Graphical method

The most concrete way to understand a pair of linear equations is to draw them. Each equation is a straight line; the solution is wherever the two lines meet.

Definitions

A linear equation in two variables has the form $a x + b y + c = 0,$ with $a, b, c$ real and at least one of $a, b$ non-zero. Its graph in the $xy$-plane is a straight line.

A pair of linear equations in two variables is two such equations considered together; a solution is an ordered pair $(x_0, y_0)$ that satisfies both.

Concept: three geometric cases

When we draw the two lines, exactly one of three things happens.

Case 1: lines intersect in one point. There is exactly one solution. The system is consistent and independent.

Case 2: lines are parallel and distinct. They never meet. There is no solution. The system is inconsistent.

Case 3: lines coincide (every point on one is on the other). Infinitely many solutions. The system is consistent and dependent.

The geometric picture matches the algebra exactly. To plot a line $a x + b y + c = 0$, find two convenient points on it (often the $x$-intercept and the $y$-intercept) and draw the line through them. Be sure to use a ruler and a uniform scale.

Algorithm for graphical solving

1. Rewrite each equation in a convenient form: solve for $y$, or find $x$- and $y$-intercepts.
2. Choose at least two (preferably three for accuracy) integer $x$-values that give integer $y$-values.
3. Plot the points carefully on graph paper.
4. Draw the lines.
5. Read off the intersection point. Sanity-check by substituting into both equations.

If the lines clearly do not meet within the visible area, check parallelism: same slope but different $y$-intercept means no solution. If both lines look identical, check whether all the points of one lie on the other; if yes, infinite solutions.

Worked examples

Example 1. Solve graphically: $x + y = 5$, $2 x - y = 4$.

For $x + y = 5$: points $(0, 5), (5, 0), (2, 3)$. For $2x - y = 4$: points $(0, -4), (2, 0), (3, 2)$.

Plot both. They cross at $(3, 2)$. Verify: $3 + 2 = 5$ ✓ and $6 - 2 = 4$ ✓.

Example 2. Solve graphically: $2 x + 3 y = 12$, $4 x + 6 y = 12$.

The second equation simplifies to $2 x + 3 y = 6$. Same left side as the first but different right side $\Rightarrow$ parallel and distinct lines. No solution.

Example 3. Solve graphically: $x - 2 y = 0$, $3 x + 4 y = 20$.

From the first, $x = 2y$. Some points: $(0, 0), (2, 1), (4, 2)$. From the second: $(0, 5), (4, 2), (20/3, 0)$.

They cross at $(4, 2)$. Verify both equations. ✓

Example 4. Solve: $x + 2y = 4$ and $2 x + 4 y = 8$.

Second is twice the first. Same line , infinite solutions, every point on $x + 2 y = 4$ works, e.g., $(0, 2), (2, 1), (4, 0)$.

Example 5. Determine graphically whether the pair $x - y = 1$, $2 x + y = 8$ is consistent.

Plot: first passes through $(0, -1), (1, 0)$; second through $(0, 8), (4, 0), (3, 2)$. They cross at $(3, 2)$. Consistent with unique solution.

Try it yourself

1. Solve graphically: $x + 2y = 6$, $3 x - y = 4$.
2. Determine graphically whether $2 x - y = 3$ and $4 x - 2 y = 7$ have a solution.
3. Solve graphically: $5 x - y = 5$, $3 x - y = 3$.
4. Solve graphically: $x + y = 7$, $2 x - 3 y = -6$.
5. Use a graph to determine the number of solutions of $x + 2 y = 4$ and $3 x + 6 y = 12$.
6. Solve graphically: $2 x + y = 6$, $4 x - 2 y = 4$.
7. Vertices of the triangle formed by $x + y = 3$, $y = 0$, $x = 0$? (Plot and read.)
8. Draw the graphs of $x - y + 1 = 0$ and $3 x + 2 y - 12 = 0$. Find the vertices of the triangle formed with the $x$-axis.
9. Find graphically the values of $x$ and $y$ that satisfy $3 x + y - 5 = 0$ and $2 x - y - 5 = 0$.
10. From a graph, is $x - 2 y = 4$, $2 x - 4 y = 12$ consistent?

Pitfalls / Insight

• Pick well-spaced integer points to plot. Three points per line are better than two.
• Use a ruler and the same scale on both axes; a sloppy plot gives a sloppy intersection.
• If both lines coincide, do not declare "no solution" , there are infinitely many.

Insight. The graphical method is the only one that shows you why there are three cases. After this, the algebraic methods are just faster ways to get the same answer.