Denominators and simultaneous solutions An algebraic fraction is defined only when its denominator is nonzero. Record these exclusions before rearranging a system. Multiplication by a denominator is reversible on the allowed domain. Outside that domain the resulting polynomial may appear true even though the original expression is undefined. Write all exclusions; clear denominators; substitute the other equation; solve; restore the partner value; reject any pair failing an exclusion or either original equation. Example For 1x+ 1y=1 and x+y=4, x,y 0, and multiplication by xy gives x+y=xy. Thus xy=4, x(4-x)=4, and the only pair is (2,2). Check: xy+ yx=2 gives (x-y)2=0, but (0,0) remains forbidden. Clearing denominators never removes the original exclusions.

An algebraic fraction is defined only when its denominator is nonzero. Record these exclusions before rearranging a system.

Multiplication by a denominator is reversible on the allowed domain. Outside that domain the resulting polynomial may appear true even though the original expression is undefined.

Write all exclusions; clear denominators; substitute the other equation; solve; restore the partner value; reject any pair failing an exclusion or either original equation.

For \(\dfrac1x+\dfrac1y=1\) and \(x+y=4\), \(x,y\ne0\), and multiplication by \(xy\) gives \(x+y=xy\). Thus \(xy=4\), \(x(4-x)=4\), and the only pair is \((2,2)\).

Check: \(\dfrac{x}{y}+\dfrac{y}{x}=2\) gives \((x-y)^2=0\), but \((0,0)\) remains forbidden. Clearing denominators never removes the original exclusions.