SOLUTION: The sum of the squares of two numbers is 4 and the difference of the squares of those two numbers is 1. Find the two numbers. I think the two equations are: Let the numbers be

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Question 765635: The sum of the squares of two numbers is 4 and the difference of the squares of those two numbers is 1. Find the two numbers.
I think the two equations are:
Let the numbers be x and y
x(square) + y(square) = 4
x(square) - y(square) = 1
That is the equations that i got but the answer is a decimal...
I'm also not sure about the answer and the equation... please help
Thanks
Found 2 solutions by mananth, Theo:
Answer by mananth(16946) (Show Source):
You can put this solution on YOUR website!
x^2+y^2=4
x^2-y^2=1
add the two equations
2x^2=5
/2
x^2=5/2
x= +/- (5/2)
plug x=(5/2) in the first equations
(5/2)^2+y^2=4
25/4 +y^2=4
y^2=4-(25/4)
y^2=-(9/4)
y=(3/2)I
x=(5/2) , y= (3/2)i

Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website!
you got the equations right.
all you had to do was solve them simultaneously.
your equations are:
x^2 + y^2 = 4 (e1)
x^2 - y^2 = 1 (e2)
add e1 to e2 to get:
2x^2 = 5
divide both sides of this equation by 2 to get:
x^2 = 5/2
substitute for x^2 in the first equation to get:
x^2 + y^2 = 4 becomes:
5/2 + y^2 = 4
solve for y^2 to get:
y^2 = 3/2
those are your answers.
x^2 = 5/2 and y^2 = 3/2
if you take the square root of 5/2, you will get 1.58113883
if you take the square root of 3/2, you will get 1.224744871
those are your solutions for x and y
x = 1.58113883
y = 1.224744871
you can then confirm by substituting for x and y in the original equations and those equations should hold true.
i checked them out and they do.