Question

Verify that -2/7 x (-3/7 x -9/14) = (-2/7 x -3/7) x -9/14

Answer 1

Answer:

applying BODMAS

B=bracket

O=of/order

D=division

M=multiplication

A=addition

S=subtract

LHS=

first solving value inside the bracket

-2/7 * (-3/7 * -9/14 )

2/7 * {(3*9)/(7*14 )}

-2/7 * (27/98) ( " - " × " -" gives +)

-(2*27)/(7*98). ( " - " × " +" gives -)

-54/686

further simplifying

-27/343

RHS

first solving value inside the bracket

(-2/7 * -3/7) * -9/14

{(2*3)/(7*7)} * -9/14 ( " - " × " -" gives +)

(6/49)* -9/14. ( " + " × " -" gives -)

-54/686.

further simplifying

-27/343

since LHS=RHS

Hence verified

Answer 2

★ 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 :-

${\sf \leadsto \dfrac{(-2)}{7} \times \bigg( \dfrac{(-3)}{7} \times \dfrac{(-9)}{14} \bigg) = \bigg( \dfrac{(-2)}{7} \times \dfrac{(-3)}{7} \bigg) \times \dfrac{(-9)}{14}}$

Let's solve the LHS and RHS separately.

LHS :-

${\sf \leadsto \dfrac{(-2)}{7} \times \bigg( \dfrac{(-3)}{7} \times \dfrac{(-9)}{14} \bigg)}$

Write both numerators and denominators with a common fraction.

${\sf \leadsto \dfrac{(-2)}{7} \times \bigg( \dfrac{(-3) \times (-9)}{7 \times 14} \bigg)}$

Multiply the numbers.

${\sf \leadsto \dfrac{(-2)}{7} \times \dfrac{27}{98}}$

Write the fraction in lowest form by cancellation method.

${\sf \leadsto \dfrac{\cancel{(-2)} \times 27}{7 \times \cancel{98}} = \dfrac{(-1) \times 27}{7 \times 49}}$

Multiply the remaining numbers.

${\sf \leadsto \dfrac{(-27)}{343} \: --- LHS}$

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RHS :-

${\sf \leadsto \bigg( \dfrac{(-2)}{7} \times \dfrac{(-3)}{7} \bigg) \times \dfrac{(-9)}{14}}$

Write both numerators and denominators with a common fraction.

${\sf \leadsto \bigg( \dfrac{(-2) \times (-3)}{7 \times 7} \bigg) \times \dfrac{(-9)}{14}}$

Multiply the numbers.

${\sf \leadsto \dfrac{6}{49} \times \dfrac{(-9)}{14}}$

Write the fraction in lowest form by cancellation method.

${\sf \leadsto \dfrac{\cancel{6} \times (-9)}{49 \times \cancel{14}} = \dfrac{3 \times (-9)}{49 \times 7}}$

Multiply the remaining numbers.

${\sf \leadsto \dfrac{(-27)}{343} \: --- RHS}$

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Comparison :-

Now, let's compare the results of both LHS and RHS.

${\sf \leadsto \dfrac{(-27)}{343} \: and \: \dfrac{(-27)}{343}}$

As we can see that the both fractions are same. So,

${\sf \leadsto \dfrac{(-27)}{343} = \dfrac{(-27)}{343}}$

So,

${\sf \leadsto LHS = RHS}$

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Hence verified !!