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

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

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

solve for

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

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

Solution steps

x = - 1 - y^{2}

Divide both sides by

- 1

\frac{x}{- 1} = \frac{- 1 - y ^{2}}{- 1}

Simplify

- x = 1 - y^{2}

Square both sides:

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

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

Solve

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

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

Verify Solutions:

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

The solutions are

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

x - 1 = 1

x - 1 = 5

5 - 1 + q = 2 q + 3

3 x - 14 - 8 = 0

x^{\frac{4}{5}} = 16

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