solve system of equations:
y = 2x^3 - 2x^2
y = -6x^2 +30x
I know (0,0) is one point.
How do I get others?
OK, I got it.
y=y, so set the equations equal
2x^3-2x^2=-6x^2+30x
2x^3+4x^2-30x=0
2x(x^2+2x-15)=0
2x(x-3)(x+5)=0
x=0/ x=3/x=-5
find the y for each of those three points.
To find additional points on the graph of the given system of equations, you need to find the values of x and y that satisfy both equations simultaneously. There are several methods to solve a system of equations, but in this case, we'll use the substitution method.
Step 1: Set the two equations equal to each other.
2x^3 - 2x^2 = -6x^2 + 30x.
Step 2: Simplify the equation.
2x^3 - 2x^2 + 6x^2 - 30x = 0.
2x^3 + 4x^2 - 30x = 0.
Step 3: Factor out the greatest common factor.
2x(x^2 + 2x - 15) = 0.
Step 4: Solve for x by setting each factor equal to zero.
2x = 0 --> x = 0.
x^2 + 2x - 15 = 0.
Step 5: Solve the quadratic equation for x by factoring or using the quadratic formula.
(x + 5)(x - 3) = 0.
So, either x + 5 = 0 or x - 3 = 0.
x + 5 = 0 --> x = -5.
x - 3 = 0 --> x = 3.
Now that you have three values of x (0, -5, 3), substitute them into either of the original equations to find the corresponding values of y.
For x = 0:
y = 2(0)^3 - 2(0)^2
y = 0 - 0
y = 0.
So, the point (0, 0) is indeed a solution.
For x = -5:
y = 2(-5)^3 - 2(-5)^2
y = 2(-125) - 2(25)
y = -250 - 50
y = -300.
So, the point (-5, -300) is another solution.
For x = 3:
y = 2(3)^3 - 2(3)^2
y = 2(27) - 2(9)
y = 54 - 18
y = 36.
Thus, the point (3, 36) is also a solution.
In summary, the system of equations has three solutions: (0, 0), (-5, -300), and (3, 36).