One positive integer is 6 less than twice another. The sum of their squares is 680 . Find the integers.

Answer 1

Call them x and y. Their squares are x^2 and y^2.

x = 2y - 6

x^2 + y^2 = 680 (substitute for x from the first equation.)

(2y - 6)^2 + y^2 = 680

(2y - 6)^2 = 4y^2 - 24y + 36

4y^2 - 24y + 36 + y^2 = 680

5y^2 - 24y - 644 = 0

5 * -644 = -3220

Find two numbers that multiply to -3220 and add to -24

-3220 = -2 * 1610

-2 * 2 * 805

-2 * 2 * 5 * 161

-2 * 2 * 5 * 7 * 23

Any of the numbers can be negative.

-2 * 23 = -46 and 2*5*7 = 70. 70 - 46 = 24

-24 is needed, so 46 and -70

(y -70/5) (y+46/5)

(y -14)(y + 56/4)

The question stated only positive numbers. y-14=0

y = 14

x = 2y - 6 = 2*14 - 6 = 28 - 6 = 22

14^2 + 22^2 = 680 Check

x = 22, y = 14

Answer 2

Let "x" be the first positive integer and "y" be the second positive integer:

List system equations

x = 2y-6

x²+y² = 680

Substitute "x" for "y" in 2nd equation

(2y-6)²+y² = 680

4y²-24y+36+y² = 680

5y²-24y-644=0

5y²+46y-70y-644=0

(5y+46)(y-14) = 0

y = -46/5, 14 (positive integer solution only)

Substitute y=14 into original linear equation

x = 2y-6 = 2(14)-6 = 28-6 = 22

Thus, the two positive integers are 14 and 22