Maths Problemo?

Question

The sum of the squares of two natural numbers is 34. If the first number is one less than twice the second number, find the numbers.
HEEEEEEEEEEEELP!

Answer 1

Always start out by choosing a variable to represent something in the problem. Suppose you let x = the first nat # and y = the second nat #. Then x^2 + y^2 = 34 and x = 2y - 1. Using substitution we get (2y - 1)^2 + y^2 = 34
4y^2 - 4y +1 + y^2 = 34
5y^2 - 4y - 33 = 0
(5y + 11)(y - 3) = 0
5y + 11 =0 or y - 3 = 0
y = -11/5 or y = 3
Since -11/5 is not a natural # and 3 is a natural number, it follows that y = 3
x = 2(3) - 1 = 5

Check: 5^2 + 3^2 = 25 + 9 = 34 and 5 = 2(3) - 1

Answer 2

5 & 3
5 squared is 25, 3 squared is 9
25 + 9=34
5 is one less than twice the number 3 (twice the number 3 is 6)

Answer 3

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Answer 4

numbers are 5 and 3

3*3-1=5 (the value of the 2nd number)

3x3 = 9
5*5 = 25
25+9=34

Answer 5

a^2+b^2=34
a=2b-1
=> ( 4b^2-4b+1 ) + b^2 = 34
=> 5b^2-4b-33 = 0
=> b = 3
=> a = 5
Answer is 5 and 3

Answer 6

3 & 5 =)

Answer 7

I can't give you an answer bc I don't remember how to solve the problem... but here is a site i found that might help!

http://www.webmath.com/

GOOD LUCK

Answer 8

is it like algerbra or algerbra 1

Answer 9

i dont have any clue