what is the solution of the linear-quadratic system of equations?
y=x^2-2x+1
y-2x=1
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To find the solution of the linear-quadratic system of equations, we need to solve the equations simultaneously.
First, we have the equation:
y = x^2 - 2x + 1 ------(1)
And the second equation:
y - 2x = 1 ------(2)
We can substitute the value of y from equation (2) into equation (1) to eliminate y:
y - 2x = 1
Therefore, we get:
x^2 - 2x + 1 - 2x = 1
x^2 - 4x + 1 = 1
x^2 - 4x = 0
We can factor out an x:
x(x - 4) = 0
This implies that either x = 0 or x - 4 = 0.
So, we have two possible solutions for x:
x = 0 or x = 4
Now, let's substitute these values of x back into one of the original equations to find the corresponding y-values:
For x = 0:
y = (0)^2 - 2(0) + 1
y = 0 - 0 + 1
y = 1
So, one solution is (x, y) = (0, 1)
For x = 4:
y = (4)^2 - 2(4) + 1
y = 16 - 8 + 1
y = 9
So, another solution is (x, y) = (4, 9)
Therefore, the solution to the linear-quadratic system of equations is:
(x, y) = (0, 1) and (4, 9)