Solutions as Common Points
A system imposes both equations at once: a solution must satisfy the first equation and the second equation. In the plane, each equation can be viewed as a set of points. The solutions of the system are therefore the points that belong to both sets. When both equations define lines, this gives a geometric picture with three possibilities: the lines meet at one point, they are the same line, or they do not meet.
Write a linear equation in two variables as \(ax+by=c\), where \(a,b,c\in\mathbb{R}\). Its graph is the set of ordered pairs \((x,y)\in\mathbb{R}^2\) that make the equation true. The coefficients of \(x\) and \(y\) matter: when at least one of \(a,b\) is nonzero, the graph is a line. When both are zero, the equation is instead either true at every point or false at every point.
For example, \(0x+0y=0\) is satisfied by every ordered pair, because both sides equal zero for every \(x\) and \(y\). In contrast, \(0x+0y=4\) is satisfied by no ordered pair. Neither equation describes a line. Keeping these cases separate matters when drawing conclusions about the geometry of a system.
The Solution Set Is an Intersection
Consider the system \(a_1x+b_1y=c_1\), \(a_2x+b_2y=c_2\), and let \(G_1,G_2\) be the graphs of the two equations. A solution is an ordered pair that satisfies both equations, not just one of them. This observation gives a set identity that holds whether or not the graphs are lines.
Proof. Let \(S\) be the solution set. An ordered pair \((x,y)\) belongs to \(S\) exactly when it satisfies the first equation and the second equation. By the definition of each graph, this means exactly that \((x,y)\in G_1\) and \((x,y)\in G_2\), or equivalently \((x,y)\in G_1\cap G_2\). Thus every element of \(S\) belongs to \(G_1\cap G_2\), and every element of \(G_1\cap G_2\) belongs to \(S\). Therefore \(S=G_1\cap G_2\). \(\square\)
If both equations have at least one nonzero variable coefficient, each graph is a line. Their intersection then tells us how many solutions the system has: no common point means no solution, one common point means one solution, and a shared line means infinitely many solutions. The line condition is essential for this interpretation. An equation whose graph is the whole plane does not contribute a line to compare with the other graph.
Describing a Line by a Point and a Direction
A useful way to reason about line intersections is to describe the points on a line using one known point and a direction. Suppose \(ax+by=c\) with \((a,b)\ne(0,0)\), and let \(p=(x_0,y_0)\) be a point on its graph. The vector \((-b,a)\) points along the line: adding a multiple of it to \(p\) keeps the equation true.
Proof. First, for every \(t\in\mathbb{R}\), the point \(p+t(-b,a)=(x_0-bt,y_0+at)\) satisfies the equation, since \(a(x_0-bt)+b(y_0+at)=ax_0+by_0-abt+abt=c\). Conversely, let \((x,y)\) be any point on the line. Then \(r=x-x_0\) and \(s=y-y_0\) satisfy \(ar+bs=0\). If \(a\ne0\), set \(t=s/a\). The relation gives \(r=-bs/a=-bt\), so \((r,s)=t(-b,a)\). If \(a=0\), then \(b\ne0\), and \(ar+bs=0\) gives \(s=0\). Set \(t=-r/b\); then \((r,s)=t(-b,0)=t(-b,a)\). In either case \((x,y)=p+t(-b,a)\), so every point on the line has the stated form. \(\square\)
This description gives a direction vector for any line from its equation. Two nonzero vectors in \(\mathbb{R}^2\) are proportional when one is a nonzero scalar multiple of the other. Proportional direction vectors point along the same direction; nonproportional direction vectors point in different directions.
Proof. Write \(L_1=p+\{s\mathbf{d}_1:s\in\mathbb{R}\}\) and \(L_2=q+\{t\mathbf{d}_2:t\in\mathbb{R}\}\) using the proposition, where both direction vectors are nonzero. Suppose first that a point \(r\) belongs to both lines. Rewriting each line from \(r\), we have \(L_1=r+\{s\mathbf{d}_1:s\in\mathbb{R}\}\) and \(L_2=r+\{t\mathbf{d}_2:t\in\mathbb{R}\}\). If \(\mathbf{d}_2=k\mathbf{d}_1\) for some \(k\ne0\), the scalar multiples of the two direction vectors form the same set, so \(L_1=L_2\). If the direction vectors are not proportional, any common point has the form \(r+s\mathbf{d}_1=r+t\mathbf{d}_2\), hence \(s\mathbf{d}_1=t\mathbf{d}_2\). Nonproportionality and the fact that both vectors are nonzero force \(s=t=0\), so \(r\) is the only common point.
Now suppose the lines are disjoint. If their direction vectors were not proportional, they would form two independent directions spanning \(\mathbb{R}^2\). In particular, there would be scalars \(s,t\) with \(q-p=s\mathbf{d}_1-t\mathbf{d}_2\). Rearranging gives \(p+s\mathbf{d}_1=q+t\mathbf{d}_2\), a point on both lines, contradicting disjointness. Thus the directions are proportional. \(\square\)
Three Geometric Possibilities
Worked Example: Two Lines Meet at One Point
Consider the system \(2x+y=5\), \(x-y=1\). The point \((2,1)\) belongs to both graphs: \(2(2)+1=5\), and \(2-1=1\). Thus it is a solution of the system.
The direction vectors from the equations are \((-1,2)\) and \((1,1)\). These are not proportional: there is no scalar \(k\) for which \((1,1)=k(-1,2)\), since the first coordinate would require \(k=-1\) while the second would require \(k=\tfrac12\). By the theorem, the lines have exactly one common point. The system therefore has exactly one solution, \((2,1)\).
Check: Substitution gives \(2(2)+1=5\) and \(2-1=1\), so the point satisfies both original equations.
Worked Example: Two Equations Describe the Same Line
Consider \(2x-3y=6\) and \(-4x+6y=-12\). Multiplying every term of the first equation by \(-2\) gives the second equation, so the two equations have exactly the same graph.
For a description of all their common points, take \(p=(3,0)\), which satisfies both equations, and use the direction vector \((3,2)\) from the first equation. The common line is \((x,y)=(3,0)+t(3,2)=(3+3t,2t)\), where \(t\in\mathbb{R}\).
Check: For every \(t\), \(2(3+3t)-3(2t)=6+6t-6t=6\). Also, \(-4(3+3t)+6(2t)=-12-12t+12t=-12\). Every point in the description satisfies both equations, and the line proposition ensures the description includes every point on the first line. Since the graphs are equal, it includes every common point.
Worked Example: Two Disjoint Lines
Consider \(2x+y=3\) and \(2x+y=7\). Both equations have the same direction vector \((-1,2)\). One point on the first line is \((0,3)\), and one point on the second is \((0,7)\), so their parameter descriptions are \((0,3)+t(-1,2)\) and \((0,7)+t(-1,2)\).
The lines cannot share a point: a common point would make the same number \(2x+y\) equal to both \(3\) and \(7\), which is impossible because \(3\ne7\). They are distinct, disjoint lines with the same direction.
Check: Every point \((-t,3+2t)\) on the first line gives \(2(-t)+(3+2t)=3\). Every point \((-t,7+2t)\) on the second gives \(2(-t)+(7+2t)=7\). The two points with the same parameter differ by \((0,4)\), so these parameter descriptions do not produce a common point: their first coordinates agree only when the parameters agree, and then their second coordinates differ by \(4\).
When an Equation Does Not Define a Line
The solution-set identity still holds for equations with zero variable coefficients, but the three-line classification does not apply to them. For instance, \(0x+0y=0\) has graph \(\mathbb{R}^2\), not a line. Intersecting it with the graph of \(x=0\) gives the entire line \(x=0\), even though the two graphs do not coincide: one is the whole plane and the other is a line.
Worked Example: A Whole Plane and a Line
Consider \(0x+0y=0\) together with \(x=0\). The first equation is true for every \((x,y)\in\mathbb{R}^2\); the second is true exactly for points on the vertical line with \(x=0\). Their common solution set is therefore \(\{(0,y):y\in\mathbb{R}\}\).
Check: For any point \((0,y)\), the first left-hand side is \(0(0)+0y=0\), and the second left-hand side is \(0\). Conversely, a common solution must satisfy \(x=0\), while the first equation places no further restriction on \(y\). Thus the stated set is exactly the solution set. The graphs are not coincident lines: the first graph is the whole plane.
Common Pitfall: Assuming Every Equation Gives a Line
A tempting but incorrect argument is: “The system \(0x+0y=0\), \(x=0\) has infinitely many solutions, so its two lines must coincide.” The conclusion misidentifies the first graph: \(0x+0y=0\) is satisfied at every point, so its graph is the whole plane. The second graph is only the line \(x=0\). They are not two coincident lines, even though their intersection contains infinitely many points.
Before classifying a system by line intersections, check each equation’s coefficients of \(x\) and \(y\). If at least one is nonzero, its graph is a line. If both are zero, check the right-hand side: \(0x+0y=0\) gives the whole plane, while \(0x+0y=c\) with \(c\ne0\) gives the empty set. This check prevents a valid statement about lines from being applied to a graph that is not a line.
| Graphs of the two equations | Common points | Number of solutions |
|---|---|---|
| Lines meeting at one point | One point | One |
| Coincident lines | The entire line | Infinitely many |
| Distinct disjoint lines | No points | None |
| One graph is the whole plane | The other graph | Depends on the other graph |
| One graph is empty | No points | None |
Check Your Understanding
Use the graphs and the equations together to answer each question. When you claim that a point is a solution, check it in both equations.
- For \(x+y=4\) and \(x-y=0\), verify that \((2,2)\) is a common point. What do the direction vectors say about whether it is the only one?
- Explain why \(3x+6y=9\) and \(x+2y=3\) have the same graph.
- Do \(x-2y=1\) and \(x-2y=5\) have a common point? Justify your answer from the equations.
- What is the graph of \(0x+0y=0\)? What is the graph of \(0x+0y=-1\)?
- Give a reason the statement “infinitely many solutions means the two graphs are coincident lines” needs a hypothesis about the equations.
