SOLUTION: find two numbers whose sum is 12 and whose difference between their squares is 48

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Question 557420: find two numbers whose sum is 12 and whose difference between their squares is 48
Answer by rajagopalan(174) About Me  (Show Source):
You can put this solution on YOUR website!
sum is 12
difference between their squares is 48
let the numbers be x and y
we form eqn 1 with x+y=12
we form eqn 2 with (x^2-y^2)=48 or (x+y)*(x-y)=48
divide eqn 2 by eqn 1
we get (x+y)*(x-y)/(x+y)=(x-y)=48/12=4
x+y=12...eqn 1
x-y=4....eqn 3
adding eqns 1 and 3 we get 2x=16
x=16/2=8
naturally y=4
verify X^2-Y^2=64-16=48 ok
cheers