Question

Using appropriate property of rational numbers, simplify the following.

$\frac{2}{5}\times \left(\frac{-3}{7}\right)-\frac{1}{6}\times \frac{3}{2}+\frac{1}{14}\times \frac{2}{5}$

Answer 1

We have,

$\frac{2}{5}\times (-\frac{3}{7})-\frac{1}{6}\times \frac{3}{2}+\frac{1}{14}\times \frac{2}{5}$

$=\frac{2}{5}\times (-\frac{3}{7})+\frac{1}{14}\times \frac{2}{5}-\frac{1}{6}\times \frac{3}{2}$ [Using, commutative property on last two terms]

$=\frac{2}{5}\times (-\frac{3}{7}+\frac{1}{14})-\frac{1}{6}\times \frac{3}{2}$ [Using, Distributive property: $a\times (b+c)=(a\times b)+(a\times c)$]

$=\frac{2}{5}\times \left(\frac{-6+1}{14}\right)-\frac{1}{6}\times \frac{3}{2}$

$=\frac{2}{5}\times \left(\frac{-5}{14}\right)-\frac{1}{6}\times \frac{3}{2}$

$=\left(\frac{-1}{7}\right)-\frac{1}{2}\times \frac{1}{2}$

$=\frac{-1}{7}-\frac{1}{4}$

$=\frac{-4-7}{28}$

$=\frac{-11}{28}$

$\therefore \frac{2}{5}\times (-\frac{3}{7})-\frac{1}{6}\times \frac{3}{2}+\frac{1}{14}\times \frac{2}{5}=\frac{-11}{28}$