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Solving Rational Equations with Unlike Denominators - Problem 3

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 equation for x, there’s a couple of different strategies you could use. First I’m going to show you an incorrect strategy because this happens a lot in my Math class and it drives me crazy. A lot of times student recognize that they need to have a common denominator and these denominators are pretty close, so this is an incorrect thing to do. Students who just say, "Oh I’m just going to add x here. Now my denominators look the same." You guys you have to be really careful with that we can’t just go around adding stuff to the top and bottom of a fraction. We can multiply the top and bottom by something and that doesn’t change this mathematical value.

Like for example if I were to multiply this by x over x which wouldn’t be useful in my problem, but if I did multiply by x over x, that’s okay because I’m really just multiplying by 1. I can’t just go adding x to the top and bottom, that’s mathematically inaccurate. So if I wanted to solve this problem correctly, one method I could use would be to find the common denominator of 3 times the quantity x plus 3. I could go through and multiply this fraction by 3 over 3, multiply this fraction by x plus 3 over x plus 3 and proceed from there. I’m going to show you a different way. Either way is fine, now both work out correctly.

What I’m going to do is cross multiply because for me that’s a little bit easier than trying to do with the common denominator. You can cross multiply only if you have two equal fractions. So in order to get two equal fractions, I’m going to add this bad boy to both sides of the equals x plus 2 over 3 is going to be equal to positive now x plus 7 over x plus 3.

Okay cross multiplying means you write an equality of the product of the diagonals so I’ll have x plus 2 times x plus 3 is going to be equal to the product of 3 times x plus 7 and there is just a straight forward solving. It’s going to be a little tricky because I have a quadratic over here meaning an x² term, but it’s still the same process that you guys have been working on for a long time. Set that thing equal to 0, and then you can either factor, complete the square, use the quadratic formula graph it however you want.

So I’m going to have x² plus 2x and then minus 15 is equal 0. I like factoring especially when the leading coefficient is 1, 2 numbers that multiply to -15 and add to +2. That will be +5 and -3 set each quantity equal to 0 and I’ll get x is equal to 3 and x is equal to -5. I think those are my solutions, I’m going to go back and make sure neither one of them would be an excluded value.

If I put 3 into the denominator here, it’s still not making my denominator 0 so that’s okay and if I put -5 in there, y denominator isn’t 0 so that’s okay. That tells me these are my two solutions. If I wanted to check my work what I would do is go back and substitute 3 into both fractions and make sure their difference is 0 then I would substitute -5 into both fractions, and make sure their difference is 0. I’ll leave that one for you guys to try on your own.

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