What is the answer to solvefor y,x=-sqrt(4-y^2) ?

The answer to solvefor y,x=-sqrt(4-y^2) is y=sqrt(-x^2+4) {-2<= x<= 0},y=-sqrt(-x^2+4) {-2<= x<= 0}

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solvefor y,x=-sqrt(4-y^2)

solve for

y, x = - 4 - y^{2}

y = - x^{2} + 4 {- 2 \leq x \leq 0}, y = - - x^{2} + 4 {- 2 \leq x \leq 0}

Solution steps

x = - 4 - y^{2}

Divide both sides by

- 1

\frac{x}{- 1} = \frac{- 4 - y ^{2}}{- 1}

Simplify

- x = 4 - y^{2}

Square both sides:

x^{2} = 4 - y^{2}

x^{2} = 4 - y^{2}

Solve

x^{2} = 4 - y^{2} : y = - x^{2} + 4, y = - - x^{2} + 4

y = - x^{2} + 4, y = - - x^{2} + 4

Verify Solutions:

y = - x^{2} + 4 {- 2 \leq x \leq 0}, y = - - x^{2} + 4 {- 2 \leq x \leq 0}

The solutions are

y = - x^{2} + 4 {- 2 \leq x \leq 0}, y = - - x^{2} + 4 {- 2 \leq x \leq 0}

x + 1 = 5

y, x = y^{\frac{2}{3}}

\frac{4}{x - 1} = 2

3 x - 2 + 2 - x = 0

What is the answer to solvefor y,x=-sqrt(4-y^2) ?
The answer to solvefor y,x=-sqrt(4-y^2) is y=sqrt(-x^2+4) {-2<= x<= 0},y=-sqrt(-x^2+4) {-2<= x<= 0}