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

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

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

solve for

x, y = 1 + 2 + 3 x

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

Solution steps

y = 1 + 2 + 3 x

Switch sides

1 + 2 + 3 x = y

Move

1

to the right side

2 + 3 x = y - 1

Square both sides:

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

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

Solve

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

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

Verify Solutions:

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

The solution is

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

x^{0.3} x = 6 x - 2

4 a^{2} - 4 a + 1 + [a + 4] = a + 7

11 - x = x - 5

5 x + 6 + 7 = 3

2 x + 5 + 3 x - 14 = 9

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