SOLUTION: If two positive numbers are squared, their sum is 521. If the first number is 2 less than 2 times the second number, what is the smaller number? Im not sure how to approach this

Algebra ->  Problems-with-consecutive-odd-even-integers -> SOLUTION: If two positive numbers are squared, their sum is 521. If the first number is 2 less than 2 times the second number, what is the smaller number? Im not sure how to approach this      Log On


   



Question 913520: If two positive numbers are squared, their sum is 521. If the first number is 2 less than 2 times the second number, what is the smaller number?
Im not sure how to approach this question on how to answer it because Im not sure where to start. Help would be appreciated! thank you!

Answer by CubeyThePenguin(3113) About Me  (Show Source):
You can put this solution on YOUR website!
x^2 + y^2 = 521
x = 2y - 2

Substitute x = 2y - 2 into the first equation.

(2y - 2)^2 + y^2 = 521
4y^2 - 8y + 4 + y^2 = 521
5y^2 - 8y - 517 = 0
(5y + 47)(y - 11) = 0

y is positive, so y = 11 and x = 20.

The smaller integer = y = 11.