What is the answer to solvefor y,x=sqrt(-y^2+16) ?

The answer to solvefor y,x=sqrt(-y^2+16) is y=sqrt(-x^2+16) {0<= x<= 4},y=-sqrt(-x^2+16) {0<= x<= 4}

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solvefor y,x=sqrt(-y^2+16)

solve for

y, x = - y^{2} + 16

y = - x^{2} + 16 {0 \leq x \leq 4}, y = - - x^{2} + 16 {0 \leq x \leq 4}

Solution steps

x = - y^{2} + 16

Square both sides:

x^{2} = - y^{2} + 16

x^{2} = - y^{2} + 16

Solve

x^{2} = - y^{2} + 16 : y = - x^{2} + 16, y = - - x^{2} + 16

y = - x^{2} + 16, y = - - x^{2} + 16

Verify Solutions:

y = - x^{2} + 16 {0 \leq x \leq 4}, y = - - x^{2} + 16 {0 \leq x \leq 4}

The solutions are

y = - x^{2} + 16 {0 \leq x \leq 4}, y = - - x^{2} + 16 {0 \leq x \leq 4}

7 - X = 5 X - 29

7 - x - 19 + x = - 2

2 x + 5 - 3 = 0

x^{\frac{1}{2}} = \frac{1}{3}

What is the answer to solvefor y,x=sqrt(-y^2+16) ?
The answer to solvefor y,x=sqrt(-y^2+16) is y=sqrt(-x^2+16) {0<= x<= 4},y=-sqrt(-x^2+16) {0<= x<= 4}