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jiaming
Posts:
1
From:
Singapore
Registered:
10/11/11
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Simultaneous equation and factorisation
Posted:
Oct 11, 2011 8:58 AM
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A positive whole number has two digits
When the two digits are reversed, a new number is formed.
The difference between the squares of the tow numbers is 2376.
The sum of the two numbers is 66 times the difference between the digits of the original number.
Find the two numbers.
Is there a way to solve it by simple factorisation or it has to be solved using simultaneous equations (e.g. let x be 10 and y be 5 etc)
The algebra one is like this, but can it be solved by simple factorisation?
Alrighty, so we set X and Y to be the two digits of the number. Let X>Y. So our first number, if we assume it is the larger of the two, is 10x+y and its reversal is x+10y
The sum of the two numbers is 66 times the difference between the digits
So that gives us
10x+y + x+10y = 66(x-y)
11(x+y) = 66(x-y)
(x+y)=6(x-y) -> Eq A
The difference between the squares is 2376, so we have
(10x+y)^2 - (x+10y)^2 = 2376
100x^2 + 20xy + y^2 - x^2 - 20xy - 100y^2 = 2376
99x^2 - 99y^2 = 2376
x^2 - y^2 = 24
(x+y)*(x-y) = 24
substitute Eq A
6(x-y)*(x-y) = 24
(x-y)^2 = 4
x-y = 2 (remember we defined x>y)
Plug that into Eq A
x+y = 12
(x-y) + (x+y) = 2 + 12
2x = 14, x = 7, ergo y = 5
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