What is the answer to solvefor y,x=1+sqrt(2+3y) ?

The answer to solvefor y,x=1+sqrt(2+3y) is y=(x^2-2x-1)/3 {x>= 1}

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solvefor y,x=1+sqrt(2+3y)

solve for

y, x = 1 + 2 + 3 y

y = \frac{x ^{2} - 2 x - 1}{3} {x \geq 1}

Solution steps

x = 1 + 2 + 3 y

Switch sides

1 + 2 + 3 y = x

Move

1

to the right side

2 + 3 y = x - 1

Square both sides:

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

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

Solve

2 + 3 y = x^{2} - 2 x + 1 : y = \frac{x ^{2} - 2 x - 1}{3}

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

Verify Solutions:

y = \frac{x ^{2} - 2 x - 1}{3} {x \geq 1}

The solution is

y = \frac{x ^{2} - 2 x - 1}{3} {x \geq 1}

x + 72 = x

500 = 3 x + 10 x + 140

1 + x - 3 = 0

y - 3 = - 5 y + 65

x - 2 - 8 = 0

What is the answer to solvefor y,x=1+sqrt(2+3y) ?
The answer to solvefor y,x=1+sqrt(2+3y) is y=(x^2-2x-1)/3 {x>= 1}