## Algebra: A Combined Approach (4th Edition)

$2,6$
lets call the numbers $x, y$ $x+y = 8$ $x\times y = 12$ $x+y = 8$ Therefore $x=8- y$ $x\times y =12$ Therefore $(8-y)\times y = 12$ $8y - y^2 = 12$ $- y^2 + 8y - 12= 0$ $- (y - 6) (y-2)$ $y = 6, 2$ $x = 2, 6$ Therefore the numbers are $2,6$