Solve the following equations algebraically and check.
y=2x^2+2x+3
y=x+3
To solve the equations algebraically, we need to find the values of x and y that satisfy both equations. Let's begin by setting the two equations equal to each other:
2x^2 + 2x + 3 = x + 3
Next, simplify the equation by combining like terms:
2x^2 + x = 0
Now, factor out the common factor, which in this case is x:
x(2x + 1) = 0
To find the values of x, set each factor equal to zero and solve individually:
x = 0 or 2x + 1 = 0
If x = 0, substitute this value back into either of the original equations to find the corresponding y value:
y = 2(0)^2 + 2(0) + 3
y = 0 + 0 + 3
y = 3
If 2x + 1 = 0, we can solve for x:
2x = -1
x = -1/2
Now, substitute this value back into either of the original equations to find the corresponding y value:
y = (-1/2) + 3
y = 5/2
Therefore, the solutions for the system of equations are (x, y) = (0, 3) and (x, y) = (-1/2, 5/2).
To check if these solutions are correct, substitute them into both original equations and verify if the left-hand side matches the right-hand side.
Checking the first equation:
For (x, y) = (0, 3):
Left-hand side: y = 2(0)^2 + 2(0) + 3 = 3
Right-hand side: x + 3 = 0 + 3 = 3
Both sides are equal, so the first solution checks out.
Checking the second equation:
For (x, y) = (-1/2, 5/2):
Left-hand side: y = (-1/2) + 3 = 5/2
Right-hand side: x + 3 = (-1/2) + 3 = 5/2
Both sides are equal, so the second solution checks out as well.
Therefore, the solutions are correct for both equations.