1 & -4
You have two equations:
x+y = -3
x*y = -4
and two unknowns (x & y), so this is solvable. One way, is to solve for y in the 1st equation,
x+y = -3 ===> y = -3 - x
and plug it into the 2nd:
-4 = x*y = x*(-3 - x) = -3x - x2
==> x2 + 3x - 4 = 0
Then use the quadratic formula to solve for x:
x = 1/2 * ( -3 +- Sqrt( 32 - 4*(-4) ) )
= 1/2 * ( -3 +- Sqrt( 9 + 16 ) ) = 1/2 * ( -3 +- 5 )
= (for positive root) 1/2 * 2 = 1
= (for negative root) 1/2 * -8 = -4
So x can be either 1 or -4 (and y = -3-x will be -4 or 1). Yay math!
The numbers are 3 and 4
The numbers are: 9 and -6
9
7 & 3
-3
There are no two real numbers that do. Using complex numbers, these two do: (-3/2 + i√151/2) & (-3/2 - i√151/2) Two numbers that add to -3 and multiply to -40 are -8 & 5 Two numbers that add to 3 and multiply to -40 are 8 & -5 Two complex numbers that add to 3 and multiply to 40 are (3/2 + i√151/2) & (3/2 - i√151/2)
83
-1
9 and 6 but subtract not add
The numbers are 3 and 4
The numbers are: 9 and -6
9
-1
109
-10 and -3
7 & 3
-3+1