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[SOLVED] Canceling fractions when more than 1 variable in an equation


New member
May 31, 2012
EDIT: If you're reading this and are still learning algebra basics, IGNORE this. I made a wrong assumption, thanks to MarkFL for pointing that out!

So far, I was led by my own assumptions to believe that this:

\(\displaystyle \frac{1}{5} + \frac{y}{2} = 7\)

could be turned into this:
\(\displaystyle \frac{2}{10} + \frac{5y}{10} = \frac{70}{10}\)

to then cancel the denominators all across and get this:
\(\displaystyle 2 + 5y = 70\)

This seemed to work fine and did wonders until some of my answers did not match the textbook's answer section where equations had more than one variable on the same side. I just want to know: is this a rule of some sort or are my now corrected assumptions (can cancel out as long as different variables are on opposite side) invalid still? (I'll re-ask at the end of this post in case this post gets across as messy).

I'll demonstrate what I now believe to be wrong then what I believe to be right :

-- Believed Wrong: -------
\(\displaystyle \frac{x}{5} + \frac{y}{2} = 7\)

is transformed into:
\(\displaystyle \frac{2x}{10} + \frac{5y}{10} = \frac{70}{10}\)

cancels out to:
\(\displaystyle 2x + 5y = 70\)

-- Believed Right: ---------
\(\displaystyle \frac{x}{5} + \frac{y}{2} = 7\)

1st, make sure different different variables are transfered on different sides BEFORE canceling out method :
\(\displaystyle \frac{y}{2} = \frac{-x}{5} + 7\)

\(\displaystyle \frac{5y}{10} = \frac{-2x}{10} + \frac{70}{10}\)

\(\displaystyle 5y = -2x + 70\)

(which you'd then do y = other side over 5, etc, I'm only illustrating the canceling out part)

So just to recap, is it right to believe that canceling out using the "same denominator" method requires that no two or more different variables be on the same side of the equation?
Last edited:


Staff member
Feb 24, 2012
Your equations b3 and c4 are equivalent, just arranged differently. They are both correct as well. It doesn't matter if there is more than one variable on one side of an equation, you can always multiply both sides of an equation by a non-zero constant and get an equivalent equation.


New member
May 31, 2012
Your equations b3 and c4 are equivalent, just arranged differently.
Hmm, you're right. Thanks a lot for the reply by the way.

I'll have to re-check my notes tonight (GMT-5 here so 7:28am) and see from the exact equations and not make them coincide like the ones I wrote off the bat above.

I'll leave as unsolved for the moment but will add where I screwed up with the actual data latest tomorrow.