What is the answer to solvefor x,cxy+c^2x+4=0 ?

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

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solvefor x,cxy+c^2x+4=0

solve for

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

x = - \frac{4}{c ( y + c )}; c \neq = 0, c \neq = - y

Solution steps

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

Move

4

to the right side

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

Factor

c x y + c^{2} x : c x (y + c)

c x (y + c) = - 4

Divide both sides by

c (y + c); c \neq = 0, c \neq = - y

x = - \frac{4}{c ( y + c )}; c \neq = 0, c \neq = - y

- 9 = - \frac{6}{5} u

4 x - 1 + 2 x - 1 = 5 x

6 (q - 7) - 3 (4 - q) = 0

\frac{x}{4.1} = - 2

4 t + 9 = - 8 t - 13

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