Graphical method of solution

A linear equation $ax+by+c=0$ is satisfied by infinitely many pairs $(x,y)$, and plotting all of them traces a straight line. Given two such equations,

where: $a_1,b_1,c_1$ and $a_2,b_2,c_2$ are the real coefficients of the first and second equation, and $x,y$ are the unknowns — solving the pair graphically means drawing both lines on one set of axes and reading off whatever they share. There are exactly three geometric outcomes, and each corresponds to a distinct relationship between the coefficients:

• Intersecting lines. The lines cross at exactly one point, giving a unique solution — the coordinates of that crossing. This is the generic case, and it happens whenever the lines are not parallel, i.e. $\dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2}$.
• Parallel lines. The lines have the same steepness but never meet, so the pair has no solution. This happens when $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2}$.
• Coincident lines. Both equations describe the same line, so every point on it is a solution — infinitely many solutions. This happens when $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}$.

To actually draw a line from $ax+by+c=0$, rewrite it as $y=\dfrac{-c-ax}{b}$ (with $b\neq0$), choose two or three convenient values of $x$, compute the matching $y$-values, plot those points, and join them with a ruler. Two points fix a line; a third is a cheap check against arithmetic slips.

Solve $x+y=5$ and $2x-y=4$ graphically, and use the ratio test to confirm the type of solution before checking the picture.

Step 1. Build a table for $x+y=5$, i.e. $y=5-x$: at $x=0,\ y=5$; at $x=5,\ y=0$; at $x=2,\ y=3$.

Step 2. Build a table for $2x-y=4$, i.e. $y=2x-4$: at $x=0,\ y=-4$; at $x=2,\ y=0$; at $x=3,\ y=2$.

Step 3. Apply the ratio test before plotting. Here $a_1=1,\ b_1=1$ for the first line and $a_2=2,\ b_2=-1$ for the second, so $\dfrac{a_1}{a_2}=\dfrac12$ and $\dfrac{b_1}{b_2}=-1$. Since $\dfrac12\neq-1$, the lines must intersect at exactly one point — a unique solution is guaranteed before any graph is drawn.

Step 4. Plot both sets of points on the same axes and draw the two lines; they cross at $(3,2)$, exactly as the ratio test predicted.

Step 5. Verify by substitution: $3+2=5$ checks against the first equation, and $2(3)-2=4$ checks against the second. Both hold, so the intersection point is confirmed algebraically, not just read off the graph. The unique solution is $x=3,\ y=2$.

Competitive stretch

The same reasoning explains why $2x+3y=9$ and $4x+6y=18$ never even need a graph: here $\dfrac{a_1}{a_2}=\dfrac{2}{4}=\dfrac12$, $\dfrac{b_1}{b_2}=\dfrac{3}{6}=\dfrac12$, and $\dfrac{c_1}{c_2}=\dfrac{9}{18}=\dfrac12$ — all three ratios equal $\dfrac12$, so the ratio test alone confirms coincident lines and infinitely many solutions, matching the fact that the second equation is just the first multiplied by 2.

• Plotting only two points per line — a single arithmetic slip can still look like a valid straight line. Always add a third point as a self-check.
• Treating parallel lines as "not solved yet." No solution is a complete, correct final answer in itself; there is nothing further to find.
• Reading the intersection point off an imprecisely drawn graph without checking it. Always substitute the coordinates back into both original equations before finalising an answer.
• Skipping the ratio test and jumping straight to plotting. Computing $\dfrac{a_1}{a_2}, \dfrac{b_1}{b_2}, \dfrac{c_1}{c_2}$ first tells you what kind of picture to expect, saving time and catching setup errors early.

Ask the tutor: if you are unsure how to pick convenient $x$-values for your table, or want to check whether your graph's intersection point is exact, ask the AI tutor to walk through your specific pair of equations step by step.