Question 30339: Quesion:
Solve x^2+y^2=9
x+y=3
(a) (3, -3) (-3, -3)
(b) (0, -3) (-3, 0)
(c) (0,0) (3, -3)
(d) 0, 3) (3, 0)
Answer by sdmmadam@yahoo.com(530)   (Show Source):
You can put this solution on YOUR website! If the given problem is a 1 mark question then you should not try the traditional mathematical analysis given below. You must just try the validity of the choices by the trial and error method.
You observe that the choice d) holds
Solve x^2+y^2=9 ----(1)
x+y = 3 ----(2)
Squaring (2)
(x+y)^2 = 9
x^2+y^2 +2xy = 9
(x^2+y^2) +2xy = 9
9 +2xy =9 (using (1))
2xy = 9 -9
2xy = 0
xy =0/2 = 0
xy = 0 ----(*) implies atleast one of x or y is 0
We have by formula
(x-y)^2 = (x+y)^2-4xy
(x-y)^2 = 3^2-0 (using (1) and (2) )
(x-y)^2 =9
(x-y) = +3 or -3 ----(**)
Consider x-y = 3 ----(3)
We have x+y = 3 ----(2)
(3)+(2) gives (x+x) = (3+3) =6
2x=6
x=6/2 = 3
x= 3 in (3) or in (2) implies y= 0
Answer: x = 3 and y=0
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Note: Supposing we consider the other value for (x-y)in (**)
That is x-y =-3 ----(4)
We have x+y = 3 ----(2)
(4)+(2) gives
2x =0 giving x= 0 and putting x=0 in (2) we get y=3
So the answer is x=0 and y= 3
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Just the role of x and y interchanged.
Of the given choices
the choice d) holds
Note: We don't even have to consider the formula
(x-y)^2 = (x+y)^2-4xy
as in other problems and then find (x-y) for solving the values of x and y
because from (*) we have already concluded that one of x or y is ZERO
and so if we take y= 0 and put it in (2),we get (it orally even) that x=3
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