Problem with getting variable by itself in fraction.

Question

I have a problem that looks something like this:

The difference of the quotient of a number and $-2$ from $12$ is $15$.

So I started off like this:

$12-\displaystyle\frac{x}{-2}=15$

Then I subtracted the $12$ from both sides to get:

$-\displaystyle\frac{x}{-2}=3$

Where would I go on from here?

score 3 · Accepted Answer · answered Jan 29 '13 at 23:47

3

You can multiply the numerator and denominator by $-1$ to clear the signs, then multiply by $2$ to clear the denominator.

answered Jan 29 '13 at 23:47

Ross Millikan

374,822

amWhy · Answer 2 · 2013-01-30T00:28:07.743

2

$$-\frac{x}{-2} = 3 \;\; \implies \;\;\not - \frac{x}{\not - 2} = 3\; \;\implies\;\;\frac{x}{2} = 3 \;\;\implies\;\; x = 2\cdot 3 = 6$$

Note that $\;-\dfrac{x}{2} = \dfrac{-x}{-2} = \dfrac{-1\cdot x}{-1\cdot 2}$ and now we can cancel $-1$ from both numerator and denomintor, or else mutltiply the numerator and denominator by $-1$; either way, it leaves us with just $\dfrac{x}{2}$.

edited Jan 30 '13 at 00:28

answered Jan 29 '13 at 23:49

amWhy

209,954

Hmm, thank you. However, what was done to convert the -x/-2 to x/2? – user60161 Jan 29 '13 at 23:56
@user60161: That was the multiply numerator and denominator by $-1$. Remember that minus times minus gives plus. – Ross Millikan Jan 30 '13 at 00:04
$\dfrac {-x}{-2} = \dfrac{-1\cdot -x}{-1\cdot -2} = \dfrac{x}{2}$. Multiplying by $\dfrac{-1}{-1} = 1$ does not change the value of the fraction. – amWhy Jan 30 '13 at 00:05
1

You can also think of it as factoring out $-1$ from both numerator and denominator, and canceling. – amWhy Jan 30 '13 at 00:06
Thank you, I understand it much better now! – user60161 Jan 30 '13 at 00:08
You're very welcome. Are you clear about the fact that $-\dfrac{x}{-2} = \dfrac{-x}{-2} = \dfrac{x}{2}$? – amWhy Jan 30 '13 at 00:16
You're welcome too, amWhy +. ;-) – Mikasa Jan 30 '13 at 06:45

Problem with getting variable by itself in fraction.

2 Answers2