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

The answer to solvefor x,y-2=sqrt(x+1) is x=y^2-4y+3 {y>= 2}

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

solve for

x, y - 2 = x + 1

x = y^{2} - 4 y + 3 {y \geq 2}

Solution steps

y - 2 = x + 1

Square both sides:

y^{2} - 4 y + 4 = x + 1

y^{2} - 4 y + 4 = x + 1

Solve

y^{2} - 4 y + 4 = x + 1 : x = y^{2} - 4 y + 3

x = y^{2} - 4 y + 3

Verify Solutions:

x = y^{2} - 4 y + 3 {y \geq 2}

The solution is

x = y^{2} - 4 y + 3 {y \geq 2}

6 - 2 x + 5 = 0

3 - 2 x - x = 0

x - 4 = x - 4

x - 3 = 0

2 x + 1 - x = 1

What is the answer to solvefor x,y-2=sqrt(x+1) ?
The answer to solvefor x,y-2=sqrt(x+1) is x=y^2-4y+3 {y>= 2}