Question

Fill in the blank:

The standard form of $\frac{-8}{-36}$ is ………..

Answer 1

Given rational number is $\frac{-8}{-36}$.

Here numerator and denominator both are negative so we have to multiply both numerator and denominator by $-1$.

$\frac{-8\times \left(-1\right)}{-36\times \left(-1\right)}=\frac{8}{36}$

Here numerator is $8$ and denominator is $36$.

Prime factors of $8$ and $36$ are as follows:

$$\begin{gathered} 8=2\times 2\times 2 \\ 36=2\times 2\times 3\times 3 \end{gathered}$$

HCF of $8$ and $36$ is $4$.

Now divide the numerator and denominator of $\frac{8}{36}$ by HCF $4$, we get:

$\frac{8÷4}{36÷4}=\frac{2}{9}$

Therefore the standard form of $\frac{-8}{-36}$ is $\frac{2}{9}$.