What is the answer to solvefor x,2y=sqrt(2x-x+1+4) ?

The answer to solvefor x,2y=sqrt(2x-x+1+4) is x=4y^2-5 {y>= 0}

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solvefor x,2y=sqrt(2x-x+1+4)

solve for

x, 2 y = 2 x - x + 1 + 4

x = 4 y^{2} - 5 {y \geq 0}

Solution steps

2 y = 2 x - x + 1 + 4

Square both sides:

4 y^{2} = x + 5

4 y^{2} = x + 5

Solve

4 y^{2} = x + 5 : x = 4 y^{2} - 5

x = 4 y^{2} - 5

Verify Solutions:

x = 4 y^{2} - 5 {y \geq 0}

The solution is

x = 4 y^{2} - 5 {y \geq 0}

2 x + 22 + 9 = x + 8

\frac{9}{x} = (4 - x)^{2}

2 x = 2 x - 4

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

(x^{2} + 5) (x + 2) = x

What is the answer to solvefor x,2y=sqrt(2x-x+1+4) ?
The answer to solvefor x,2y=sqrt(2x-x+1+4) is x=4y^2-5 {y>= 0}