To solve the system of equations:
- ( x + 2y = 1 ) (Equation 1)
- ( 2x + y^2 = -1 ) (Equation 2)
We can solve this system using substitution or elimination. Let's use substitution by first solving Equation 1 for ( x ):
From Equation 1:
[
x = 1 - 2y
]
Now, substitute this expression for ( x ) into Equation 2:
[
2(1 - 2y) + y^2 = -1
]
Now simplify this:
[
2 - 4y + y^2 = -1
]
Rearranging gives:
[
y^2 - 4y + 2 + 1 = 0
]
[
y^2 - 4y + 3 = 0
]
Now we can factor or use the quadratic formula:
The quadratic can be factored as:
[
(y - 3)(y - 1) = 0
]
Setting each factor to zero gives:
[
y - 3 = 0 \quad \text{or} \quad y - 1 = 0
]
So,
[
y = 3 \quad \text{or} \quad y = 1
]
Now we can find the corresponding ( x ) values for each ( y ).
For ( y = 3 ):
[
x = 1 - 2(3) = 1 - 6 = -5
]
So one solution is ( (x, y) = (-5, 3) ).
For ( y = 1 ):
[
x = 1 - 2(1) = 1 - 2 = -1
]
So the other solution is ( (x, y) = (-1, 1) ).
Thus, the solutions to the system of equations are:
[
(x, y) = (-5, 3) \quad \text{and} \quad (x, y) = (-1, 1)
]
