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) {0<= x<= 2},y=-sqrt(-x^2+4) {0<= x<= 2}

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

solve for

y, x = 4 - y^{2}

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

Solution steps

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

The solutions are

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

2 x - 6 - x - 5 = x - 1

2 x + 8 = 3 x - 2

2 x + 8 = - 5

- 6 + x + 10 = 2

15 = 6 + 3 w - 9

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) {0<= x<= 2},y=-sqrt(-x^2+4) {0<= x<= 2}