Question

the product if two rational numbers -16/9.If one of the no. is -4/3, find the value​

Answer 1

Step-by-step explanation:

product of two no. s =-16/9

one no. =-4/3

let other no. =x

ATS

x. -4/3 =-16/9

x =-16/9÷4/3

x=-16/9×-3/4

x=4/3

Answer 2

★ How to do :-

Here, we are given with a rational number that should be multiplied by an other number. We are also given with the answer obtained while multiplying those two fractions. But, we aren't given with the second number that the first number should be multiplied with. We are asked to find the same. To find the answer, we make use of some other concepts such as variables and the transaction of numbers from one hand side to the other. While we are using this method, the sign of the appropriate number changes. We can also check out our answer by verification method. So, let's solve!!

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➤ Solution :-

Let the other number be y.

${\sf \leadsto \dfrac{(-4)}{3} \times y = \dfrac{(-16)}{9}}$

Shift the fraction on LHS to RHS.

${\sf \leadsto y = \dfrac{(-16)}{9} \div \dfrac{(-4)}{3}}$

Take the reciprocal of second fraction and multiply both fractions.

${\sf \leadsto y = \dfrac{(-16)}{9} \times \dfrac{3}{(-4)}}$

Write the fraction in lowest form by cancellation method.

${\sf \leadsto y = \dfrac{\dfrac{(-16)} \times 3}{9 \times \cancel{(-4)}} = \dfrac{4 \times 3}{9 \times 1}}$

Multiply the remaining numbers.

${\sf \leadsto y = \dfrac{12}{9}}$

Write the obtained fraction in lowest form by cancellation method.

${\sf \leadsto y = \cancel \dfrac{12}{9} = \dfrac{4}{3}}$

Write the fraction in the form of mixed fraction.

${\sf \leadsto \dfrac{4}{3} = 1 \dfrac{1}{3}}$

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${\red{\underline{\boxed{\bf So, \: the \: other \: fraction \: is \: 1 \dfrac{1}{3}.}}}}$

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Verification :-

${\sf \leadsto \dfrac{(-4)}{3} \times y = \dfrac{(-16)}{9}}$

Substitute the value of y.

${\sf \leadsto \dfrac{(-4)}{3} \times 1 \dfrac{1}{3} = \dfrac{(-16)}{9}}$

Write the second fraction in the form of improper fraction.

${\sf \leadsto \dfrac{(-4)}{3} \times \dfrac{4}{3} = \dfrac{(-16)}{9}}$

Write both numerators and denominators with a common fraction.

${\sf \leadsto \dfrac{(-4) \times 4}{3 \times 3} = \dfrac{(-16)}{9}}$

Multiply the numbers.

${\sf \leadsto \dfrac{(-15)}{9} = \dfrac{(-16)}{9}}$

So,

${\sf \leadsto LHS = RHS}$

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Hence verified !!