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Solving Rational Equations with Unlike Denominators  Problem 4
Alissa Fong
Alissa Fong
MA, Stanford University
Teaching in the San Francisco Bay Area
Alissa is currently a teacher in the San Francisco Bay Area and Brightstorm users love her clear, concise explanations of tough concepts
When trying to solve this rational equation, a lot of students look at this left hand side and they think about wanting to cross multiply. Cross multiply is possible if you have two equal fractions, so you’d have to take this left hand side and combine it into one fraction in order to cross multiply. Well that’s a good strategy except it doesn’t really look good in this problem because these denominators are really, really different. Let me show you a different method that would be a little easier.
Looking at this third fraction, I’m checking out the denominator here and thinking about what it would look like if it were factored. There’s a lot of guessing and checking that goes into factoring, so I’ll tell you guys I tried this on my own before we started this video tape and it factors like these 3x plus 2 times 2x take away 3. Does that ring bells in your head at all?
Check it out you guys, that’s our two bottoms from the left hand side that’s good. That means that really easily I can get three fractions that all have the same denominators and then I could cancel out those denominators. Mathematically what I would do is multiply all three fractions by this product so that the denominators were eliminated.
Let me show you what that means. So I’m going to get all three fractions to have this denominator. This guy is all set. This first fraction has the 3x plus 2 piece. It still needs the 2x minus 3, so I’m going to multiply top and bottom by 2x minus 3. This second fraction has the 2x minus 3 business, but it needs 3x plus 2 so I’m going to multiply top and bottom by 3x plus 2. Be really careful with parentheses.
Now I have 3 fractions that all have the same denominators, which means I can get rid of the denominators. I don’t even have to deal with them anymore. I’m going to rewrite this problem only looking at the tops or the numerators. My first fraction the numerator looks like x times 2x take away 3, my second fraction tells me x plus 2 times 3x plus 2 and my last fraction all I need is the 4x piece, just the numerators. From here on out it’s just combining like terms and Foiling then eventually we’ll solve for x and check for excluded values.
Distributing here I’ll have 2x² take away 3x. I’m going to FOIL a little bit in my head you’ve got to trust me on this one, outers I’ll have 2x inners I’ll have 6 more x so altogether that’s 8x and it’s equal to 4x's. I’m still just simplifying the left hand side. Combining like terms I’ll have, oh sorry guys I forgot a 3 right there my fault. My first should be 3x² it's okay it will still work. Combining like terms over here I’m going to have 5x² plus 5x plus 4 is equal to 4x.
Next thing I’m going to do is subtract 4X's from both sides so that this trinomial side equal to 0. 5x² plus 9x plus 4 equals 0. Now I’m ready to solve for x. I can factor, use the quadratic formula, I could graph it, complete the square, those are all options for me. I like factoring so I’m going to try to find 2 numbers that when I plug them in, I get my middle term of 9x. My number needs to multiply to 4, so I’m going try +1 and +4. When I FOIL I’ll have 5x plus 4 more good that’s 9 Xs.
Okay so there is my factored form. In order to solve for x I said each one of those products equal to 0 using the zero product property. Here I’ll have x is equal to 4/5 that’s one of my answers I think. My other answer is going to come from solving this guy for x, x is equal to 1.
Okay, once I think I have the answers I might be wrong. I might have done all the right Math here, but this might not be both solutions and the way I check is by making sure they’re not excluded values. An excluded value is any x value that would make one of these denominators equal to 0. So I’m not going to do all the Math, I’m just going to check in my head and make sure they’re not 0 like if I multiply this by 4/5, it’s not going to add up to 0 when I add 2. Same thing here if I stick 4/5 there and subtract 3, I’m not going to get 0. Same thing here 4/5 I’m not in danger of having my differences add up to 0.
So that guy I think is an answer, let’s check 1. If I put 1 in there it’s not going to be 0 good. 1 in there not 1 in there also not 0. That’s how I check and make sure neither one of these are excluded values.
Guys these problems are really tricky, there’s tons of places to make errors, so it’s a good idea once you do your homework to look in the back of the book. Check your answers it’s not cheating, it’s a good way to make sure you’re doing these processes correctly. If you got them incorrect, my guess is you just lost some negative sign somewhere or something. Write out all of your work like I did with all of this stuff. Use lots of space in your paper and that will help your teacher or your tutor identify where your misconceptions are.
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Alissa Fong
M.A. in Secondary Mathematics, Stanford University
B.S., Stanford University
Alissa has a quirky sense of humor and a relatable personality that make it easy for students to pay attention and understand the material. She has all the math tips and tricks students are looking for.
Your tutorials are good and you have a personality as well. I hope you have more advanced college level stuff, because I like the way you teach.”
Thanks alot for such great lectures... I never found learning this easier ever before... keep up the great work.... :)”
You seem so kind, it's awesome. Easier to learn from people who seem to be rooting for ya!' thanks”
Sample Problems (4)
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Solving Rational Equations with Unlike Denominators
Problem 1 8,313 viewsSolve:
x + 3 = x + 2 x + 1 x − 2 
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Problem 2 5,636 viewsSolve:
3x = 5 + 6 x − 2 x − 2 
Solving Rational Equations with Unlike Denominators
Problem 3 5,609 viewsSolve:
x + 2 − x + 7 = 0 3 x + 3 
Solving Rational Equations with Unlike Denominators
Problem 4 4,711 viewsSolve:
x + x + 2 = 4x 3x + 2 2x − 3 6x² − 5x − 6
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