The sum of numerator and denominator of a fraction is 4 less than twice the denominator. If each of the numerator and denominator is decreased by 1, the fraction becomes ¹⁄₃. Find the fraction.

 The sum of numerator and denominator of a fraction is 4 less than twice the denominator. If each of the numerator and denominator is decreased by 1, the fraction becomes ¹⁄₃. Find the fraction.

Let numerator = x, denominator = y. Fraction = x⁄y

Given

sum of numerator and denominator of a fraction is 4 less than twice the denominator

 x + y = 2y − 4

⇒ x = y − 4  --- (1)

Given

If each of the numerator and denominator is decreased by 1, the fraction becomes ¹⁄₃

 (x − 1)⁄(y − 1) = ¹⁄₃

⇒ 3(x − 1) = y − 1

⇒ 3x − 3 = y − 1

⇒ 3x = y + 2  --- (2)

Solve using substitution method

Substitute (1) into (2):

3(y − 4) = y + 2

3y − 12 = y + 2

2y = 14

y = 7 

 Then 

x = y − 4 

x = 7 -4

x= 3  

x =3 

y=7

 The fraction is ³⁄₇

linear equations in two unknowns

formation of linear equations and solution by substitution method

cbse 10th maths previous years question papers 

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