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**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:

a1

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

could be turned into this:

a2

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

to then cancel the denominators all across and get this:

a3

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

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:**-------

b1

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

is transformed into:

b2

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

cancels out to:

b3

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

**-- Believed Right:**---------

c1

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

1st, make sure different different variables are transfered on different sides BEFORE canceling out method :

c2

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

c3

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

c4

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

*(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?

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