Question

the sum of the numerator and denominator of a fraction is 12vif 1 is added to numerator and 3 is to denominator the difference will become -4 find the fraction​

Answer 1

EXPLANATION.

Let the numerator of fraction be = x

Let the denominator of fraction be = y

Sum of the numerator and denominator of a

fraction = 12.

=> x + y = 12 .......(1)

If 1 is added to numerator and 3 is to denominator

The fraction will become = -4.

=> x + 1 / y + 3 = -4

=> x + 1 = -4 ( y + 3 )

=> x + 1 = -4y - 12

=> x + 4y = -13 ......(2)

From equation (1) and (2) we get,

=> -3y = 25

=> y = -25/3

put the value of y = -25/3 in equation (1)

=> x - 25/3 = 12

=> x = 12 + 25/3

=> x = 36 + 25 / 3

=> x = 61/3

Therefore,

=> Fraction be = x/y = 61/3 / - 25/3 = -61/25.

Answer 2

Correct Question :-

The sum of the numerator and denominator of a fraction is 12. if 1 is added to numerator and 3 is to denominator then, the difference will become -4. Find the fraction.

Solution :-

☯️ Let,

The Numerator be x.

The Denominator be y.

According To The Question,

$\sf{x + y = 12.....(1)}$

$\to\sf{ \dfrac{x + 1}{y + 3} = - 4}$

$\to \sf{x + 1 = - 4(y + 3)}$

$\to\sf{x + 1 = - 4y - 12}$

$\to\sf{x + 4y = - 13.....(2)}$

From 1] and 2],

$\dashrightarrow\sf{ - 3y = 25}$

$\dashrightarrow\boxed{\sf{y = - \dfrac{25}{3} }}$

Now,

$\dashrightarrow\sf{x - \dfrac{25}{3} = 12}$

$\dashrightarrow\boxed{\sf{x = \dfrac{61}{3}} }$

Now, Fraction :-

$\boxed{\bf\pink{ \dfrac{x}{y} = - \dfrac{61}{25}}}$