Two-Parameter Descriptions of Solution Sets

From One Condition to Two Parameters

A triple \((x,y,z)\) has three coordinates, but a consistent system may impose only one effective linear condition on them. In that case, two coordinates can be chosen freely; the remaining coordinate is then determined by the equation. The two free choices are recorded by parameters, giving a compact description of every solution triple.

For example, in \(2x-y+3z=7\), we can choose \(y\) and \(z\) freely and solve for \(x\). Let \(y=s\) and \(z=t\), where \(s,t\in\mathbb{R}\). Then \(2x=7+s-3t\), so \(x=\tfrac72+\tfrac12s-\tfrac32t\). This suggests the description \((x,y,z)=(\tfrac72+\tfrac12s-\tfrac32t,s,t)\). A complete description must do two things: every triple it produces must satisfy the equation, and every solution of the equation must be produced by some choice of \(s\) and \(t\).

The choice of which variables to assign to \(s\) and \(t\) is often convenient rather than unique. The essential step is to solve for one variable whose coefficient is nonzero. Once that variable is expressed in terms of the other two, both parameters may vary independently over all real numbers; no additional restriction on them is imposed by that one equation.

A General Result and Its Proof

Proof. First, choose any \(s,t\in\mathbb{R}\) and set \(x=\tfrac da-\tfrac ba s-\tfrac ca t\), \(y=s\), and \(z=t\). Substitution gives \(ax+by+cz=a(\tfrac da-\tfrac ba s-\tfrac ca t)+bs+ct=d-bs-ct+bs+ct=d\). Thus every triple produced by the formula is a solution. Conversely, let \((x,y,z)\) be any solution. Set \(s=y\) and \(t=z\). Since \(ax+by+cz=d\) and \(a\ne0\), solving for \(x\) gives \(x=\tfrac da-\tfrac ba y-\tfrac ca z=\tfrac da-\tfrac ba s-\tfrac ca t\). The triple therefore has the stated form. Both inclusions hold, so the formula describes exactly the solution set. \(\square\)

This proof establishes the two directions that are needed for an exact description. Substitution proves that the formula does not produce extraneous triples; solving the equation for an arbitrary solution proves that it does not omit any. The condition \(a\ne0\) matters because dividing by \(a\) is valid only then. If instead \(b\ne0\) or \(c\ne0\), solve for \(y\) or \(z\) and choose the other two coordinates as parameters.

Proof. If at least one coefficient is nonzero, choose a variable with a nonzero coefficient and solve for that variable. The other two coordinates can be assigned arbitrary real values, and the calculation in the theorem shows that this gives all solutions. If all three coefficients are zero, the equation becomes \(0=d\). When \(d\ne0\), this is false for every triple; when \(d=0\), it is true for every triple. \(\square\)

Worked Examples

What the Parameters Record

In the first example, \(s\) records the chosen second coordinate and \(t\) records the chosen third coordinate. The first coordinate is not chosen independently: its value has to balance the equation. The same interpretation applies to the second example even though three equations are written down. Because the last two equations repeat the first condition, they do not place any additional restriction on the triples.

The parameters also distinguish the triples in these descriptions. When the formula has the form \((a+bs+ct,s,t)\), the parameter values can be read directly from the second and third coordinates. More generally, when two coordinates are chosen as the parameters, those coordinates identify the parameter pair. This makes the description useful both for generating solutions and for showing how a given solution belongs to the family.

Proof. Suppose the parameter pairs \((s,t)\) and \((s',t')\) produce the same triple. Their second coordinates must be equal, so \(s=s'\); their third coordinates must be equal, so \(t=t'\). Therefore the parameter pairs were not different after all. \(\square\)

A parameterization is sometimes written as a single vector expression, and sometimes coordinate by coordinate. Both forms say the same thing. For example, the expression \(\mathbf{p}+s\mathbf{u}+t\mathbf{v}\) means that each coordinate of the solution triple is calculated from the corresponding coordinates of \(\mathbf{p}\), \(s\mathbf{u}\), and \(t\mathbf{v}\). Writing the coordinates explicitly can make the substitution check easier.

Common Pitfall: Reusing One Parameter

Consider the equation \(x+y+z=0\). A tempting but incorrect description is \((x,y,z)=(-t,t,t)\). Substitution gives \(-t+t+t=t\), which equals zero only when \(t=0\). Thus, except for the zero triple, the proposed formula does not even produce solutions. It also reuses a single parameter, forcing \(y=z\), although the equation does not require that condition.

To correct the description, choose two independent parameters: \(y=s\) and \(z=t\). Then \(x=-s-t\), so

Substitution verifies the equation: \((-s-t)+s+t=0\). The triple \((0,1,-1)\) is a solution, and it is obtained by \(s=1\), \(t=-1\). This example shows why two parameters must be allowed to vary independently: tying them together can omit valid solutions, and a separate substitution check is needed to catch formulas that do not satisfy the equation.