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The sum of the squares of two natural numbers is 58. The difference of their squares is 40. What are the two natural numbers?

Answer 1

The numbers are #7# and #3#.

We let the numbers be #x# and #y#.

#{(x^2 + y^2 = 58), (x^2 - y^2 = 40):}#

We can solve this easily using elimination, noticing that the first #y^2# is positive and the second is negative. We are left with:

#2x^2 = 98#

#x^2 = 49#

#x = +-7#

However, since it is stated that the numbers are natural, that's to say greater than #0#, #x = + 7#.

Now, solving for #y#, we get:

#7^2 + y^2 = 58#

#y^2 = 9#

#y = 3#

Hopefully this helps!

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.