Answer:
(x = 1/7, y = 2) and (x = 1, y = 2).
Explanation:
To solve the system of equations y - 2x = 0 and x² - 2xy + 4x + 2y = y² + 1, we will first manipulate one of the equations to express one variable in terms of the other and then substitute it into the other equation.
Step 1: Solve y - 2x = 0 for y.
y = 2x
Step 2: Substitute y = 2x into the second equation x² - 2xy + 4x + 2y = y² + 1.
x² - 2x(2x) + 4x + 2(2x) = (2x)² + 1
x² - 4x² + 4x + 4x = 4x² + 1
-3x² + 8x = 4x² + 1
Step 3: Bring all terms to one side of the equation to form a quadratic equation.
-3x² - 4x² + 8x - 1 = 0
-7x² + 8x - 1 = 0
Step 4: Solve the quadratic equation for x. We can use the quadratic formula: x = (-b ± √(b² - 4ac)) / 2a.
In this case, a = -7, b = 8, and c = -1.
x = [-(8) ± √((8)² - 4(-7)(-1))] / 2(-7)
x = [-8 ± √(64 - 28)] / -14
x = [-8 ± √36] / -14
x = [-8 ± 6] / -14
Step 5: Find the two possible values of x.
a) x = (-8 + 6) / -14
x = -2 / -14
x = 1/7
b) x = (-8 - 6) / -14
x = -14 / -14
x = 1
So, the two possible values of x are x = 1/7 and x = 1.
Step 6: Find the corresponding values of y using y = 2x (from Step 1).
a) When x = 1/7,
y = 2(1/7)
y = 2/7
b) When x = 1,
y = 2(1)
y = 2
So, the two possible solutions to the system of equations are (x = 1/7, y = 2) and (x = 1, y = 2).