What is the answer to solvefor x,3y^3+3xy+4x=0 ?

The answer to solvefor x,3y^3+3xy+4x=0 is x=-(3y^3)/(3y+4);\quad y\ne-4/3

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solvefor x,3y^3+3xy+4x=0

solve for

x, 3 y^{3} + 3 x y + 4 x = 0

x = - \frac{3 y ^{3}}{3 y + 4}; y \neq = - \frac{4}{3}

Solution steps

3 y^{3} + 3 x y + 4 x = 0

Move

3 y^{3}

to the right side

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

Factor

3 x y + 4 x : x (3 y + 4)

x (3 y + 4) = - 3 y^{3}

Divide both sides by

3 y + 4; y \neq = - \frac{4}{3}

x = - \frac{3 y ^{3}}{3 y + 4}; y \neq = - \frac{4}{3}

5 a (5 a + x) = 2 a (2 a - x)

7 (x - 1) + 4 = 4 x - 3

y + 2 + 3 y = 2 y - 6

2 lo g_{6} (6) + 4 x = 10

8 s - 1 = 6 s - 5

What is the answer to solvefor x,3y^3+3xy+4x=0 ?
The answer to solvefor x,3y^3+3xy+4x=0 is x=-(3y^3)/(3y+4);\quad y\ne-4/3