Cartesian plane and linear graphs

The plane and its addresses. Two perpendicular number lines — the horizontal $x$-axis and the vertical $y$-axis — cross at the origin, $(0, 0)$. Every point on the plane gets a unique address, an ordered pair $(x, y)$: the first number, the abscissa, is the horizontal step from the origin (right if positive, left if negative); the second, the ordinate, is the vertical step (up if positive, down if negative). The two axes cut the plane into four quadrants, labelled I through IV going anticlockwise from the top-right, and the sign pattern of $(x, y)$ tells you instantly which one a point sits in — both positive in I, both negative in III, and so on.

What makes an equation "linear." A linear equation in two variables has the general form

where: $a$ and $b$ are fixed numbers (not both zero) that multiply the variables, $c$ is a fixed constant, and $x, y$ are the two unknowns. For a fixed $a, b, c$, this single equation is satisfied by infinitely many pairs $(x, y)$ — choose any $x$ and there is exactly one matching $y$ that keeps the equation true (and vice versa). Plotting every one of those pairs is what produces the linear graph, and "linear" is precisely the promise that this picture will always be a straight line, never a curve.

Three-step recipe. Drawing the graph of a linear equation only ever takes three moves: choose two or three convenient values of $x$ (values that keep the arithmetic simple); substitute each into the equation and solve for the matching $y$, building a small table of solutions; then plot the resulting points on the grid and join them with a ruler, extending the line a little past the outer points in both directions.

A linear equation such as $x + y = 6$ forces a perfectly rigid trade-off between $x$ and $y$: whatever $x$ goes up by, $y$ must go down by exactly the same amount, so that the sum stays fixed at $6$. Geometrically, a constant rate of trade-off between horizontal and vertical movement is exactly what a straight line is — every step you take along it changes $x$ and $y$ in the same fixed proportion, never curving away. That is why tabulating any two solutions and joining them with a ruler always reproduces the full graph correctly: the line has no wiggle room to deviate between (or beyond) the points you plotted, which is also why a third, independently computed point is such a powerful check — if it does not fall on the line through the first two, an arithmetic slip has been made somewhere. This same idea explains why the graphical method solves simultaneous equations. Each linear equation's graph is the complete set of $(x, y)$ pairs that make that equation true. A point where two such graphs cross is, by definition, on both lines at once — so its coordinates satisfy both equations simultaneously. Two distinct straight lines cross at exactly one point (unless they happen to be parallel), so reading off that single crossing point is equivalent to solving the pair of equations by elimination or substitution, just done visually instead of algebraically.

Solve the pair of simultaneous linear equations $x + y = 5$ and $x - y = 1$ graphically, and verify the answer algebraically.

Step 1. Tabulate solutions for the first equation, $y = 5 - x$, choosing convenient values: $x = 0 \Rightarrow y = 5$; $x = 2 \Rightarrow y = 3$; $x = 5 \Rightarrow y = 0$.

Step 2. Tabulate solutions for the second equation, $y = x - 1$: $x = 0 \Rightarrow y = -1$; $x = 1 \Rightarrow y = 0$; $x = 4 \Rightarrow y = 3$.

Step 3. Plot both sets of points on the same grid and join each set with a ruler. The first line runs from $(0, 5)$ down through $(2, 3)$ to $(5, 0)$; the second runs from $(0, -1)$ up through $(1, 0)$ to $(4, 3)$. Reading the grid carefully, the two lines cross at the point $(3, 2)$.

Step 4. Verify the crossing point algebraically by substituting $(3, 2)$ into both original equations: $3 + 2 = 5$ ✓ and $3 - 2 = 1$ ✓. Both equations are satisfied, confirming the graphical reading was accurate.

Answer: the solution is $x = 3, y = 2$.

Competitive stretch

Before drawing anything, a quick check often narrows down where to look: since both equations use only positive coefficients on the left summing or differing to positive constants, and the crossing point of $x+y=5$ and $x-y=1$ must satisfy $x = \frac{5+1}{2} = 3$ and $y = \frac{5-1}{2} = 2$ by simply adding and subtracting the two equations — the same answer the graph gives, obtained without plotting a single point. This "add and subtract" shortcut is worth knowing as a fast sanity check, even though the syllabus method for this stage is the graphical one.

• Plotting coordinates in the wrong order. In $(x, y)$, the first number is the horizontal move and the second is the vertical move — plotting $(3, 5)$ as "5 across, 3 up" is one of the most common reversal errors.
• Using only two points to draw the line. Two points are technically enough, but a single arithmetic slip in either one goes unnoticed. Always compute a third point as a check, as shown in the core idea figure.
• Trusting a graph reading without an algebraic check. Grid readings can be slightly off if the line is not drawn perfectly straight; substituting the coordinates back into the original equation(s), as in Step 4, is the reliable final check.
• Misreading the grid scale. If each square on the graph paper represents 2 units rather than 1, reading coordinates directly off the squares without checking the scale gives answers off by a constant factor.

Ask the tutor: if the crossing point on your hand-drawn graph looks slightly off from a clean whole number, ask your tutor to redraw both lines with you on graph paper and confirm the reading by substituting the coordinates back into both original equations.