specialj96
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- Joined
- Sep 5, 2010
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Problem is 2/A - 3/Y = 1/B solve for Y
specialj96 said:Problem is 2/A - 3/Y = 1/B solve for Y
\(\displaystyle \frac{2}{A} - \frac{3}{Y} \:=\:\frac{1}{B} \qquad\text{ Solve for }Y.\)
soroban said:Hello, specialj96!
\(\displaystyle \frac{2}{A} - \frac{3}{Y} \:=\:\frac{1}{B} \qquad\text{ Solve for }Y.\)
Evidently, no one hates fractions as much as I do. ... Denis does .... ask him (he is the one sitting in the corner for being "wise guy" - trying send Matt to the corner...)
I always eliminate them immediately if possible.
\(\displaystyle \text{We have: }\;\frac{2}{A} - \frac{3}{Y} \:=\:\frac{1}{B}\)
\(\displaystyle \text{Multiply through by }ABY:\;\;\rlap{\:/}ABY\left(\frac{2}{\rlap{\:/}A}\right) - AB\rlap{/}Y\left(\frac{3}{\rlap{/}Y}\right) \:=\:A\rlap{\:/}BY\left(\frac{1}{\rlap{\:/}B}\right)\)
. . . . . . . . . .\(\displaystyle \text{and we have: }\qquad 2BY\quad -\quad 3AB \qquad=\qquad AY\)
\(\displaystyle \text{Re-arrange terms: }\;2BY - AY \;=\;3AB\)
. . . . . . . \(\displaystyle \text{Factor: }\;(2B - A)Y \;=\;3AB\)
\(\displaystyle \text{Therefore: }\;Y \;=\;\frac{3AB}{2B-A}\)