What is the answer to solvefor x,sqrt(3x)+y=0 ?

The answer to solvefor x,sqrt(3x)+y=0 is x=(y^2)/3 {y<= 0}

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solvefor x,sqrt(3x)+y=0

solve for

x, 3 x + y = 0

x = \frac{y ^{2}}{3} {y \leq 0}

Solution steps

3 x + y = 0

Move

y

to the right side

3 x = - y

Square both sides:

3 x = y^{2}

3 x = y^{2}

Solve

3 x = y^{2} : x = \frac{y ^{2}}{3}

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

Verify Solutions:

x = \frac{y ^{2}}{3} {y \leq 0}

The solution is

x = \frac{y ^{2}}{3} {y \leq 0}

4 x + 9 - 6 x = 3

x - 16 = x + 4

7 x - 3 x = 4

6 x^{2} - 50 = 4 x - 10

What is the answer to solvefor x,sqrt(3x)+y=0 ?
The answer to solvefor x,sqrt(3x)+y=0 is x=(y^2)/3 {y<= 0}