# Solving Rational Equations with Unlike Denominators - Problem 1

###### Transcript

Here I’m asked to solve for x and it’s not too bad because I have two equal fractions. I could go through and multiply both sides by x plus 1 excuse me multiply this side my x plus 1 over x plus 1, multiply that side by x minus 2, or x minus 2 so I’d have the same denominators, but what would be a little bit quicker is to do what we call cross multiplying. That’s where you write the diagonals as products and set those 2 products equal to each other like x plus 1 times x plus 2 that’s my first diagonal is equal to x plus 3 times x minus 2.

Make sure you set these 2 products equal. Sometimes I see students that try to write this as another fraction or something, the whole point of cross multiplying was so that we don’t have any fractions left. Then go through and FOIL on both sides I’m kind of doing this shortcut style, so you have to trust me that these are the foiled out trinomials then what I want to do is solve for x. So if I go through and subtract x² from both sides, those would be gone so I’m left with 3x pus 2 equals x take away 6 subtract x from both sides and then subtract 2 from both sides. I’m just solving so I get answer x equals -4.

I think that’s correct. Before I move on I want to make sure first of all that -4 isn’t an excluded value. It would be an excluded value if -4 for x would make one of my denominators equal to 0. That’s not a problem here.

The last thing I’m going to do is check my solution. To check I’m going to go back and see if when I substitute in -4 for both these fractions I get equal fractions. -4 plus 3 on top will be -1 -4 plus 1 would be -3. Let’s see if that’s equal to here. -4 plus 2 would be -2. -4 take away 2 is -6 and indeed those are equal fractions so I know I did the work correctly.

So guys I wanted to show you this problem because I think it’s a great example of how to cross multiply things. In this case we were lucky because our x² terms were eliminated and from there it’s just basic solving.