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

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

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

solve for

y, x = 2 - y^{2}

y = - x^{2} + 2 {0 < x < 2}, y = - - x^{2} + 2 {0 < x < 2}

Solution steps

x = 2 - y^{2}

Square both sides:

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

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

Solve

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

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

Verify Solutions:

y = - x^{2} + 2 {0 < x < 2}, y = - - x^{2} + 2 {0 < x < 2}

The solutions are

y = - x^{2} + 2 {0 < x < 2}, y = - - x^{2} + 2 {0 < x < 2}

1.24 = 2 π \times \frac{0.381}{g}

6 - k = 4

x + 58 - 9 = - 2

x + 8 = 5

5 y - 21 - y - 5 = 2

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