What is the answer to solvefor x,y^2+3xy+4=0 ?

The answer to solvefor x,y^2+3xy+4=0 is x=(-y^2-4)/(3y);\quad y\ne 0

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solvefor x,y^2+3xy+4=0

solve for

x, y^{2} + 3 x y + 4 = 0

x = \frac{- y ^{2} - 4}{3 y}; y \neq = 0

Solution steps

y^{2} + 3 x y + 4 = 0

Move

y^{2}

to the right side

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

Move

4

to the right side

3 x y = - y^{2} - 4

Divide both sides by

3 y; y \neq = 0

x = \frac{- y ^{2} - 4}{3 y}; y \neq = 0

x, 3 x + y + 7 = 0

12 \cdot k = 48

\frac{2 x - 3}{5} - \frac{4 x - 5}{10} = \frac{3 x + 1}{2} - 3

9.9 = 3.1 - 3.4 x

g, lo g_{10} (5) x^{3} - lo g_{10} (x^{2}) = 2

What is the answer to solvefor x,y^2+3xy+4=0 ?
The answer to solvefor x,y^2+3xy+4=0 is x=(-y^2-4)/(3y);\quad y\ne 0