I need to solve a pair of equations where one is linear and the other is quadratic.
2x2 + 3xy + y2 = 0
3x + y = 2
last 2 in 2x2 is square root two, same after y2.
help please!!!!!!!!!!!!!!!
just substitute the linear into the quadratic:
y = 2-3x
2x^2 + 3x(2-3x) + (2-3x)^2 = 0
2x^2 + 6x - 9x^2 + 4 - 12x + 9x^2 = 0
2x^2 - 6x + 4 = 0
2(x-1)(x-2) = 0
solutions: (1,-1)(2,-4)
or,
2x^2 + 3xy + y^2 = 0
(2x+y)(x+y) = 0
y = -2x or y = -x
3x+(-2x) = 2
x = 2, so y = -4
3x+(-x) = 2
x = 1, so y = -1
To solve this system of equations, we can either use substitution or elimination. Let's use the substitution method.
Starting with the first equation:
2x^2 + 3xy + y^2 = 0
We can rewrite this equation as:
(2x^2 + 3xy + y^2) = 0
(x + y)(2x + y) = 0
Thus, we have two possibilities:
1) x + y = 0
2) 2x + y = 0
Now, let's solve the second equation:
3x + y = 2
We can rearrange this equation to express y in terms of x:
y = 2 - 3x
Now, we can substitute this expression for y into the two possibilities for the first equation.
For possibility 1: x + y = 0
Substituting y = 2 - 3x, we get:
x + (2 - 3x) = 0
-2x + 2 = 0
-2x = -2
x = 1
Substituting x = 1 into y = 2 - 3x, we get:
y = 2 - 3(1)
y = 2 - 3
y = -1
Thus, the solution to this possibility is (x, y) = (1, -1).
For possibility 2: 2x + y = 0
Substituting y = 2 - 3x, we get:
2x + (2 - 3x) = 0
2 - x = 0
x = 2
Substituting x = 2 into y = 2 - 3x, we get:
y = 2 - 3(2)
y = 2 - 6
y = -4
Thus, the solution to this possibility is (x, y) = (2, -4).
Therefore, the system of equations has two solutions: (1, -1) and (2, -4).