In a rational number, twice the numerator is 2 more than the denominator. If 3 is added to both the numerator and the denominator the new fraction is 2/3. Find the original fraction?
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Let the numerator be x and the denominator be y.
2x=y+2
∴2x-y=2 - (1)
x+3/y+3=2/3
3x+9=2y+6 [cross multiplication]
3x-2y=6-9
∴3x-2y= - 3 -(2)
(2) - (1)→
3x - 2y= - 3
-
4x- 2y= 4 [ multiplying by 2, so that the co-ordinate of y of both equation will be same]
→ - x = - 7
∴ x= 7 - (3)
applying (3) in (1)
2*7- y=2
14- y=2
-y=2- 14
-y = -12
∴y=12
hence, the required fraction is 7/12
2x=y+2
∴2x-y=2 - (1)
x+3/y+3=2/3
3x+9=2y+6 [cross multiplication]
3x-2y=6-9
∴3x-2y= - 3 -(2)
(2) - (1)→
3x - 2y= - 3
-
4x- 2y= 4 [ multiplying by 2, so that the co-ordinate of y of both equation will be same]
→ - x = - 7
∴ x= 7 - (3)
applying (3) in (1)
2*7- y=2
14- y=2
-y=2- 14
-y = -12
∴y=12
hence, the required fraction is 7/12
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