Integers — MATHS STD 7 — Question

 We know that two rational numbers are equal if they have the same standard form.
$\frac{2}{-3}=\frac{2\times(-1)}{-3\times(-1)}=\frac{-2}{3}$
The standard form of $\frac{2}{-3}\text{ is }\frac{-2}{3}$
Consider the rational number $\frac{-6}{9}$
$HCF$ of $6$ and $9 = 3$
Dividing the numerator and denominator of $\frac{-6}{9}$ by $3,$
We have:
$\frac{-6}{9}=\frac{-6\div3}{9\div3}=\frac{-2}{3}$
So, the rational number $\frac{-6}{9}$ is equal to $\frac{2}{-3}$
It can be checked that:
Standard form of $\frac{14}{-18}=\frac{-7}{9}$
Standard form of $\frac{3}{-2}=\frac{-3}{2}$
Hence, the correct answer is option $(b).$