One positive integer is 1 less than twice another. The sum of their squares is 157. Find the integers.

11 and 6

Step-by-step explanation:

Let the integers are x and y.

We have:

x = 2y - 1x² + y² = 157

Substitute x into the second equation and solve for y:

(2y - 1)² + y² = 1574y² - 4y + 1 + y² = 1575y² - 4y - 156 = 0D = (-4)² - 4*5*(-156) = 3136y = (4 ± √3136)/10 = (4 ± 56)/10y = 6 is the only integer solution

Find x:

x = 2*6 - 1 = 11

The answer simplified would be. -6300x

Idk what the answer is bc i don't get it