Coincident Lines and Infinitely Many Solutions

When Two Equations Describe the Same Line

Two equations in \(x\) and \(y\) can look different while expressing exactly the same constraint. For instance, multiplying every term of an equation by the same nonzero number does not change which ordered pairs satisfy it. When two equations describe the same line, a point on that line satisfies both equations. Since a line contains infinitely many points, the system has infinitely many solutions—not just two or three special answers.

Throughout, \(x,y\in\mathbb{R}\). A linear equation \(ax+by=c\), with \(a\) and \(b\) not both zero, describes the set of points \((x,y)\) that satisfy the equation. We call that set its line. A solution of a system of two equations is an ordered pair that satisfies both equations at once; therefore, the system’s solutions are the points common to the two lines.

The nonzero condition matters. Multiplying an equation by a nonzero number is reversible: divide by that number to recover the original equation. Multiplying by zero is not reversible; it turns the equation into \(0=0\), which is true for every point and does not preserve the original constraint.

Scaling Preserves Every Solution

Proof. Suppose first that \((x,y)\) satisfies \(ax+by=c\). Multiplying both sides by \(k\) gives \(kax+kby=kc\), so the point satisfies the scaled equation. Conversely, if \(kax+kby=kc\), division of both sides by \(k\), which is permitted because \(k\ne0\), gives \(ax+by=c\). Thus each equation has the same solution set. A point solves the two-equation system exactly when it satisfies both equations. Since the equations have the same solution set, that happens exactly when the point lies on their common line. \(\square\)

This result explains both the geometry and the algebra. Geometrically, the equations mark the same set of points. Algebraically, satisfying either equation automatically means satisfying the other. The second equation does not impose a new restriction on the possible points.

Why a Line Gives Infinitely Many Solutions

The example’s free choice of \(x\) is one way to see the pattern. There is also a direct geometric-algebraic argument that works for any equation \(ax+by=c\) with \((a,b)\ne(0,0)\). First choose a point \((x_0,y_0)\) on the line. Such a point exists: if \(a\ne0\), take \(x_0=c/a\) and \(y_0=0\); if \(a=0\), then \(b\ne0\), and we can take \(x_0=0\) and \(y_0=c/b\).

Now consider the direction \((-b,a)\). It is not the zero pair, because \(a\) and \(b\) are not both zero. For every real \(t\), form the point \((x_0-tb,\ y_0+ta)\). Substitution shows that it remains on the line:

The two terms involving \(t\) cancel. Moreover, these points are distinct for distinct values of \(t\): if two of them were equal, then \((t_1-t_2)(-b,a)=(0,0)\). Since \((-b,a)\ne(0,0)\), at least one of its coordinates is nonzero, forcing \(t_1=t_2\).

Proof. Choose a point \((x_0,y_0)\) on the line as above. For each \(t\in\mathbb{R}\), the point \((x_0-tb,y_0+ta)\) satisfies the equation, because its left-hand side is \(ax_0+by_0-tab+tba=c\). If the points for \(t_1\) and \(t_2\) are equal, then \((t_1-t_2)(-b,a)=(0,0)\). At least one coordinate of \((-b,a)\) is nonzero, so \(t_1=t_2\). Thus distinct real values give distinct points, and there are infinitely many such points. When two equations describe the same line, every point on it satisfies both, so all these distinct points solve the system. \(\square\)

A Test for Coincident Lines

Scaling gives a convenient way to recognize coincident lines. In the setting of two genuine line equations, it is also an exact test: their solution sets agree precisely when one whole equation is a nonzero scalar multiple of the other. The constants are part of this test; matching only the coefficients of \(x\) and \(y\) is not enough.

Proof. If such a \(k\) exists, the second equation is a nonzero scalar multiple of the first, so the scaling theorem shows that their solution sets agree.

For the converse, assume the solution sets agree. Choose a point \((x_0,y_0)\) on the first line; the construction above guarantees such a point. The vector \((-b,a)\) is nonzero, and for every real \(t\), the point \((x_0-tb,y_0+ta)\) lies on the first line. Because the solution sets agree, it also satisfies the second equation. Substituting \(t=0\) and \(t=1\) into that equation gives, respectively,

Subtracting the first equality from the second gives \(-db+ea=0\), or \(ea=db\). If \(a\ne0\), let \(k=d/a\). Then \(e=db/a=kb\). Also \(k\ne0\): if \(d=0\), the relation \(ea=db\) gives \(e=0\), contradicting \((d,e)\ne(0,0)\). If \(a=0\), then \(b\ne0\), and \(ea=db\) forces \(d=0\). Since \((d,e)\ne(0,0)\), we have \(e\ne0\); setting \(k=e/b\) gives \(d=ka=0\) and \(e=kb\), with \(k\ne0\). In either case, \((d,e)=k(a,b)\). Finally, using the shared point \((x_0,y_0)\), we obtain \(f=dx_0+ey_0=k(ax_0+by_0)=kc\). All three entries, including the constant, are proportional. \(\square\)

Common Mistake: Scaling Only the Variable Terms

A frequent error is to notice that the variable coefficients are proportional and conclude that the equations describe the same line without checking the constants. For example, compare \(x+2y=3\) and \(2x+4y=8\). The coefficients of \(x\) and \(y\) in the second equation are twice those in the first, but doubling the entire first equation gives \(2x+4y=6\), not \(2x+4y=8\).

The difference can be checked with a point. The point \((3,0)\) satisfies the first equation, since \(3+2(0)=3\). It does not satisfy the second, since \(2(3)+4(0)=6\ne8\). Therefore, the equations do not have the same solution set. The correction is to compare all three entries—both variable coefficients and the constant—and require one common nonzero multiplier.

Another error is to allow a zero multiplier. For example, multiplying \(x+y=1\) by \(0\) produces \(0=0\). The point \((0,0)\) satisfies \(0=0\), but it does not satisfy \(x+y=1\). The zero multiplier has erased the constraint, so it cannot be used to establish coincident lines.

Why Coincidence Matters

When two equations describe the same line, the system contains the same constraint twice. Solving the system does not mean finding a single crossing point: every point on the shared line works. The proportionality test makes this visible before a long calculation. Check that a single nonzero number scales both coefficients and the right-hand side; then the equations are equivalent, and every point satisfying either one satisfies both.