1. x = 31
2. y = 5 + 23
3.
4. y = -7
5. x = -2 + 13
I think it's a^2 b^2 + 6ab + 9
9514 1404 393
Answer:
(3, 1)
Step-by-step explanation:
We assume you want the solution to the system ...
The second equation gives a nice expression for x, so we can use that in the first equation.
2(y+2) -3y = 3 . . . . substitute for x in the first equation
2y +4 -3y = 3 . . . . . eliminate parentheses
-y = -1 . . . . . . . . . . . collect terms, subtract 4
y = 1 . . . . . . . . . . . . multiply by -1
x = 1 +2 = 3 . . . . . . substitute for y in the second equation
The solution is (x, y) = (3, 1).
Quadratic formula is x=(-b+-sqrt(b^2-4ac))/(2a)
a= 1, b=-4, and c=3
Therefore, x=(-(-4)+-sqrt((-4)^2-4(1)(3))/(2(1))
Solve for both the plus and minus equations; the answers will be the roots to the quadratic equation.