What is the answer to solvefor y,x=1+sqrt(1-y^2) ?

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

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solvefor y,x=1+sqrt(1-y^2)

solve for

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

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

Solution steps

x = 1 + 1 - y^{2}

Switch sides

1 + 1 - y^{2} = x

Move

1

to the right side

1 - y^{2} = x - 1

Square both sides:

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

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

Solve

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

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

Verify Solutions:

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

The solutions are

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

4 3 x + 7 = 4 2 x + 13

\frac{5 x - 12}{3 x ^{\frac{1}{3}}} = 0

2 x + x = 4

t + 7 t = 0

x^{2} - 6 x + 9 = 4

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