Substitution and paired solutions A simultaneous solution must satisfy both original equations. A linear relation lets one variable be replaced by an equal expression. Substitution preserves both conditions: a root of the resulting quadratic supplies one candidate, then its partner is fixed by the linear equation. A repeated root produces one pair. Isolate the simpler variable; substitute with brackets; solve the quadratic; back-substitute each root separately; test both equations and any contextual restrictions. Example For y=x+1 and xy=6, x(x+1)=6, so (x+3)(x-2)=0. The pairs are (-3,-2) and (2,3). Check: the values of x and y cannot be freely recombined. In the example, (2,-2) fails both conditions. For a length, retain only positive values.

A simultaneous solution must satisfy both original equations. A linear relation lets one variable be replaced by an equal expression.

Substitution preserves both conditions: a root of the resulting quadratic supplies one candidate, then its partner is fixed by the linear equation. A repeated root produces one pair.

Isolate the simpler variable; substitute with brackets; solve the quadratic; back-substitute each root separately; test both equations and any contextual restrictions.

For \(y=x+1\) and \(xy=6\), \(x(x+1)=6\), so \((x+3)(x-2)=0\). The pairs are \((-3,-2)\) and \((2,3)\).

Check: the values of \(x\) and \(y\) cannot be freely recombined. In the example, \((2,-2)\) fails both conditions. For a length, retain only positive values.