# *to "get rid" of a fraction multiply by the...?

Mar 7, 2018

Multiply by the value in the denominator of the fraction

#### Explanation:

Let's say that you had the following equation $\setminus \frac{2}{3} x = 21$. You could possibly divide both sides by $\setminus \frac{2}{3}$, although I don't think solving it through that method is quite as pleasant as working with integers. Therefore, you could multiply both sides by the denominator of the fraction (which is 3) to "get rid of" the fraction.

$3 \setminus \times \setminus \frac{2}{3}$
You can also view this as $\setminus \frac{3}{1} \setminus \times \setminus \frac{2}{3}$, and from that, you can see that the 3 in the numerator of the first fraction and the 3 of the denominator of the second fraction can cancel each other out (think about it: \frac{3}{3} = 1).

So we know that $3 \setminus \times \setminus \frac{2}{3} = 2$

Since you multiplied the left side of the equation by 3, you have to do that to the right side of the equation too.

$2 x = 63$
$x = \setminus \frac{63}{2}$

The equation wasn't so "beautiful" because we still got a fraction as the value for $x$, but I hope you understood how you answer your question.

Mar 7, 2018

Multiply By The Reciprocal

#### Explanation:

A few examples...

1) $\frac{5}{6} \cdot \frac{6}{5} = \textcolor{red}{1}$

2) $\frac{9}{20} \cdot \frac{20}{9} = \textcolor{red}{1}$

3) $\frac{9999}{5} \cdot \frac{5}{999} = \textcolor{red}{1}$

No matter the fraction, turning it "upside down" (flipping its numerator/denominator) then multiplying by the same fraction will usually give you a value = 1

But there are some more advanced cases where this does not always occur. Especially when dealing with variables...

Let's try something a bit harder...say you're given two fractions to divide:

(8x^5y)/(25z^6) ÷ color(blue)((20xy^4)/(15z^3))

As usual, multiply by the reciprocal of the divisor...

$\frac{8 {x}^{5} y}{25 {z}^{6}} \cdot \textcolor{b l u e}{\frac{15 {z}^{3}}{20 x {y}^{4}}}$... Multiply both sides together

$\frac{120 {x}^{5} y {z}^{3}}{500 x {y}^{4} {z}^{6}}$ ..."Divide" by canceling out common terms

$\textcolor{red}{\frac{6 {x}^{4}}{25 {y}^{3} {z}^{3}}}$