Question

10. The sum of the numerator and denominator of a fraction is 3 less than twice the
denominator. If the numerator and denominator are decreased by 1, the numerator
becomes half the denominator. Determine the fraction.
[CBSE 2001C, 2010)​

Answer 1

Answer:

the numerator be 'x' and denominator be ' y'

then,

the sum of the numerator and denominator of a fraction is 3 less than the twice is x+y=2y-3 :- x-y+3=0 - equation 1

if the numerator and denominator is decreased by 1 then

x-1/y-1=1/2 :- 2(x-1) = 1(y-1) :- 2x-y-1=0 -equation 2

then equation 1 = equation 2

x-y+3=2x-y-1 :- x=4

from the equation 1 

x-y+3=0 :- 4-y+3=0 :- y=7

there fore the fraction is 3/7 

pls mark me as brainliest

thank you

Answer 2

Step-by-step explanation:

Hey..... good morning..❤

Here ur answers.......✌

The fraction is 4/7

Step-by-step explanation:

The given problem is on linear equations with two variables say x and y.

Let the fraction required be x/y.

Sum of numerator and denominator = x + y

Given x + y is 3 less than twice the denominator → x + y = 2y - 3

x - y +3 = 0 → (1)

Also, if numerator and denominator are decreased by 1 → (x - 1), (y - 1)

The numerator becomes half of the denominator  

2x - 2 = y – 1

2x - y =2 -1 =1

2x - y = 1 → (2)

Subtracting (2) and (1) gives x - y + 3 - (2x - y - 1) = 0

x - y + 3 – 2x + y + 1 = 0

-x + 4 =0

x = 4

Substituting x value in (1) gives 4 - y + 3 = 0

y = 7

Therefore x = 4 and y = 7;

The fraction required is 4/7

Hope may help uuu....❤