Simple algebra question, solve for the variable

May 2010
9
0
I'm a little rusty here (or possibly losing my mind at this point), but actually this question involves a bigger problem (substitution rule/calculus) but I'm trying to understand a simple algebra concept first. See the following:

y = 2;
x = 10;

x/y = 5

Ok. I want to solve for y now. Exactly how is this done in algebra? What do you divide or multiply both sides by? I know the answer is y = 1/5x, but I'm unsure as to how one would get to that.

The only way I could see it is to solve for x first:

x = 5y, and then divide 5 on both sides: x/5 = y. There must be an easier and non redundant way, correct?
 
Nov 2009
927
260
Wellington
Uhm, you just said it, y = 2 :D

But to solve \(\displaystyle \frac{x}{y} = 5\), you can rearrange as follows :

>> For \(\displaystyle x\) :

\(\displaystyle \frac{x}{y} = 5\)

\(\displaystyle \frac{x}{y} \times y = 5 \times y\)

\(\displaystyle \frac{x \times y}{y} = 5 \times y\)

\(\displaystyle \frac{x}{1} = 5 \times y\) (cancel out the \(\displaystyle y\))

\(\displaystyle x = 5 \times y\)

\(\displaystyle x = 5y\)

>> For \(\displaystyle y\) :

\(\displaystyle \frac{x}{y} = 5\)

\(\displaystyle \frac{y}{x} = \frac{1}{5}\) (inverse on both sides)

\(\displaystyle \frac{y}{x} \times x = \frac{1}{5} \times x\)

\(\displaystyle \frac{xy}{x} = \frac{x}{5}\)

\(\displaystyle \frac{y}{1} = \frac{x}{5}\) (cancelling out)

\(\displaystyle y = \frac{x}{5}\)

\(\displaystyle y = \frac{1}{5} x\)

This is how I would do it (the long way of course, in real situations I would spare most steps). Does it make sense ?
 
  • Like
Reactions: dwsmith

dwsmith

MHF Hall of Honor
Mar 2010
3,093
582
Florida
Uhm, you just said it, y = 2 :D

But to solve \(\displaystyle \frac{x}{y} = 5\), you can rearrange as follows :

>> For \(\displaystyle x\) :

\(\displaystyle \frac{x}{y} = 5\)

\(\displaystyle \frac{x}{y} \times y = 5 \times y\)

\(\displaystyle \frac{x \times y}{y} = 5 \times y\)

\(\displaystyle \frac{x}{1} = 5 \times y\) (cancel out the \(\displaystyle y\))

\(\displaystyle x = 5 \times y\)

\(\displaystyle x = 5y\)

>> For \(\displaystyle y\) :

\(\displaystyle \frac{x}{y} = 5\)

\(\displaystyle \frac{y}{x} = \frac{1}{5}\) (inverse on both sides)

\(\displaystyle \frac{y}{x} \times x = \frac{1}{5} \times x\)

\(\displaystyle \frac{xy}{x} = \frac{x}{5}\)

\(\displaystyle \frac{y}{1} = \frac{x}{5}\) (cancelling out)

\(\displaystyle y = \frac{x}{5}\)

\(\displaystyle y = \frac{1}{5} x\)

This is how I would do it (the long way of course, in real situations I would spare most steps). Does it make sense ?
You went all out.
 
  • Like
Reactions: Bacterius