Okay we're doing Rational Equations in my Algebra class. And I know that you have to find a common denominator but I'm not sure what to do after that. Here's an example equation:
When given expressions like these, you can take the multiple of the denominators as the common denominator. so the common denominator for 3/(x + 1) and 1/(x - 2) is (x + 1)(x - 2)
Here's how to do the problem. There are two mainstream methods, the first is the one you are interested in.
3/(x + 1) - 1/(x - 2) = 1/(x^2 - x - 2)
3(x - 2) - 1(x + 1)................1
-------------------- = ---------------
.....(x + 1)(x - 2)..........(x^2 - x - 2)
=> ........3x - 6 - x - 1..................1
........-------------------- = ---------------
.............(x + 1)(x - 2)..........(x^2 - x - 2)
=> ............2x - 7.......................1
........-------------------- = ---------------
.............(x + 1)(x - 2)..........(x^2 - x - 2)
notice that the denominators are the same, so the numerators must be the same.
so 2x - 7 = 1
=> 2x = 1 + 7
=> x = 4
The alternate method is this:
notice that 1/(x^2 - x - 2) = 1/(x + 1)(x - 2) .............so the bottom can factor into two expressions, each of which are the denominator of a fraction on the other side of the equal sign. We can multiply by the denominator (x + 1)(x - 2)
so we get:
(x + 1)(x - 2)[3/(x + 1) - 1/(x - 2)] = (x + 1)(x - 2)[1/(x^2 - x - 2)]
=> 3(x - 2) - 1(x + 1) = 1
=> 3x - 6 - x -1 = 1
=> 2x - 7 = 1 ........................notice this is the same equation we ended up solving in the previous method
=> x = 4
In terms of easiness, i'd say both are equal, however, I would suggest the second one, it is shorter and quicker to do, and hence a better method if you are taking a test. But if a question (or your professor) specifically mentions "common denominator", then you must do the first method.
I dont suspect either method would be difficult for you to learn however. if you get both of them, then whatever you are required to do you can do it. Do you understand both methods?
Notice that in the first method 3(x-2) - 1(x+1) was the numerator of the combined fraction. essentially the second method wipes out all denominators so you don't have to work with fractions anymore but just a linear equation. Here's what happens:
we have 3/(x + 1) - 1/(x - 2) = 1/(x^2 - x - 2), and we also know (x^2 - x - 2) = (x + 1)(x - 2), so we can write the equation as:
3/(x + 1) - 1/(x - 2) = 1/(x + 1)(x - 2)................got that part, right?
okay, now we notice that the denominator on the right has factors that appear in the denominators on the left. So we go: ok, if i can wipe out the denominator on the right, it will also wipe out those on the left, so we multiply through by the denominator on the right:
(x + 1)(x - 2)[ 3/(x + 1) - 1/(x - 2)] = (x + 1)(x - 2)[1/(x + 1)(x - 2)] .....still with me?
so expanding the brackets we get:
(x + 1)(x - 2)(3/(x + 1)) - (x + 1)(x - 2)(1/(x - 2)) = (x + 1)(x - 2) * 1/(x + 1)(x - 2)
We immediately see that the denominator on the right cancels with what we multiply by.
We also see that the (x + 1) of what we multiplied by cancels the (x + 1) in the denominator of 3/(x+1). so we are left with (x - 2)*3. similarly, the (x - 2) cancels the denominator of -1/(x - 2), so we are left with (x + 1)* (-1)
so the equation becomes:
3(x - 2) - (x + 1) = 1
and we just solve for x from there.
got it?
yes, that is how the equation is set up. then you know you have to multiply each term in the bracket by what's on the outside. so the (x + 1)'s cancel in the first term and the (x - 2)'s cancel in the second term, and we are left with 3(x - 2) and -(x + 1).
the denominator of the second fraction in brackets is x - 2