Question

-13/20+11/14+-5/7+7/10​

Answer 1

★ How to do :-

Here, we are given with four fractions in which we are asked to add all of them together. We can find the sum of these fractions by a concept. Thus concept is called as the rearrangement of fractions. By this property, the place of a fraction changes whereas it's value doesn't change. This property can only used while adding or multiplying the fractions or integers. We shouldn't use this in subtraction and division. Here, we are given with addition so that we can use it. We can decide the fractions by looking into the denominators. So, let's solve!!

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➤ Solution :-

${\tt \leadsto \dfrac{(-13)}{20} + \dfrac{11}{14} + \dfrac{(-5)}{7} + \dfrac{7}{10}}$

Rearrange the fractions as we want.

${\tt \leadsto \bigg( \dfrac{(-13)}{20} + \dfrac{7}{10} \bigg) + \bigg( \dfrac{11}{14} + \dfrac{(-5)}{7} \bigg)}$

LCM of 20 and 10 is 20.

LCM of 14 and 7 is 14.

${\tt \leadsto \bigg( \dfrac{(-13)}{20} + \dfrac{7 \times 2}{10 \times 2} \bigg) + \bigg( \dfrac{11}{14} + \dfrac{(-5) \times 2}{7 \times 2} \bigg)}$

Multiply the numerator and denominator of both fractions.

${\tt \leadsto \bigg( \dfrac{(-13)}{20} + \dfrac{14}{20} \bigg) + \bigg( \dfrac{11}{14} + \dfrac{(-10)}{14} \bigg)}$

Write both numerators with a common denominator.

${\tt \leadsto \bigg( \dfrac{(-13) + 14}{20} \bigg) + \bigg( \dfrac{11 + (-10)}{14} \bigg)}$

Add the numerators of both fractions.

${\tt \leadsto \dfrac{1}{20} + \dfrac{1}{14}}$

Now, add the remaining numbers.

LCM of 20 and 14 is 140.

${\tt \leadsto \dfrac{1 \times 7}{20 \times 7} + \dfrac{1 \times 10}{14 \times 10}}$

Multiply the numerators and denominators of both fractions.

${\tt \leadsto \dfrac{7}{140} + \dfrac{10}{140}}$

Now, add these fractions.

${\tt \leadsto \dfrac{7 + 10}{140} = \dfrac{17}{140}}$

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${\red{\underline{\boxed{\bf So, \: the \: answer \: obtained \: is \: \dfrac{17}{140}.}}}}$