A rational expression does not start with an "equals" sign. It is not a full equation. A rational equation does start out as an equation, complete with "equals" sign.
If you are faced with a single rational expression (polynomial in the numerator and polynomial in the denominator) and you want to simplify it, factor everything and cancel factors that are in both the numerator and denominator. (Once you start looking at graphs of rational functions, those "canceled" factors are what lead to holes in the graph.)
For instance: Say you started with (x+2)/(x2+5x+6). Factor the denominator. Now the rational expression looks like (x+2)/[(x+2)(x+3)]. The (x+2) factor "cancels" and you are left with 1/(x+3) which is in simplest terms.
If you are faced with the sum or difference of two rational expressions, find the least common denominator. Multiply each term by "1" to make the denominators the same, combine like terms in the numerator, and simplify.
For instance: Say you started with 1/(x2+5x+6) -1/(x2+7x+10). In order to find the least common denominator, factor both denominators. Now you have 1/[(x+2)(x+3)] -1/[(x+5)(x+2)], and the least common denominator is (x+2)(x+3)(x+5) (Do you see why?). The first term is missing the (x+5) factor in the denominator, so multiply it by (x+5)/(x+5) (That's what I mean by multiply by "1"). Multiply the second term by (x+3)/(x+3). Now you have something that looks more complicated, but the denominators are the same, so we'll put it all together over a single denominator.
1(x+5) - 1(x+3)
Now simplify the numerator, and you get
Now the (x+2) factor cancels, and you get 1/[(x+3)(x+5)], which cannot be simplified further.
When dealing with rational expressions, your "answer" will still probably be a rational expression, just a simpler one. When dealing with rational equations, your answer will be of the form x=____.
Say you started with (x+2)/(x+3) = 2(x+2)/(x+5). This is a rational equation, and nothing can be factored further at this point. The least common denominator is (x+3)(x+5). You can multiply both sides of the equation by the common denominator, and when you do, factors will (for many problems) cancel.
(x+3)(x+5) [(x+2)/(x+3)] = [2(x+2)/(x+5)] (x+3)(x+5)
That leaves (x+5)(x+2) = 2(x+2)(x+3). We have left rational expressions and are back to friendly polynomials. Distribute
and combine like terms. Since it's a quadratic now, move everything to one side.
To solve this quadratic, factor it:
which has solutions x=-2 and x=-1. Always double check that the solutions work. When checking rational functions, make sure they don't create a zero in one of the original denominators. (When checking solutions to radical functions, completely check each solution you find. At some step, you probably squared a variable expression, canceling out a potential negative.)
Note that when solving a rational equation, every step of the process is an equation. We start with a balance, and it's still a balance at each step. The only point at which that mental image gets tricky is the last step where we figured out what x could be. That step just answers the question "What values of x make this equation a true statement?"
Also note that clearly rational equations can have more than a single solution.
I'm not sure exactly what 5 steps you are referring to in your question. The concept is: escape ratio land by multiplying both sides of the equation by the common denominator between all rational expressions in the equation. Then proceed to solve using whatever tools you have at your disposal. (Most likely, you are now dealing with a quadratic or linear equation.) Check your solutions.