Asked by John
Solve the system of equations
y=x^2-2x+1
y-2x=1
y=x^2-2x+1
y-2x=1
Answers
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Answered by
GPT 3.5
To solve the system of equations, we can substitute the value of y from the second equation into the first equation:
y - 2x = 1
Subtracting 1 from both sides:
y - 1 - 2x = 0
Substituting y = 2x - 1 into the first equation:
2x - 1 - 2x = x^2 - 2x + 1
Simplifying:
-1 = x^2 - 2x + 1
Rearranging and combining like terms:
x^2 - 2x + 2 = 0
This is a quadratic equation. To solve for x, we can use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
Here, a = 1, b = -2, and c = 2. Plugging these values into the quadratic formula:
x = (-(-2) ± √((-2)^2 - 4(1)(2))) / (2(1))
x = (2 ± √(4 - 8)) / 2
x = (2 ± √(-4)) / 2
Since √(-4) is not a real number, the quadratic equation does not have any real solutions. Therefore, the system of equations is inconsistent and has no solution.
y - 2x = 1
Subtracting 1 from both sides:
y - 1 - 2x = 0
Substituting y = 2x - 1 into the first equation:
2x - 1 - 2x = x^2 - 2x + 1
Simplifying:
-1 = x^2 - 2x + 1
Rearranging and combining like terms:
x^2 - 2x + 2 = 0
This is a quadratic equation. To solve for x, we can use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
Here, a = 1, b = -2, and c = 2. Plugging these values into the quadratic formula:
x = (-(-2) ± √((-2)^2 - 4(1)(2))) / (2(1))
x = (2 ± √(4 - 8)) / 2
x = (2 ± √(-4)) / 2
Since √(-4) is not a real number, the quadratic equation does not have any real solutions. Therefore, the system of equations is inconsistent and has no solution.
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