# Solving for the Indicated Variable

I am currently enrolled for a precalculus class and I got stuck with this problem. It would be helpful if I got any help; so far I tried doing it but I'm stuck! $$by-d=ay+c,\quad\text{solve for }y.$$

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First, move all the terms with $y$ to one side and all terms without to the other:

$by-ay=c+d$

Factor out the $y$

$(b-a)y=c+d$

Assuming $a \ne b$, divide (typo corrected)

$y=\frac{c+d}{b-a}$

If $a=b$ and $c=-d$, $y$ can be anything ( $0y = 0$ is true for any $y$). However, if $a=b$ and $c \neq -d$, there are no solutions for $y$ as the left side is 0, but the right side is not ($0y \neq 5$, or 3, or anything).

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+1. Since the OP seems to be a beginner, I think it might be a good idea to explain what happens if $a = b$... –  Srivatsan Sep 2 '11 at 1:27
Thanks a lot for your help but i am wondering how is it you got (b-a)y =c+d ? I understand how you had to subtract the ay but by-ay= c+d did you factor it to get (b-a)y=c+d? –  Liz Sep 2 '11 at 1:31
@Liz: That's the distributive law at work. –  Ｊ. Ｍ. Sep 2 '11 at 1:35
@Liz, yes. $by-ay$ can be factored as $(b-a)y$. This is called "distributive law", since the multiplication operation distributes over the subtraction. –  Srivatsan Sep 2 '11 at 1:36
@Srivatsan Narayanan: In your last comment, I think you want $a \ne b$. I said that in the original response. –  Ross Millikan Sep 2 '11 at 4:00

Try gathering the terms with $y$ on the same side, $$by-d=ay+c\implies by-ay=d+c$$ by subtracting $ay$ from both sides and adding $d$ to both sides of the equation. Try using the distributive law to now solve for $y$, assuming $a\neq b$.

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