What is the answer to solvefor y,x=sqrt(y^2+9) ?

The answer to solvefor y,x=sqrt(y^2+9) is y=sqrt(x^2-9) {x>= 3},y=-sqrt(x^2-9) {x>= 3}

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solvefor y,x=sqrt(y^2+9)

solve for

y, x = y^{2} + 9

y = x^{2} - 9 {x \geq 3}, y = - x^{2} - 9 {x \geq 3}

Solution steps

x = y^{2} + 9

Square both sides:

x^{2} = y^{2} + 9

x^{2} = y^{2} + 9

Solve

x^{2} = y^{2} + 9 : y = x^{2} - 9, y = - x^{2} - 9

y = x^{2} - 9, y = - x^{2} - 9

Verify Solutions:

y = x^{2} - 9 {x \geq 3}, y = - x^{2} - 9 {x \geq 3}

The solutions are

y = x^{2} - 9 {x \geq 3}, y = - x^{2} - 9 {x \geq 3}

x - 2 = \frac{10}{x - 2} - 3

- 7 = 8 - 3^{5} x^{3}

x^{2} - 3 x + 2 = x - 1

3 x^{\frac{5}{2}} - 28 = 68

x + 3 = x + 11

What is the answer to solvefor y,x=sqrt(y^2+9) ?
The answer to solvefor y,x=sqrt(y^2+9) is y=sqrt(x^2-9) {x>= 3},y=-sqrt(x^2-9) {x>= 3}