Solving rational equations is one of those weird #Math topics that #algebra students often struggle with. It builds on the concept of simplifying rational expressions that we talked about before in my previous post. The only difference is that we now have multiple rational expressions with an equal sign and a variable that we need to solve for (sounds like fun, I know, but it's actually easier than simplifying rational expressions!) They look something like this
We need to find the LCD (least common denominator) of the fractions, and then it becomes nothing more than a linear equation we can solve. Let’s try the above example for starters:
Notice the 2 is missing an x in its denominator. Let’s fix that:
Once all the denominators are the same, we can basically pretend they (the denominators) are not there and we are just left with the numerators. I’ll show you algebraically why that works, of course:
Multiply both sides by x (the common denominator):
after distributing the x, we get
If you look at each fraction, we can cancel an x from the top and bottom:
Now, if we rewrite our equation, we get:
This is exactly what the numerators of our original fraction were after we rewrote the fractions with the LCD! But now you know why we can just write the numerators alone as an equation when the denominators are all equal (we’ll do more examples so don’t worry if you’re not comfortable.)
Just to finish up:
If you plug in x=1 into the original equation, we will get 3=3 (try it out), which is true and tells us that x=1 is in fact a solution. If we get a false statement, like 1=2, then we know that particular x value is not a solution.
Let’s try another example:
If you notice, our LCD is (x-1)x and there’s two methods of solving it: 1) figure out what each fraction is missing from its denominator that would make it equal to the LCD, and then multiply top and bottom by whatever it’s missing. Let’s see how that would look:
If you notice, each denominator now matches our LCD, and we can go ahead and simplify the numerators:
Our denominators are all the same, so we can make them disappear! Then let’s just right the numerators as an equation:
At the beginning, we said x could not be 1, so our only solution is x=2.
There’s another method where you just multiply each fraction by the LCD and that is usually a cleaner way of going about things (of course, there are exceptions.) Let’s try the previous example using this method.
Our LCD is (x-1)x so let’s multiply each fraction by that:
I know this looks UGLY, but let’s see what we can cancel out from each fraction:
After cancelling out what we can, we’re left with:
(look familiar? 😉) This is exactly what we ended up with from the previous method, and we would get the same answer of x=2 if we continued solving.
Let’s do one more and try this one completely on your own until you get an answer, and then check to see how you did. Good luck! 🧐
When you’ve got your answer scroll down a bit
The LCD is (x-2)(x+7) so let’s multiply our entire equation by the LCD:
Let’s cancel out what we can from each fraction:
We’re left with the equation, which we’ll solve step-by-step:
BUT, if you look at our original equation, these are the two x values that would give us a denominator of 0, so neither of them are solutions (they’re called extraneous solutions.) Our final (annoying) answer is no solutions!