Question

One positive integer is 4 less than twice another. The sum of their squares is 100. Find the integers.

Answer 1

x = 8, y = 6

Explanation:

Let positive integer = x

another = y

x = 2y-4

x^2+ y^2 = 100

(2y-4)^2 + y^2 = 100

4y^2 - 16 y + 16 + y^2 = 100

5y^2 - 16 y + 16 - 100 = 0

5y^2 - 16y - 84 = 0

(5y + 14) ( y - 6) = 0

5y = -14 or y = 6

positive so we can ingnore 5y = -14

y = 6 in x = 2y - 4,

x = 12-4

x= 8

so, y = 6

x = 8