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

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

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

solve for

y, x = 1 + 2 y

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

Solution steps

x = 1 + 2 y

Switch sides

1 + 2 y = x

Move

1

to the right side

2 y = x - 1

Square both sides:

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

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

Solve

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

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

Verify Solutions:

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

The solution is

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

8 x + 97 - 15 = - 8

4 - 3 x = x + 8

x, x + y = 6

1 - x^{2} = 2 - x

9 x + 1 = 10 x + 9

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