Graphs of linear equations

Learning outcome

Describe lines from coordinate tables and interpret their intersections.

Concepts and assumptions

Use real coordinates. The horizontal x-axis and vertical y-axis meet perpendicularly at (0, 0). Positive directions are right and up; negative directions are left and down. Use equal one-unit intervals on both axes.

In (x, y), horizontal position comes first. A graph contains every allowed solution pair, not just tabulated samples.

For ax + by = c, a and b are not both zero. If b ≠ 0, then y = −(a/b)x + c/b. Slope measures vertical change divided by nonzero horizontal change. If b = 0, a ≠ 0: x = c/a is vertical. A line y = k is horizontal.

Two distinct points determine a line. Extend both ways: domains here are unrestricted. At x-axis intercepts y = 0; at y-axis intercepts x = 0. Intersections satisfy both equations. Distinct parallel lines have none; coincident lines share infinitely many points.

Worked examples

Example 1 — Descending line. Describe x + 2y = 6.

Rearrange: y = (6 − x)/2.

x	0	2	6
y	3	2	0

For x = 2, y = (6 − 2)/2 = 2. Join the points. It falls one unit per two units rightward, with intercepts (0, 3) and (6, 0).

Example 2 — Intersection. Compare x + y = 5 and x − y = 1.

Rewrite as y = 5 − x and y = x − 1.

x	0	3	5
y = 5 − x	5	2	0
y = x − 1	−1	2	4

One descends; the other rises. Equating gives 5 − x = x − 1, so x = 3 and y = 2. Intersection: (3, 2). Check: 3 + 2 = 5; 3 − 2 = 1.

Example 3 — Vertical line. Describe x = −2.

x	−2	−2	−2
y	−2	0	3

All points are two units left of the y-axis. The vertical line crosses only the x-axis, at (−2, 0). Its slope is undefined: horizontal change is zero.

Common mistakes

Reversing coordinates; uneven scales; drawing only a segment; assuming every line meets both axes; dividing by zero for vertical slope.

Practice questions

1. Tabulate 2x + y = 4 at x = 0, 1, 2; state its intercepts.
2. Give three points on y = −3; describe its line and intercepts.
3. Find and verify the intersection of y = x + 1 and y = −x + 5.
4. Compare y = 2x + 1 with y = 2x − 3, then with 2y = 4x + 2. Count intersections.

Worked solutions

1. Rearrange to y = 4 − 2x:

x	0	1	2
y	4	2	0

The line descends. Intercepts: (0, 4) and (2, 0).

2. Points (−2, −3), (0, −3), (2, −3) form a horizontal line three units below the x-axis. Its y-intercept is (0, −3); no x-intercept exists.

3. Equate: x + 1 = −x + 5, so 2x = 4. Thus (x, y) = (2, 3). Check: 3 = 2 + 1 and 3 = −2 + 5.

4. The first pair has slope 2 but different y-intercepts; equating gives 1 = −3: no intersection. Dividing 2y = 4x + 2 by 2 reproduces the first equation: infinitely many intersections.