Solving Systems of Equations by Graphing: Lesson Plan

A system of two linear equations asks for an (x, y) that makes both true. On a graph, each equation is a line, and the solution is where the lines cross. For y = 2x − 1 and y = −x + 5, the lines meet at (2, 3).

Learning objectives

• Solve a system of two linear equations by graphing.
• Recognise systems with one, no or infinitely many solutions from their slopes.
• Check a graphical solution algebraically, by substitution.

The opening question

"Phone plan A costs €10 a month plus €2 per GB. Plan B costs €20 plus €1 per GB. When are they the same price?" Each plan is a line; they cross at 10 GB, €30. Before that A is cheaper, after it B.

The lesson, click by click

1. Click 0: the line y = 2x − 1. Every point on it solves the first equation.
2. Click 1: the line y = −x + 5. Every point on it solves the second.
3. Click 2: the crossing point, (2, 3), solves both.
4. Click 3, the check: 2x − 1 = −x + 5, so 3x = 6, x = 2 and y = 3.
5. Click 4: parallel lines (same slope, different intercept): no solution. The same line twice: infinitely many.

Common mistakes to watch for

• Reading only one coordinate. The solution is a pair, (2, 3), not "2".
• Trusting a rough sketch. A crossing at (2.4, 3.8) looks like (2, 4) on a hand drawing. Always check algebraically.
• Rearranging wrongly. From 2x + y = 7, y = 7 − 2x, not 2x − 7.

Check questions

1. Solve 2x + y = 7 and x − y = 2. ((3, 1).)
2. How many solutions do y = x + 1 and y = x − 3 have? (None: parallel lines.)
3. In the phone-plan problem, which plan is cheaper for 5 GB? (A: €20 against €25.)

Building it in Tracé

1. Add a graph and choose Equation or inequality in x and y (circle, region…) to type 2x + y = 7 directly, without rearranging; add the second line the same way.
2. Or plot two functions, y = 2x - 1 and y = -x + 5, and mark the crossing point.
3. Tick Live slider on one slope and slide it until the lines are parallel: the solution runs off to infinity, then disappears.

Questions

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