Question

orm an equation in the following cosesi
(a) You have a fraction with numerator 2 and denominator 7. When y is added to the
numerator and x is subtracted from the denominator, the numerator and
denominator in the resultant fraction are equal​

Answer 1

Answer:

Let the fraction be ba.

Therefore, b−3a+2=1

a+2=b−3

a−b=−5              .......(1)

Also, b+3a−2=83

8a−16=3b+9

8a−3b=25

Solving these equations, we get,

a=8 and b=13

Thus, the fraction is 138.

'

Answer 2

Solution :

The given fraction is given as 2/7 .

y Is added to the numerator and x is subtracted from the denominator .

Thus , the new fraction becomes-

=> ( 2 + y )/( 7 - x)

Here , the numerator and denominator becomes equal .

So ,

[ y + 2]/[ 7 - x ] = 1/1

Cross Multiplying

y + 2 = 7 - x

=> x + y - 5 = 0 .

This is the required equation to be formed .

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Additional Information -

• Componendo Dividendo Rule

For two fractions , a/b and c/d

[ a + b ]/[ a - b ] = [ c + d ]/[ c - d ]

(a + b)² = a² + 2ab + b²

(a + b)² = (a - b)² + 4ab

(a - b)² = a² - 2ab + b²

(a - b)² = (a + b)² - 4ab

a² + b² = (a + b)² - 2ab

a² + b² = (a - b)² + 2ab

2 (a² + b²) = (a + b)² + (a - b)²

4ab = (a + b)² - (a - b)²

ab = {(a + b)/2}² - {(a-b)/2}²

(a + b + c)² = a² + b² + c² + 2(ab + bc + ca)

(a + b)³ = a³ + 3a²b + 3ab² b³

(a + b)³ = a³ + b³ + 3ab(a + b)

(a - b)³ = a³ - 3a²b + 3ab² - b³

a³ + b³ = (a + b)( a² - ab + b² )

a³ + b³ = (a + b)³ - 3ab( a + b)

a³ - b³ = (a - b)( a² + ab + b²)

a³ - b³ = (a - b)³ + 3ab ( a - b )

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