Question

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

Answer 1

$\Huge\tt{{\color{purple}{{\underline{AN}}}}{\color{orchid}{{\underline{SW}}}}{\pink{{\underline{ER}}}}{\color{lightpink}{:-}}}$

• The fraction is 4/7

$\huge\Large\underline\mathtt\purple{Step \:By\:Step \:Explanation:-}$

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)

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

Step-by-step explanation:

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