Question

write a pair of fractions whose sum is 3/11 and difference is 2/11

Answer 1

Step-by-step explanation:

Answer:

Let the fractions be a/b and x/y

According to question

a/b + x/y = 7 / 11 ------(1)

a/b - x/y = 2 / 11 -------(2)

On adding equation 1 and 2, we get

2× a/b = ( 7 + 2) / 11

=> a/b = 9 / 22

Now,

On substituting the value of a/b in equation 1, we get

9 / 22 + x/y = 7 / 11

=> x/y = 7 / 11 - 9 / 22

=> x/y = ( 14 - 9) / 22

=> x/y = 5 / 22

First fraction = 9 / 22

Second fraction = 5 / 22

Step-by-step explanation:

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Answer 2

• The two number are (5/22) and (1/22)

Given:

The sum of a pair of fractions is 3/11

And the difference of the pair of the fraction is 2/11

To find:

The two pair of fractions

Now,

Let one fraction be 'a'

And another fraction be 'b'

So,

According to question:

⇒ (a + b) = 3/11   -----(i)

⇒ (a - b) = 2/11   ------(ii)

Subtracting equation (i) from (ii), we get

  a + b = 3/11

-  (a - b = 2/11)

(-)   (+)      (-)  

--------------------------

       2b = (3/11) - (2/11)

--------------------------

Now,

⇒ 2b = 1/11

⇒ b = (1/11) ÷ 2

[By taking 2 to RHS]

⇒ b = 1/11 * 1/2

⇒ b = 1/22

We got the value of b,

By putting the value of b in equation (i)we get,

⇒ a + b = 3/11

⇒ a + 1/22 = 3/11

⇒ a = (3/11) - (1/22)

⇒ a = (6 - 1)/22

[By taking LCM = 22]

⇒ a = 5/22