denominator. We are multiplying fraction 1 by $\dfrac{x}{x}$ and fraction two by $\dfrac{x+3}{x+3}$ (because only multiplication by $1$ is allowed, otherwise the problem changes). From here we get: $\dfrac{x}{x(x+3)} + \dfrac{x+3}{x(x+3)}$. And we add these together to get $\dfrac{2x+3}{x(x+3)}$