Rewrite the expression with a common denominator.
\[\begin{aligned}&\frac{y+4x}{xy}=1\\&x=5\\&y=0\end{aligned}\]
Break down the problem into these 2 equations.
\[\frac{y+4x}{xy}=1\]
\[\frac{y+4x}{xy}=x\]
Solve the 1st equation: \(\frac{y+4x}{xy}=1\).
Multiply both sides by \(xy\).
\[y+4x=xy\]
Subtract \(4x\) from both sides.
\[y=xy-4x\]
Factor out the common term \(x\).
\[y=x(y-4)\]
Divide both sides by \(y-4\).
\[\frac{y}{y-4}=x\]
Switch sides.
\[x=\frac{y}{y-4}\]
\[x=\frac{y}{y-4}\]
Solve the 2nd equation: \(\frac{y+4x}{xy}=x\).
Multiply both sides by \(xy\).
\[y+4x=xxy\]
Use Product Rule: \({x}^{a}{x}^{b}={x}^{a+b}\).
\[y+4x={x}^{2}y\]
Move all terms to one side.
\[y+4x-{x}^{2}y=0\]
Use the Quadratic Formula.
In general, given \(a{x}^{2}+bx+c=0\), there exists two solutions where:
\[x=\frac{-b+\sqrt{{b}^{2}-4ac}}{2a},\frac{-b-\sqrt{{b}^{2}-4ac}}{2a}\]
In this case, \(a=-y\), \(b=4\) and \(c=y\).
\[{x}^{}=\frac{-4+\sqrt{{4}^{2}-4\times -yy}}{2\times -y},\frac{-4-\sqrt{{4}^{2}-4\times -yy}}{2\times -y}\]
Simplify.
\[x=\frac{-4+2\sqrt{4+{y}^{2}}}{-2y},\frac{-4-2\sqrt{4+{y}^{2}}}{-2y}\]
\[x=\frac{-4+2\sqrt{4+{y}^{2}}}{-2y},\frac{-4-2\sqrt{4+{y}^{2}}}{-2y}\]
Simplify solutions.
\[x=\frac{2-\sqrt{4+{y}^{2}}}{y},\frac{2+\sqrt{4+{y}^{2}}}{y}\]
\[x=\frac{2-\sqrt{4+{y}^{2}}}{y},\frac{2+\sqrt{4+{y}^{2}}}{y}\]
Collect all solutions.
\[x=\frac{y}{y-4},\frac{2-\sqrt{4+{y}^{2}}}{y},\frac{2+\sqrt{4+{y}^{2}}}{y}\]
x=y/(y-4),(2-sqrt(4+y^2))/y,(2+sqrt(4+y^2))/y
