# Moving variables around

Sorry for the easy question in advance, I'm trying to remember my algebra and I've come across a problem I don't quite remember the answer to:

I've gotten the original problem (trying to get the $y$ to one side): $$7\left(\frac{-9-2y}{y}\right) + 3y = -13$$ to this: $$-\frac{14y}{5} + 3y = -13 + \frac{63}{5}$$ by simply distributing the 7.

But now I'm stuck because I don't remember how to "split apart" the $y$ from $-14y/5$. Mainly it's the /5 that confuses me, I don't remember how to deal with (i.e. what the rules are) the dividing when "splitting apart" the numbers from the variable.

Thanks bunches!

• try multiplying both sides of the equation by 5. Jun 14 '11 at 4:38
• Don't apologize for an easy question, someone will find it useful and be glad that you asked it. Jun 14 '11 at 6:47

Multiply by 5.

Example:

$\frac{2}{5} \times 5 = 2$

Or more germanely,

$-\frac{14y}{5} + 3y = -13 + \frac{63}{5}$

$-14y + 15 y = -65 + 63$

$y = -2$

You don't want to "split it apart", you want to do the operation: take $3y$, and multiply it by $1$. But not just any $1$, multiply it by a well-disguised $1$, namely, $\frac{5}{5}$, so that you have two fractions with the same denominator: looking at just the left hand side of what you have, $$-\frac{14y}{5} + 3y = -\frac{14y}{5} + 3y\left(\frac{5}{5}\right) = -\frac{14y}{5} + \frac{15y}{5},$$ and now do the operation: $$-\frac{14y}{5} + \frac{15y}{5} = \frac{-14y+15y}{5} = \frac{y}{5}.$$ So instead of $$-\frac{14y}{5}+3y = -13+\frac{63}{5},$$ you can write $$\frac{y}{5} = -13+\frac{63}{5}.$$ I would suggest doing the same thing on the right side (write it as a single fraction with denominator $5$), and then multiply both sides of the equation by $5$ to cancel the denominator.

In general, if we have three numbers $a$, $b$, and $c$, then the quantity $$\frac{ab}{c}$$ really represents $$a\times b\times \frac{1}{c}.$$ Because we can rearrange and group multiplications however we want (this is commutativity and associativity), we have that $$\frac{ab}{c}=\left(a\times\frac{1}{c}\right)\times b=\left(b\times \frac{1}{c}\right)\times a=\left(a\times b\right)\times\frac{1}{c}$$ In our case, we have $a=-14$, $b=y$, and $c=5$. To separate out the $y$ from $\frac{-14y}{5}$, we choose the form above that has the $b$ on the outside, and get $$\frac{-14y}{5}=\left((-14)\times\frac{1}{5}\right)\times y=\frac{-14}{5}\times y.$$ Now you can use the distributive law, which says that for any $d$, $e$, and $f$, $$(d+e)\times f= (d\times f)+(e\times f)$$ to "group" the left side of the equation: $$\frac{-14y}{5}+3y=\left(\frac{-14}{5}\times y\right)+(3\times y)=\left(\frac{-14}{5}+3\right)\times y=\left(\frac{-14}{5}+\frac{15}{5}\right)\times y=\left(\frac{1}{5}\right)\times y=\frac{y}{5}.$$

Your specific question has been answered, so I will do the problem in what may be an easier way that minimizes manipulation of fractions. (The term "fraction" comes from broken (number), as in fracture. Broken things want to be made whole.)

First it is helpful to write things in good old fraction notation, maybe like $$7\left( \frac{-9 -2y}{5}\right)+3y=-13.$$ (The "$/$" symbol can interfere with visual processing of the information.)

To me, the natural thing to do now is to immediately multiply through by $5$, in order to get rid of denominators.

When we multiply the first messy term by $5$, this cancels the $5$ in the denominator, that's why we are doing it. So the first term becomes $7(-9-2y)$. The second term becomes $15y$, and on the other side of the equality symbol we get $-65$. So our equation becomes $$7(-9-2y)+15y=-65.$$ Process the first term. We get $$-63 -14y+15y=-65,$$ and we are essentially finished.