d y = x ^ { 4 } + x
Solve for d
\left\{\begin{matrix}d=\frac{x\left(x^{3}+1\right)}{y}\text{, }&y\neq 0\\d\in \mathrm{R}\text{, }&\left(x=0\text{ or }x=-1\right)\text{ and }y=0\end{matrix}\right.
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yd=x^{4}+x
The equation is in standard form.
\frac{yd}{y}=\frac{x^{4}+x}{y}
Divide both sides by y.
d=\frac{x^{4}+x}{y}
Dividing by y undoes the multiplication by y.
Examples
Quadratic equation
{ x } ^ { 2 } - 4 x - 5 = 0
Trigonometry
4 \sin \theta \cos \theta = 2 \sin \theta
Linear equation
y = 3x + 4
Arithmetic
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Matrix
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Simultaneous equation
\left. \begin{cases} { 8x+2y = 46 } \\ { 7x+3y = 47 } \end{cases} \right.
Differentiation
\frac { d } { d x } \frac { ( 3 x ^ { 2 } - 2 ) } { ( x - 5 ) }
Integration
\int _ { 0 } ^ { 1 } x e ^ { - x ^ { 2 } } d x
Limits
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