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Solving Rational Equations with Unlike Denominators - Problem 2 5,013 views

Teacher/Instructor 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

In order to solve this problem, one thing you could do would be combine like terms. Like I could subtract 6 over x minus 2 from both sides so that all my x minus 2s would be on the left, all my x minus 2s denominators would be on the left side of my equals. That’s kind of tricky though. I want to show you a shortcut.

If all of my terms have the exact same denominator then I can like clear the denominators, which means just erasing all of them. Here is what I mean.

I’m going to multiply this by x minus 2 over x minus 2. That’s allowed, I’m not changing the value of 5, I’m just multiplying by 1 over 1. Then what happens is that I have 3 fractions that have all x minus 2 in the denominators, so mathematically what I’m going to do is multiply everything by x minus 2.

What that means in terms of how I’m going to solve this problem is that I don’t have to look at my denominators anymore. All I’m going to look at is the tops. 3x is equal to 5 times x minus 2 plus 6. Now I’m a happy camper. That’s an easy solving problem. Go through and distribute my 5, combine like terms I’m going to subtract 2x or 5x from both sides so I’ll have -2x is equal to -4, x is equal to 2. That’s going to be my answer, but before I decide I’m done I need to check and make sure it’s not an excluded value.

If I substitute 2 into this fraction, I have bad news. If I put 2 in the denominator therefore x I’ll have 2 take away 2 which is 0, shoot! That means this is an excluded value and there are in fact no solutions to this problem. There is no number that when I substitute it in for x it makes this a true equality. That’s weird but sometimes it happens in Math. Not every problem has a solution. It’s kind of tricky.

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