specialj96
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Problem is 2/A - 3/Y = 1/B solve for Y
A little help
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D
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specialj96 said:
Matt is "almost" correct.
2/A - 3/Y = 1/B
3/Y = 2/A - 1/B
3/Y = (2*B - A)/(A*B) ... one assumption ... (2B-A)<>0
Y/3 = A*B/(2*B - A)
Y = 3*A*B/(2*B - A)
Hello, specialj96!
Evidently, no one hates fractions as much as I do.
I always eliminate them immediately if possible.
\(\displaystyle \text{We have: }\;\frac{2}{A} - \frac{3}{Y} \:=\:\frac{1}{B}\)
\(\displaystyle \text{Multiply through by }ABY:\;\;\rlap{\:/}ABY\left(\frac{2}{\rlap{\:/}A}\right) - AB\rlap{/}Y\left(\frac{3}{\rlap{/}Y}\right) \:=\:A\rlap{\:/}BY\left(\frac{1}{\rlap{\:/}B}\right)\)
. . . . . . . . . .\(\displaystyle \text{and we have: }\qquad 2BY\quad -\quad 3AB \qquad=\qquad AY\)
\(\displaystyle \text{Re-arrange terms: }\;2BY - AY \;=\;3AB\)
. . . . . . . \(\displaystyle \text{Factor: }\;(2B - A)Y \;=\;3AB\)
\(\displaystyle \text{Therefore: }\;Y \;=\;\frac{3AB}{2B-A}\)
Evidently, no one hates fractions as much as I do.
I always eliminate them immediately if possible.
\(\displaystyle \text{We have: }\;\frac{2}{A} - \frac{3}{Y} \:=\:\frac{1}{B}\)
\(\displaystyle \text{Multiply through by }ABY:\;\;\rlap{\:/}ABY\left(\frac{2}{\rlap{\:/}A}\right) - AB\rlap{/}Y\left(\frac{3}{\rlap{/}Y}\right) \:=\:A\rlap{\:/}BY\left(\frac{1}{\rlap{\:/}B}\right)\)
. . . . . . . . . .\(\displaystyle \text{and we have: }\qquad 2BY\quad -\quad 3AB \qquad=\qquad AY\)
\(\displaystyle \text{Re-arrange terms: }\;2BY - AY \;=\;3AB\)
. . . . . . . \(\displaystyle \text{Factor: }\;(2B - A)Y \;=\;3AB\)
\(\displaystyle \text{Therefore: }\;Y \;=\;\frac{3AB}{2B-A}\)
D
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soroban said: