Solve the system of equations by elimination. (2 marks)
y=2x^2-2x-3
y=-x^2-2x-3
just equate the two:
2x^2-2x-3 = -x^2-2x-3
2x^2 = -x^2
x = 0
so, y = -3
To solve the system of equations by elimination, we need to eliminate one variable by adding or subtracting the equations. Let's eliminate the y variable.
Given equations:
1) y = 2x^2 - 2x - 3
2) y = -x^2 - 2x - 3
To eliminate y, we can subtract equation 2 from equation 1:
2x^2 - 2x - 3 - (-x^2 - 2x - 3) = 0
Simplifying the equation:
2x^2 - 2x - 3 + x^2 + 2x + 3 = 0
3x^2 = 0
Now we can solve for x by dividing both sides of the equation by 3:
x^2 = 0
Taking the square root of both sides:
x = 0
Now, let's substitute the value of x back into one of the original equations (e.g., equation 1) to solve for y:
y = 2(0)^2 - 2(0) - 3
y = 0 - 0 - 3
y = -3
Therefore, the solution to the system of equations is x = 0 and y = -3.
To solve the system of equations by elimination, we'll eliminate one variable and solve for the other.
Given equations:
1) y = 2x^2 - 2x - 3
2) y = -x^2 - 2x - 3
Since both equations have "y" isolated on one side, we can set them equal to each other as follows:
2x^2 - 2x - 3 = -x^2 - 2x - 3
Now, we will simplify the equation by combining like terms:
2x^2 - x^2 - 2x - (-2x) - 3 - (-3) = 0
This simplifies to:
x^2 = 0
Taking the square root of both sides, we get:
x = 0
Now, substitute the value of x into either of the original equations to solve for y. Let's use equation 1:
y = 2(0)^2 - 2(0) - 3
y = 0 - 0 - 3
y = -3
Therefore, the solution to the system of equations is x = 0 and y = -3.