Solve pairs of simultaneous linear equations by graphing and by inspection

Learning goal: Analyze and solve pairs of simultaneous linear equations graphically. Understand that the solution to a pair of linear equations in two variables corresponds to the point or points of intersection of their graphs, because those points satisfy both equations simultaneously; use graphs to find or estimate the solution; recognize all three solution types; and solve simple cases by inspection.

A pair of simultaneous linear equations. A single linear equation in two variables, such as y = 2x − 1, has infinitely many solutions: every point on its line. A pair of simultaneous linear equations, also called a system, asks for the ordered pairs (x, y) that satisfy both equations at the same time. The word simultaneously is the whole idea: a pair that works in one equation but not the other is not a solution of the system.

Why the solution is the intersection. The graph of each equation is the set of all points that satisfy it. A point that lies on both graphs satisfies both equations, and a point that satisfies both equations lies on both graphs. So the solutions of the pair are exactly the points of intersection of the two lines. There is nothing else to look for: if the lines cross at one point, that point is the one solution; if they never cross, there is no solution; if they are the same line, every point on it is a solution.

A fully worked model. Consider y = 2x − 1 and y = −x + 5. Graph the first: intercept (0, −1), slope 2, so it passes through (1, 1), (2, 3) and (3, 5). Graph the second: intercept (0, 5), slope −1, so it passes through (1, 4), (2, 3) and (5, 0). The lines cross at (2, 3). Check both equations: 2(2) − 1 = 3 and −2 + 5 = 3. The pair (2, 3) satisfies both simultaneously, so it is the solution. Now test (4, 7), which lies on the first line since 2(4) − 1 = 7: in the second equation −4 + 5 = 1, not 7, so (4, 7) is not a solution of the pair. Being on one line is not enough.

The three solution types. Two lines in the plane can intersect in exactly one point, in no point, or in every point. One solution: the lines have different slopes, so they must cross somewhere, as with y = 2x − 1 and y = −x + 5; a vertical line such as x = −1, which has no slope, crosses every non-vertical line exactly once. No solution: the lines are parallel and distinct, which means they have the same slope and different intercepts (or are two different vertical lines, such as x = 2 and x = 5), as with y = 2x − 1 and y = 2x + 2; the second is always 3 units above the first, so they never meet. Infinitely many solutions: the two equations describe the same line, as with x + y = 4 and 3x + 3y = 12, where the second is the first multiplied by 3; every point on the line, such as (1, 3) or (0, 4), satisfies both.

Solving simple cases by inspection. Some pairs can be classified without graphing. The pair 3x + 2y = 5 and 3x + 2y = 6 has no solution because the expression 3x + 2y cannot equal 5 and 6 at the same time. The pair 2x − y = 4 and 2x − y = 7 fails the same way. The pair y = 3x − 2 and 2y = 6x − 4 has infinitely many solutions because dividing the second equation by 2 gives the first. And a pair like y = 4 and x = −1 has the single solution (−1, 4) by reading the two equations directly. Inspection works when one equation is a nonzero multiple of the other, or when the left sides match and the right sides do not.

Finding versus estimating from a graph. When the intersection lands on a grid point, as (2, 3) does, the graph gives the exact solution and substitution confirms it. When it does not, the graph gives an estimate. For y = 2x + 1 and y = −x + 6, the lines cross between x = 1 and x = 2, a little closer to 2, at a height between 4 and 5. An estimate of (1.7, 4.4) is reasonable. Substitution shows how close it is: 2(1.7) + 1 = 4.4 exactly on the first line, and −1.7 + 6 = 4.3, close to but not exactly 4.4. The exact intersection has 2x + 1 = −x + 6, so 3x = 5 and x = 5/3, with y = 2(5/3) + 1 = 13/3. Check in the second equation: −5/3 + 18/3 = 13/3. An estimate from a graph is honest as long as you say it is an estimate and confirm by substitution how close it is.

Misconceptions to avoid. A point on one line is not a solution of the system unless it is on the other line too. Two lines that do not cross inside the drawn window may still cross further out; only equal slopes with different intercepts, or two different vertical lines, guarantee that they never meet; equal slopes with the same intercept, or the same vertical line written twice, mean the two equations describe one line. And a pair with infinitely many solutions does not mean any pair (x, y) works; it means every point on that one shared line works, and a point off the line, such as (0, 0) for x + y = 4, still fails.

Worked graph with one solution: y = x + 1 and y = 3x − 3. First line: intercept (0, 1), slope 1, through (2, 3) and (4, 5). Second line: intercept (0, −3), slope 3, through (1, 0) and (2, 3). The lines share (2, 3). Check: 2 + 1 = 3 and 3(2) − 3 = 3. One solution, (2, 3). The slopes 1 and 3 differ, which is why exactly one intersection exists.

Worked no solution by inspection: y = (1/2)x + 3 and 2y = x − 2. Divide the second equation by 2: y = (1/2)x − 1. Both lines have slope 1/2 and the intercepts 3 and −1 differ, so the lines are parallel and distinct. No point satisfies both; the first line is always 4 units above the second.

Worked infinitely many by inspection: x + y = 4 and 3x + 3y = 12. Dividing the second by 3 gives x + y = 4, the same line. Every point on it is a solution: (1, 3) gives 1 + 3 = 4 and 3 + 9 = 12; (6, −2) gives 6 − 2 = 4 and 18 − 6 = 12. But (2, 5) is not a solution, because 2 + 5 = 7, not 4.

Worked estimate then exact: y = x and y = −x + 5. The graph shows the crossing between x = 2 and x = 3, on the diagonal where y = x. Estimate (2.5, 2.5). Substitution: 2.5 = 2.5 on the first line and −2.5 + 5 = 2.5 on the second, so the estimate is in fact exact. In general the algebra confirms it: x = −x + 5 gives 2x = 5 and x = 5/2.

Representation task: On one coordinate grid, graph the three lines y = 2x − 1, y = −x + 5 and y = 2x + 2, each with its intercept and one more point labeled. Mark the intersection of the first two and verify it in both equations. Explain in writing why the first and third lines will never intersect, using their slopes and intercepts. Then write a fourth equation that is the same line as y = −x + 5 but looks different, such as a multiple of it, and explain why the pair has infinitely many solutions.

Independent performance: Create three pairs of linear equations, one of each solution type, without telling a partner which is which. At least one pair must have an intersection that is not a grid point. The partner must graph each pair, classify it, find or estimate the solution, and verify by substitution in both equations, stating clearly whether the result is exact or an estimate. Then classify one more pair by inspection alone. Success criteria: every graph shows the intercept and slope used; each classification is justified by slopes and intercepts, not by appearance alone; each solution is substituted into both equations; an estimate is labeled as an estimate with the substitution showing how close it is.