What is the answer to solvefor x,sqrt(3x)-2y=4 ?

The answer to solvefor x,sqrt(3x)-2y=4 is x=(16+16y+4y^2)/3 {y>=-2}

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solvefor x,sqrt(3x)-2y=4

solve for

x, 3 x - 2 y = 4

x = \frac{16 + 16 y + 4 y ^{2}}{3} {y \geq - 2}

Solution steps

3 x - 2 y = 4

Move

2 y

to the right side

3 x = 4 + 2 y

Square both sides:

3 x = 16 + 16 y + 4 y^{2}

3 x = 16 + 16 y + 4 y^{2}

Solve

3 x = 16 + 16 y + 4 y^{2} : x = \frac{16 + 16 y + 4 y ^{2}}{3}

x = \frac{16 + 16 y + 4 y ^{2}}{3}

Verify Solutions:

x = \frac{16 + 16 y + 4 y ^{2}}{3} {y \geq - 2}

The solution is

x = \frac{16 + 16 y + 4 y ^{2}}{3} {y \geq - 2}

0 = x x^{2} - 9

x - 6 = 6

x - 6 = x

x + 2 = x - 2

10 + 3 a = a

What is the answer to solvefor x,sqrt(3x)-2y=4 ?
The answer to solvefor x,sqrt(3x)-2y=4 is x=(16+16y+4y^2)/3 {y>=-2}