by Simon
24. August 2010 18:57
Currently in the process of doing prerequisite review for my pre-calculus 7A-7B. Either the review material is more difficult than the stuff I learned in MathPath1, or I've forgotten more than I'd care to admit. Either way, the current problem that has me stumped is as follows:
So, looking at this, my first thought is to multiply both sides by the LCD.
(4t-3)(4)](\frac{1}{t-1} + \frac{t}{4t-3}) = (\frac{1}{4})[(t-1)(4t-3)(4)]$)
This leaves me with the expression:
(4t-3)(4)}{t-1} + \frac{t(t-1)(4t-3)(4)}{4t-3} = \frac{(t-1)(4t-3)(4)}{4}$)
Bits divide out nicely, leaving me with:
(4)+t(t-1)(4)=(t-1)(4t-3) $)
This multiples out and simplifies to:
And that doesn't factor nicely. Now, I could go to completing the square, but according to the computer, this whole problem is meant to be equivalent to a first-degree linear equation. So, people - what have I missed?
Incidentally, I hate fractions.
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Tags:
Fractions
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