Question

if the numerator of a certain fraction is increased by 2 and the denominator by 1 the fraction becomes equal to 5 / 8 and if the numerator and denominator are each demands by 1 the fraction becomes equal to 1 / 2 then find the fraction

Answer 1

Answer:
$Fraction\:is\: \frac{8}{15}$

Solution:

Let the fraction is
$\frac{x}{y}$
according to the question,

$\frac{x + 2}{y + 1} = \frac{5}{8} \\ \\ \\ 8(x + 2) = 5(y + 1) \\ \\ 8x + 16 = 5y + 5 \\ \\ 8x - 5y = - 11....eq1$
now another statement is

$\frac{x - 1}{y - 1} = \frac{1}{2} \\ \\ 2(x - 1) = y - 1 \\ \\ 2x - 2 - y = - 1 \\ \\ 2x - y = 1....eq2$
Now solve equations 1 and 2

multiply eq2 by 4

$8x - 5y = - 11 \\ 8x - 4y = 4 \\ - \: \: \: + \: \: \: \: \: \: - \\ - - - - - \\ - y = - 15 \\ \\ y = 15$
put the value of y in eq2

$2x - y = 1 \\ \\ 2x - 15 = 1 \\ \\ 2x = 16 \\ \\ x = 8$
So the fraction is
$\frac{8}{15}$
Hope it helps you.

Answer 2

Answer:

$\frac{x}{y}=\frac{7}{15}$

Step-by-step explanation:

Given : If the numerator of a certain fraction is increased by 2 and the denominator by 1 the fraction becomes equal to 5/8 and if the numerator and denominator are each demands by 1 the fraction becomes equal to 1/2.

To find : The fraction  ?

Solution :

Let the fraction be $\frac{x}{y}$

If the numerator of a certain fraction is increased by 2 and the denominator by 1 the fraction becomes equal to 5/8.

i.e. $\frac{x+2}{y+1}=\frac{5}{8}$

Cross multiply,

$8x+16=5y+5$

$8x-5y=-11$ .....(1)

If the numerator and denominator are each demands by 1 the fraction becomes equal to 1/2.

i.e. $\frac{x - 1}{y - 1} = \frac{1}{2}$

Cross multiply,

$2x-2=y-1$

$2x-y=1$ .....(2)

Solving equation (1) and (2),

Multiply equation (2) by 4 and subtract from (1)

$8x-4y-8x+5y=4+11$

$y=-15$

Substitute in equation (2),

$2x-(-15)=1$

$2x=1-15$

$2x=-14$

$x=-7$

The fraction became $\frac{x}{y}=\frac{-7}{-15}$

$\frac{x}{y}=\frac{7}{15}$