Question

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.

Answer 1

Let the numerator and denominator of the fraction be x and y respectively. Then the fraction is `x/y`

The sum of the numerator and denominator of the fraction is 3 less than twice the denominator. Thus, we have

`x+y=2y-3`

`⇒ x+y-2y+3=0`

`⇒ x-y+3=0`

If the numerator and denominator are decreased by 1, the numerator becomes half the denominator. Thus, we have

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

`⇒ (x-1)/(y-1)=1/2`

`⇒ 2(x-1)=y-1`

`⇒ 2x -2 =y-1`

`⇒ 2x -y -1=0`

So, we have two equations

`x-y+3=0`

`2x-y-5=0`

Here x and y are unknowns. We have to solve the above equations for x and y.

By using cross-multiplication, we have

`x/((-1)xx(-1)-(-1)xx3)=(-y)/(1xx(-1)-2xx3)=1/(1xx(-1)-2xx(-1))`

`⇒ x/(1+3)=(-y)/(-1-6)=(1)/(-1+2)`

`⇒ x/4 = (-y)/-7 =1/1`

`⇒ x/4=y/7=1`

`⇒ x = 4,y=7`
Hence, the fraction is `4/7`