How does this separation of the variable from the fraction result in $3\cdot \frac{1}{x}$ & not $\frac{3}{x}$

Question

I wish to write the following expression as the product of a whole number or a fraction and variable expression. The answer given in my textbook is as follows:

$\frac{3}{x} = \frac{3\cdot 1}{1\cdot x} = \frac{3}{1} \cdot \frac{1}{x} = 3\cdot \frac{1}{x} $

However, I am not sure how to get to $3\cdot \frac{1}{x} $. When I complete the multiplication I get to $\frac{3}{1} \cdot \frac{1}{x} = \frac{3}{x}$, which of course loops me back round!

Answer 1

As you stated we have

$\frac{3}{x} = \frac{3\cdot 1}{1\cdot x} = \frac{3}{1} \cdot \frac{1}{x}$

Now we know that $\frac{3}{1}=3$ (dividing by 1 does not change a number).

Hence, we have

$ \frac{3}{1} \cdot \frac{1}{x}=3\cdot \frac{1}{x}$

Answer 2

The key is simply to notice that

$$\frac31=3$$

Thus,

$$\frac31\cdot\frac1x=3\cdot\frac1x$$

Answer 3

What you learned about fractions with numerical numerator and denominator works just as well in algebra.

You are probably comfortable with $$ 3 \times \frac{1}{7} = \frac{3}{7}\ . $$

It's still right when you replace the $7$ by $x$ .

In general, $$ a \times \frac{b}{c} = \frac{a \times b}{c}\ . $$

As a beginning student, you're pretty safe assuming all the straightforward rules of arithmetic carry over to algebra. A little further along in your studies you may have to think about just why the rules of arithmetic work with numbers and with variables.