Question

by what rational number must 1/26 be divided to get - 8/39​

Answer 1

Answer:

1/26÷x=8/39

x= 8/39×1/26

x=4/507

Step-by-step explanation:

hope so you got your answer ☺️

Answer 2

➤ Question :-

${\tt \leadsto \dfrac{1}{26} \div x = \dfrac{(-8)}{39}}$

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★ How to do :-

Here, we are given with a fraction and the answer obtained while dividing two fractions. In that one of the fraction is given. The other number is given which is placed as a variable x. We are asked to find thr value of that variable x. Here, we are going to shift the numbers from one side to other in which we change the sign of the particular fraction. Then, we find the value of x and then to know that is a correct answer, we can verify the statement by applying the value of x. So, let's solve!!

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➤ Solution :-

${\tt \leadsto \dfrac{1}{26} \div x = \dfrac{(-8)}{39}}$

Shift the number on RHS to LHS, changing it's sign.

${\tt \leadsto x = \dfrac{1}{26} \div \dfrac{(-8)}{39}}$

Take the reciprocal of second fraction and multiply both the fractions.

${\tt \leadsto x = \dfrac{1}{26} \times \dfrac{39}{(-8)}}$

Write the denominator of first and numerator of second fraction in lowest form by cancellation method.

${\tt \leadsto x = \dfrac{1}{\cancel{26}} \times \dfrac{\cancel{39}}{(-8)} = \dfrac{1}{2} \times \dfrac{3}{(-8)}}$

Multiply the fractions now, as they cannot be cancelled further.

${\tt \leadsto x = \dfrac{1 \times 3}{2 \times (-8)} = \pink{\underline{\boxed{\tt \dfrac{3}{(-16)}}}}}$

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Verification :-

${\tt \leadsto \dfrac{1}{26} \div x = \dfrac{(-8)}{39}}$

Substitute the value of x.

${\tt \leadsto \dfrac{1}{26} \div \dfrac{3}{(-16)} = \dfrac{(-8)}{39}}$

Take the reciprocal of second fraction and multiply both the fractions on LHS.

${\tt \leadsto \dfrac{1}{26} \times \dfrac{(-16)}{3} = \dfrac{(-8)}{39}}$

Write both numerators and denominators in one fraction on LHS.

${\tt \leadsto \dfrac{1 \times (-16)}{26 \times 3} = \dfrac{(-16)}{78} = \dfrac{(-8)}{39}}$

Write the fraction in lowest form by cancellation method.

${\tt \leadsto \cancel \dfrac{(-16)}{78} = \dfrac{(-8)}{39}}$

Now,

${\tt \leadsto \dfrac{(-8)}{39} = \dfrac{(-8)}{39}}$

So,

${\sf \leadsto LHS = RHS}$

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Hence verified !!