Rewrite the expression with a common denominator.
\[\begin{aligned}&\frac{y+5x}{xy}=3\\&x=2\\&y=0\end{aligned}\]
Break down the problem into these 2 equations.
\[\frac{y+5x}{xy}=3\]
\[\frac{y+5x}{xy}=x\]
Solve the 1st equation: \(\frac{y+5x}{xy}=3\).
Multiply both sides by \(xy\).
\[y+5x=3xy\]
Subtract \(5x\) from both sides.
\[y=3xy-5x\]
Factor out the common term \(x\).
\[y=x(3y-5)\]
Divide both sides by \(3y-5\).
\[\frac{y}{3y-5}=x\]
Switch sides.
\[x=\frac{y}{3y-5}\]
\[x=\frac{y}{3y-5}\]
Solve the 2nd equation: \(\frac{y+5x}{xy}=x\).
Multiply both sides by \(xy\).
\[y+5x=xxy\]
Use Product Rule: \({x}^{a}{x}^{b}={x}^{a+b}\).
\[y+5x={x}^{2}y\]
Move all terms to one side.
\[y+5x-{x}^{2}y=0\]
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, \(a=-y\), \(b=5\) and \(c=y\).
\[{x}^{}=\frac{-5+\sqrt{{5}^{2}-4\times -yy}}{2\times -y},\frac{-5-\sqrt{{5}^{2}-4\times -yy}}{2\times -y}\]
Simplify.
\[x=\frac{-5+\sqrt{25+4{y}^{2}}}{-2y},\frac{-5-\sqrt{25+4{y}^{2}}}{-2y}\]
\[x=\frac{-5+\sqrt{25+4{y}^{2}}}{-2y},\frac{-5-\sqrt{25+4{y}^{2}}}{-2y}\]
Simplify solutions.
\[x=-\frac{-5+\sqrt{25+4{y}^{2}}}{2y},-\frac{-5-\sqrt{25+4{y}^{2}}}{2y}\]
\[x=-\frac{-5+\sqrt{25+4{y}^{2}}}{2y},-\frac{-5-\sqrt{25+4{y}^{2}}}{2y}\]
Collect all solutions.
\[x=\frac{y}{3y-5},-\frac{-5+\sqrt{25+4{y}^{2}}}{2y},-\frac{-5-\sqrt{25+4{y}^{2}}}{2y}\]
x=y/(3*y-5),-(-5+sqrt(25+4*y^2))/(2*y),-(-5-sqrt(25+4*y^2))/(2*y)
