College Intermidate Algebra Problem: Solve for X

Elvenwarrior2001

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Sep 9, 2011
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I'll begin with the problem, mention what steps I've taken, and then summarize.

Solve equation:
(x+1)/(2x2+11x-6) - (x)/(2x2-7x+3) = (1)/(x2+3x-18)

Since there are quadratic equations in the denominators I figured I'd factor them first. I came up with:

(x+1)/(x+6)(2x-1) - (x)/(2x-1)(x-3) = (1)/(x+6)(x-3)

From there I decided eliminated the denominator by multiplying each term by the LCD: (x+6)(2x-1)(x-3)

After eliminating the denominator I added like terms and came up with the following quadratic equation: x2-5x-1

x2-5x-1 can not be factored or simplified. Given that the answer is supposed to be a whole number or a simple fraction I figured this was obviously not the answer. Granted I could theoretically say the answer is 5 +/- the squareroot of 21 all over 2, but this is not an acceptable answer.

Here's where I'd like to mention that this is for online homework, i.e. I enter my answer into the required field in the form they want me to. For these types of problems they want it as a whole number or simple fraction. In fact, unlike most problems, there's note even an option to use the squareroot sign, or the plus or minus sign for that matter. I plugged in "NO SOLUTION" which is what you put if there's, of course, no solution; it told me I was wrong.

Is there something I'm doing wrong? This is supposed to come out as a whole number or simple fraction like I said, and I've run the numbers 10 or 15 times, and I even took the liberty of running it through a computer program just for kicks and the computer said "No solution," which I found to be hilarious.

Ideas?
 
eliminated the denominator by multiplying each term by the LCD: (x+6)(2x-1)(x-3)

Multiply each ratio by the LCD, not each term.


came up with the following quadratic equation: x2-5x-1

This is not an equation; equations always contain an equal sign.

After you multiply each ratio by the LCD and cancel the factors in all of the denominators, you should arrive at the last equation appearing in Denis' post above.
 
Thanks a lot! It makes sense now. I spent so long making the same mistake again. Thanks helping me get beyond that step! :)
 
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