# Verify:- 1.-1/3+(4/9+-8/13)=(-1/3+4/9)+-8/13​

Verify:- 1.-1/3+(4/9+-8/13)=(-1/3+4/9)+-8/13​

1. ### ★ How to do :-

Here, we are given with some of the fractions on the left side and the same three fractions on the right side. But on the LHS, the last two fractions are grouped in brackets whereas on the RHS, the first two fractions are grouped in brackets. If this is the thing in simplifying of fractions then here the concept used is the associative property. This concept is only used for addition and multiplication of fractions and integers. This cannot be used in subtraction and division. This rule always comes as equal when used for addition and multiplication. So, let’s solve!!

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### ➤ Solution :-

$${\tt \leadsto \dfrac{(-1)}{3} + \bigg( \dfrac{4}{9} + \dfrac{(-8)}{13} \bigg) = \bigg( \dfrac{(-1)}{3} + \dfrac{4}{9} \bigg) + \dfrac{(-8)}{13}}$$

Let’s solve the LHS and RHS separately.

### LHS :–

$${\tt \leadsto \dfrac{(-1)}{3} + \bigg( \dfrac{4}{9} + \dfrac{(-8)}{13} \bigg)}$$

LCM of 13 and 9 is 117.

$${\tt \leadsto \dfrac{(-1)}{3} + \bigg( \dfrac{4 \times 13}{9 \times 13} + \dfrac{(-8) \times 9}{13 \times 9} \bigg)}$$

Multiply the numerators and denominators in the bracket.

$${\tt \leadsto \dfrac{(-1)}{3} + \bigg( \dfrac{52}{117} + \dfrac{(-72)}{117} \bigg)}$$

Write both numerators with a common denominator.

$${\tt \leadsto \dfrac{(-1)}{3} + \bigg( \dfrac{52 + (-72)}{9} \bigg)}$$

Write the second number in the numerator with one sign.

$${\tt \leadsto \dfrac{(-1)}{3} + \bigg( \dfrac{52 – 72}{117} \bigg)}$$

Subtract the numbers in bracket.

$${\tt \leadsto \dfrac{(-1)}{3} + \dfrac{(-20)}{117}}$$

LCM of 3 and 117 is 117.

$${\tt \leadsto \dfrac{(-1) \times 39}{3 \times 39} + \dfrac{(-20)}{117}}$$

Multiply the numerators and denominators of first fraction.

$${\tt \leadsto \dfrac{(-39)}{117} + \dfrac{(-20)}{117}}$$

Write both numerators with a common denominator.

$${\tt \leadsto \dfrac{(-39) + (-20)}{3}}$$

Write the second number in numerator with one sign.

$${\tt \leadsto \dfrac{(-39) – 20}{3} = \dfrac{(-59)}{117}}$$

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### RHS :–

$${\tt \leadsto \bigg( \dfrac{(-1)}{3} + \dfrac{4}{9} \bigg) + \dfrac{(-8)}{13}}$$

LCM of 3 and 9 is 9.

$${\tt \leadsto \bigg( \dfrac{(-1) \times 3}{3 \times 3} + \dfrac{4}{9} \bigg) + \dfrac{(-8)}{13}}$$

Multiply the numerator and denominator of first fraction.

$${\tt \leadsto \bigg( \dfrac{(-3)}{9} + \dfrac{4}{9} \bigg) + \dfrac{(-8)}{13}}$$

Write both numerators with a common denominator.

$${\tt \leadsto \bigg( \dfrac{(-3) + 4}{9} \bigg) + \dfrac{(-8)}{13}}$$

$${\tt \leadsto \dfrac{1}{9} + \dfrac{(-8)}{13}}$$

LCM of 9 and 13 is 117.

$${\tt \leadsto \dfrac{1 \times 13}{9 \times 13} + \dfrac{(-8) \times 9}{13 \times 9}}$$

Multiply the numerators and denominators of both fractions.

$${\tt \leadsto \dfrac{13}{117} + \dfrac{(-72)}{117}}$$

Write both numerators with a common denominator.

$${\tt \leadsto \dfrac{13 + (-72)}{117} = \dfrac{13 – 72}{117}}$$

Subtract the numbers in the bracket.

$${\tt \leadsto \dfrac{(-59)}{9}}$$

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Let’s compare them now.

$${\tt \leadsto \dfrac{(-59)}{117} \: \: \sf and \: \: \tt \dfrac{(-59)}{117}}$$

We can see that they are like fractions, we can compare them easily.

$${\tt \leadsto \dfrac{(-59}{117} = \dfrac{(-59)}{117}}$$

So,

$${\sf \leadsto LHS = RHS}$$

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$${\red{\underline{\boxed{\bf So, \: the \: both \: pairs \: are \: equal.}}}}$$