Question

13. The sum of denominator
and numerator of a fraction is
3 less than twice the
denominator. If each of the
numerator and denominator is
denominator is decreased by
1, the fraction becomes 1/2.
Find the fraction.​

Answer 1

Answer:

The fraction is 4/7

Step-by-step explanation:

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

x - 1 = \frac{1}{2} (y - 1)x−1=

2

1

(y−1)

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

Answer 2

Let the numerator be x and the denominator be y

Sum of numerator and denominator = x + y

Given x + y is 3 less than twice the denominator

So,

=> x + y = 2y - 3

=> x - y +3 = 0 ...(1)

Also, if numerator and denominator are decreased by 1, The numerator becomes half of the denominator

=>(x-1) = 1/2(y-1)

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;

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