what is the solution of the linear-quadratic system of equations?
y = x^2 - 2x + 1
y - 2x = 1
To find the solution of the linear-quadratic system of equations, we can substitute y from the first equation into the second equation:
x^2 - 2x + 1 - 2x = 1
Simplifying this equation, we get:
x^2 - 4x + 1 = 1
Subtracting 1 from both sides:
x^2 - 4x = 0
Factoring out x, we have:
x(x - 4) = 0
From the Zero Product Property, we can set each factor equal to zero:
x = 0 or x - 4 = 0
Solving for x, we find:
x = 0 or x = 4
Now, substituting these x-values into either of the original equations, we can find the corresponding y-values.
For x = 0:
y = 0^2 - 2(0) + 1
y = 1
So, one solution is (0, 1).
For x = 4:
y = 4^2 - 2(4) + 1
y = 16 - 8 + 1
y = 9
Therefore, another solution is (4, 9).
Hence, the solution of the linear-quadratic system of equations is (0, 1) and (4, 9).