Question

The sum of denominator and numerator of a
fraction is 3 less than twice the denominator. for each of the numerator and denominator is decreased by 1 the fraction becomes find
the fraction?​

Answer 1

Answer:

let the numerator of the fraction be x and let the denominator be y

Step-by-step explanation:

according to the question-

x+y=2y-3

2y-y=x+3

y=x+3

x-1/y-1=

sorry your question is not complete.

Answer 2

Answer:

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....

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