Question

find using Distributivi
(9/16×4/12)+(9/16×-3/9)​

Answer 1

★ How to do :-

Here, we are given with some fractions in which we are asked to simplify those fractions using the distributive law. So, here, we have two fraction which are same or common. We take the fraction repeating twice as a common fraction. We have other two fraction and also we are given to add them. The method to solve this is, we should multiply the common fraction with the sum of other two factions. We should find the sum because there's subtraction sign in between two brackets. So, let's solve!!

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➤ Solution :-

${\tt \leadsto \bigg( \dfrac{9}{16} \times \dfrac{4}{12} \bigg) + \bigg( \dfrac{9}{16} \times \dfrac{(-3)}{9} \bigg)}$

Take the fraction as common and add the fractions in bracket.

${\tt \leadsto \dfrac{9}{16} \times \bigg( \dfrac{4}{12} + \dfrac{(-3)}{9} \bigg)}$

LCM of 12 and 9 is 36.

${\tt \leadsto \dfrac{9}{16} \times \bigg( \dfrac{4 \times 3}{12 \times 3} + \dfrac{(-3) \times 4}{9 \times 4} \bigg)}$

Multiply the numerator and denominator of both fraction in the bracket.

${\tt \leadsto \dfrac{9}{16} \bigg( \dfrac{12}{36} + \dfrac{(-12)}{36} \bigg)}$

Now, add the fractions in the bracket.

${\tt \leadsto \dfrac{9}{16} \times \bigg( \dfrac{12 + (-12)}{36} \bigg)}$

Add the numbers in the numerator.

${\tt \leadsto \dfrac{9}{16} \times 0}$

We know that any number multiplied by zero gives zero. So,

${\tt \leadsto \dfrac{9 \times 0}{16} = \pink{\underline{\boxed{\tt 0}}}}$

$\Huge\therefore$ The answer obtained while simplified is Zero(0).