-1
$\begingroup$

For example, $$\frac{\left(\frac{1}{x-1}\right)}{x-1}$$ I thought you would just need to multiply the top of the fraction, not both.

  • 2
    Because you don't want to change the value of the fraction, so you multiply it by 1. This is the same as multiplying by $(x - 1) / (x - 1)$ and trying it has a clear advantage. – 2017-02-05
  • 0
    You could also just rewrite it as $\frac{(\frac{1}{x-1})}{x-1}=\frac{1}{x-1}\cdot \frac{1}{x-1}$ by noting that $\frac{a}{b}=a\cdot \frac{1}{b}$ – 2017-02-05
  • 0
    You want to multiply it by ONE so that it's value is unchanged. And if $x\ne 1$ then $\frac {x-1}{x-1}$ IS one. – 2017-02-05

3 Answers 3

1

I believe you wish to simplify the fraction, not solve it (since you solve equations, and this expression is missing an $=$ sign).

Remember when you're dividing one fraction by another fraction, multiply by the reciprocal of the fraction in the larger fraction's denominator:

$$\frac{\big(\frac{1}{x-1}\big)}{x-1}=\frac{\big(\frac{1}{x-1}\big)}{\big(\frac{x-1}{1}\big)}=\frac{1}{(x-1)}\frac{1}{(x-1)}=\frac{1}{(x-1)^2}$$

provided $x \not= 1.$

1

A compound fraction is a fraction which contains fractions in the numerator or the denominator (or both).

The most efficient way to simplify (turn it into an ordinary fraction) a complex fraction is to multiply both the numerator and the denominator by the least common denominator of the minor fractions (the fractions which are contained in the numerator or denominator).

Since both the numerator and the denominator are multiplied by the same expression, effectively we are multiplying the original expression by $1$, so the resulting expression will be equivalent to the original expression, except it will no longer be a complex fraction--it will be a simple fraction.

In the case of the complex fraction

\begin{equation} \frac{\left(\frac{1}{x-1}\right)}{x-1} \end{equation}

only the numerator contains a minor fraction, namely $\frac{1}{x-1}$ so the LCD of the minor fractions is $x-1$. We multiply both the numerator and denominator by $x-1$:

\begin{eqnarray} \frac{\frac{1}{x-1}}{x-1}\cdot\frac{x-1}{x-1}&=\frac{1}{(x-1)^2}& \end{eqnarray}

which turns it into an ordinary fraction--a fraction containing no fractions in the numerator or denominator.

0

There are two cases -

Case 1-

You have one fraction in that fraction numerator is further fraction.

$$\frac{\frac{1}{x-1}}{x-1}$$

Then make denominator also fraction by writing 1 in its denominator. Like this.

$$\frac{\frac{1}{x-1}}{\frac{x-1}1}$$

Then reciprocal the fraction in denominator and multiply with the fraction in numerator.

$$\frac{1}{x-1} × \frac{x-1}{1}$$

Case 2-

You have one fraction in that fraction denominator is further fraction.

$$\frac{x-1}{\frac{1}{x-1}}$$

Then make numerator also fraction by writing 1 in its denominator. Like this.

$$\frac{\frac{x-1}{1}}{\frac1{x-1}}$$

Then reciprocal the fraction in denominator and multiply with the fraction in numerator.

$$\frac{x-1}1 × \frac{x-1}1$$