SOLUTION: Find two integers whose difference is 4 and whose square differ by 72

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Question 1052973: Find two integers whose difference is 4 and whose square differ by 72
Found 2 solutions by ikleyn, MathTherapy:
Answer by ikleyn(52778) About Me  (Show Source):
You can put this solution on YOUR website!
.
Answer. 11 and 7.

Solution 

x%5E2+-+y%5E2 = 72  --->

(x-y)*(x+y) = 72.

But x-y = 4, according to the condition. Hence,

4*(x+y) = 72  --->

x + y = 72%2F4 = 18.

So, you have two equations:

x + y = 18,
x - y =  4.

Add them.

From this point, please complete the solution on your own.


Answer by MathTherapy(10551) About Me  (Show Source):
You can put this solution on YOUR website!

Find two integers whose difference is 4 and whose square differ by 72
Let larger and smaller integers be L and S, respectively

We then get: system%28L+-+S+=+4%2C+L%5E2+-+S%5E2+=+72%29

Simplify to get: system%28L+-+S+=+4%2C+L+%2B+S+=+18%29

You should be able to finish this!