What is the answer to solvefor y,x=sqrt(2y-y^2) ?

The answer to solvefor y,x=sqrt(2y-y^2) is y=1-sqrt(-x^2+1) {0<= x<= 1},y=sqrt(-x^2+1)+1 {0<= x<= 1}

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solvefor y,x=sqrt(2y-y^2)

solve for

y, x = 2 y - y^{2}

y = 1 - - x^{2} + 1 {0 \leq x \leq 1}, y = - x^{2} + 1 + 1 {0 \leq x \leq 1}

Solution steps

x = 2 y - y^{2}

Square both sides:

x^{2} = 2 y - y^{2}

x^{2} = 2 y - y^{2}

Solve

x^{2} = 2 y - y^{2} : y = 1 - - x^{2} + 1, y = - x^{2} + 1 + 1

y = 1 - - x^{2} + 1, y = - x^{2} + 1 + 1

Verify Solutions:

y = 1 - - x^{2} + 1 {0 \leq x \leq 1}, y = - x^{2} + 1 + 1 {0 \leq x \leq 1}

The solutions are

y = 1 - - x^{2} + 1 {0 \leq x \leq 1}, y = - x^{2} + 1 + 1 {0 \leq x \leq 1}

x - 5 = 23 - x

2 x + 5 - x + 14 = - 1

10 - 4 y = 5

5 - x - 3 = 0

What is the answer to solvefor y,x=sqrt(2y-y^2) ?
The answer to solvefor y,x=sqrt(2y-y^2) is y=1-sqrt(-x^2+1) {0<= x<= 1},y=sqrt(-x^2+1)+1 {0<= x<= 1}