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

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

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

solve for

y, x = 2 y - 1

y = \frac{x ^{2} + 1}{2} {x \geq 0}

Solution steps

x = 2 y - 1

Square both sides:

x^{2} = 2 y - 1

x^{2} = 2 y - 1

Solve

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

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

Verify Solutions:

y = \frac{x ^{2} + 1}{2} {x \geq 0}

The solution is

y = \frac{x ^{2} + 1}{2} {x \geq 0}

(x - 3)^{\frac{7}{5}} = 128

- 2 x - 3 + x + 3 = 3

x^{0.2} = 5

4 x = x - 8

x, 2 y + 6 = x - 3

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