Question

verify {x+y}+z =x+{y+z],where x = 1/2,y=2/3,z=-1/5

Answer 1

★ How to do :-

Here, we are given with some of the fractions on LHS of the statement and the same three fractions in the RHS of the statement. On LHS, the last two fractions are grouped by brackets and on the RHS, the first two fractions are grouped in the brackets. If we are given with the fractions by grouping in this format, then this property can be classified as the associative property. This property can only be done with the multiplication and addition of fractions. It cannot be done in the subtraction and division of fractions or integers. This also cannot be done by the negative integers. As in this question, we are given with the multiplication it's possible with this property. But, we should verify the statement. So, let's solve!!

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➤ Solution :-

${\sf \leadsto (x + y) + z = x + (y + z)}$

Substitute the values of x, y and z.

${\sf \leadsto \bigg( \dfrac{1}{2} + \dfrac{2}{3} \bigg) + \dfrac{(-1)}{5} = \dfrac{1}{2} + \bigg( \dfrac{2}{3} + \dfrac{(-1)}{5} \bigg)}$

Let's solve the LHS and RHS separately.

LHS :-

${\sf \leadsto \bigg( \dfrac{1}{2} + \dfrac{2}{3} \bigg) + \dfrac{(-1)}{5}}$

First let's solve the fractions in bracket

LCM of 2 and 3 is 6.

${\sf \leadsto \bigg( \dfrac{1 \times 3}{2 \times 3} + \dfrac{2 \times 2}{3 \times 2} \bigg) + \dfrac{(-1)}{5}}$

Multiply the numerators and denominators of both fractions.

${\sf \leadsto \bigg( \dfrac{3}{6} + \dfrac{4}{6} \bigg) + \dfrac{(-1)}{5}}$

Write both numerators with a common denominator.

${\sf \leadsto \bigg( \dfrac{3 + 4}{6} \bigg) + \dfrac{(-1)}{5}}$

Add the numbers.

${\sf \leadsto \dfrac{7}{6} + \dfrac{(-1)}{5}}$

LCM of 6 and 5 is 30.

${\sf \leadsto \dfrac{7 \times 5}{6 \times 5} + \dfrac{(-1) \times 6}{5 \times 6}}$

Multiply the numerators and denominators of both fractions.

${\sf \leadsto \dfrac{35}{30} + \dfrac{(-6)}{30}}$

Write the second number with one sign.

${\sf \leadsto \dfrac{35 + (-6)}{30} + \dfrac{35 - 6}{30}}$

Subtract the numbers to get the value of LHS

${\sf \leadsto \dfrac{29}{30} \: --- LHS}$

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RHS :-

${\sf \leadsto \dfrac{1}{2} + \bigg( \dfrac{2}{3} + \dfrac{(-1)}{5} \bigg)}$

First let's solve the numbers in bracket.

LCM of 3 and 5 is 15.

${\sf \leadsto \dfrac{1}{2} + \bigg( \dfrac{2 \times 5}{3 \times 5} + \dfrac{(-1) \times 3}{5 \times 3} \bigg)}$

Multiply the numerators and denominators of both fractions in bracket.

${\sf \leadsto \dfrac{1}{2} + \bigg( \dfrac{10}{15} + \dfrac{(-3)}{15} \bigg)}$

Write both numerators with a common denominator.

${\sf \leadsto \dfrac{1}{2} + \bigg( \dfrac{10 + (-3)}{15} \bigg)}$

Write the second number in numerator with one sign.

${\sf \leadsto \dfrac{1}{2} + \bigg( \dfrac{10 - 3}{15} \bigg)}$

Subtract the numbers.

${\sf \leadsto \dfrac{1}{2} + \dfrac{7}{15}}$

LCM of 2 and 15 is 30.

${\sf \leadsto \dfrac{1 \times 15}{2 \times 15} + \dfrac{7 \times 2}{15 \times 2}}$

Multiply the numerators and denominators of both fractions.

${\sf \leadsto \dfrac{15}{30} + \dfrac{14}{30} = \dfrac{15 + 14}{30}}$

Add the numbers to get the value of RHS.

${\sf \leadsto \dfrac{29}{30} \: --- RHS}$

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Let's compare the results of both LHS and RHS.

Comparison :-

${\sf \leadsto \dfrac{29}{30} \: and \: \dfrac{29}{30}}$

As we can see that they both are equal. So,

${\sf \leadsto \dfrac{29}{30} = \dfrac{29}{30}}$

So,

${\sf \leadsto LHS = RHS}$

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Hence verified !!