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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M.A. in Secondary Mathematics, Stanford University B.S., Stanford University
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