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! $by-d=ay+c,\quad\text{solve for }y.$
Solving for the Indicated Variable
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algebra-precalculus
2 Answers
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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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0@Liz, if you're already happy with Ross's answer, you might want to click the green checkmark on the left of his answer. – 2011-09-02
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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 $a\neq b$.