Complex fraction

Question

my book uses a technique to write fractions in a way that I am not familiar with.

How does $$\frac{\frac{1}{243}-1}{-\frac{2}{3}}$$ equate to $$\frac{3\left(1-\frac{1}{243}\right)}{2}$$ Thanks in advance.

Answer 1

The numerator and denominator of the fraction were multiplied by $-3$. That clears the complex fraction in the denominator and inverts the binomial in the numerator.

Answer 2

You multiply numerator and denominator both by $-3$ ("expand the fraction by a factor of $-3$"). This operation is well-known not to change the value of a fraction.

Answer 3

Division means multiplication of the reciprocal, so $$(\frac1{243}-1)\div (-\frac23)\implies(\frac1{243}-1)\times(-\frac32).$$

As well, a binomial involving subtraction can be rewritten by factoring out a negative: $$(a-b)=-(b-a).$$

Combining the two ideas, we have \begin{align} (\frac1{243}-1)\times(-\frac32)&=(1-\frac1{243})\times(\frac32)\\ &=\frac32\times(1-\frac1{243})\\ &=\frac{3(1-\frac1{243})}{2} \end{align}