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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.
Solution
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`
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