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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