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shubh110
@shubh110
September 2021
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Two numbers differ by 3 and their product is 504 . find the number
Answers & Comments
raj962
Let two numbers be x and y
x - y = 3 ------(1)
and xy = 504 ------(2)
then from first x = y + 3 -------(3)
Putting value of eq(3) in eq (2)
(y + 3) y = 504
y2 + 3y - 504 = 0
y2 + ( 24 - 21 ) y -504 = 0
y2 + 24y -21y -504 = 0
y ( y + 24 ) - 21 ( y + 24 ) = 0
( y - 21 ) ( y + 24 )
y = 21 or y = -24
then putting this value of y in eq(1)
we get x = 24 or x = - 21
3 votes
Thanks 7
surendarrajawat
Hey
I know it's a little messed up. Sorry for that.
Hope it helps :)
4 votes
Thanks 6
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Answers & Comments
x - y = 3 ------(1)
and xy = 504 ------(2)
then from first x = y + 3 -------(3)
Putting value of eq(3) in eq (2)
(y + 3) y = 504
y2 + 3y - 504 = 0
y2 + ( 24 - 21 ) y -504 = 0
y2 + 24y -21y -504 = 0
y ( y + 24 ) - 21 ( y + 24 ) = 0
( y - 21 ) ( y + 24 )
y = 21 or y = -24
then putting this value of y in eq(1)
we get x = 24 or x = - 21
I know it's a little messed up. Sorry for that.
Hope it helps :)