(x-y)/y = z Trying to solve for y But I can't remember how to do it!!! Help!
B BVC New member Joined Jan 28, 2011 Messages 2 Jan 28, 2011 #1 (x-y)/y = z Trying to solve for y But I can't remember how to do it!!! Help!
D Deleted member 4993 Guest Jan 28, 2011 #2 BVC said: (x-y)/y = z Trying to solve for y But I can't remember how to do it!!! Help! Click to expand... Okay - this one I'll do: One way of doing this: \(\displaystyle \frac{x-y}{y} \ = \ z\) \(\displaystyle \frac{x}{y} - \frac{y}{y}\ = \ z\) \(\displaystyle \frac{x}{y} - 1 \ = \ z\) \(\displaystyle \frac{x}{y} \ = \ z \ + \ 1\) .... multiply both sides by y \(\displaystyle x \ = \ y \cdot (z \ + \ 1)\) ...... divide both sides by (z+1) assuming (z+1) is not equal to 0 \(\displaystyle y \ = \ \frac{x}{z + 1} \\)
BVC said: (x-y)/y = z Trying to solve for y But I can't remember how to do it!!! Help! Click to expand... Okay - this one I'll do: One way of doing this: \(\displaystyle \frac{x-y}{y} \ = \ z\) \(\displaystyle \frac{x}{y} - \frac{y}{y}\ = \ z\) \(\displaystyle \frac{x}{y} - 1 \ = \ z\) \(\displaystyle \frac{x}{y} \ = \ z \ + \ 1\) .... multiply both sides by y \(\displaystyle x \ = \ y \cdot (z \ + \ 1)\) ...... divide both sides by (z+1) assuming (z+1) is not equal to 0 \(\displaystyle y \ = \ \frac{x}{z + 1} \\)
B BVC New member Joined Jan 28, 2011 Messages 2 Jan 28, 2011 #3 Thank you very much - I really appreciate your fast response !
