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

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

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

solve for

x, y = 2 x + 1

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

Solution steps

y = 2 x + 1

Square both sides:

y^{2} = 2 x + 1

y^{2} = 2 x + 1

Solve

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

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

Verify Solutions:

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

The solution is

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

9 x + 145 = 2 x + 20

a + 5 = a + 44

x - 5 + 2 x - 9 = 2 + 4 x - 36

θ, r = 1 - 2 θ

x + 3 - 2 = x

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