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Solve xy^1+4y=x³y²?

1 Answer

May 12, 2018

$y = \frac{x + 4}{x^{3}}, x \neq 0$ $y=(x+4)/x^3,x!=0$

Explanation:

$x y^{1} + 4 y = x^{3} y^{2}$ $xy^1+4y=x^3y^2$

Factor $y$ $y$ out of the left side.
$y (x + 4) = x^{3} y^{2}$ $y(x+4)=x^3y^2$

Isolate for $y$ $y$ .
$\frac{x + 4}{x^{3}} = \frac{y^{2}}{y}$ $(x+4)/x^3=(y^2)/y$

$y = \frac{x + 4}{x^{3}}, x \neq 0$ $y=(x+4)/x^3,x!=0$

Note: since this is a rational expression, don't forget to state restrictions to variables in the denominator.

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