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Linear systems by graphing

Algebra · level 6, lesson 4 of 6

A question to start

Two phone plans. Plan A costs 30 dollars a month plus 2 dollars per gigabyte of data. Plan B costs only 10 dollars a month, but 4 dollars per gigabyte. If you hardly use any data, Plan B is cheaper. If you use a lot, Plan A is cheaper. Where exactly is the switch-over point, and how much do the plans cost there?

Each plan on its own is a linear relation, like the taxi in level 5. The question asks about both at once: is there a number of gigabytes where the two costs are equal? Two equations about the same two unknowns are called a system, and the simplest way to understand one is to draw it. (The answer is 10 gigabytes and 50 dollars; see the first drawing.)

The idea

A system of linear equations

Call the data used \(g\) (in gigabytes) and the monthly cost \(y\) (in dollars). The two plans give two equations:

\[y = 2g + 30 \qquad\text{and}\qquad y = 4g + 10.\]

A solution of the system is a pair of values, \(g\) and \(y\), that makes both equations true at the same time. For example, \(g = 5\), \(y = 40\) makes the first equation true (\(2 \times 5 + 30 = 40\)) but not the second (\(4 \times 5 + 10 = 30\)), so it is not a solution of the system. It is a point on Plan A's line but not on Plan B's.

Every equation is a line; every solution is a point

The graph of one linear equation is a straight line, and every point on the line is a pair that makes that equation true. So a pair that makes both equations true is a point that lies on both lines: a point where they cross. Solving a system by graphing means drawing both lines and reading off where they meet.

Plan APlan B 1050 3010 gigabytesdollars

The lines cross at \((10, 50)\). Check in both equations: \(2 \times 10 + 30 = 50\) ✓ and \(4 \times 10 + 10 = 50\) ✓. So at 10 gigabytes both plans cost 50 dollars. The graph tells us more than the single point, too: to the left of 10, Plan B's line is lower (cheaper); to the right, Plan A's is.

Drawing a line quickly

Two methods, depending on how the equation is written:

  • Slope and intercept, for equations like \(y = 2x + 3\): start at \((0, 3)\) on the \(y\)-axis, then step 1 across and 2 up, and again. Join the points.
  • Intercepts, for equations like \(2x + 3y = 12\): put \(x = 0\) to find where it crosses the \(y\)-axis (\(3y = 12\), so \((0, 4)\)), and put \(y = 0\) to find where it crosses the \(x\)-axis (\(2x = 12\), so \((6, 0)\)). Join them.

Two points fix a line, but a third point is a good check: if the three do not line up, one of them is wrong.

Three possible pictures

Two straight lines in a plane can be arranged in only three ways:

cross onceone solution parallelno solution same lineinfinitely many
  1. They cross at one point. This happens when the slopes are different. The system has exactly one solution.
  2. They are parallel. Same slope, different intercepts: the lines run side by side forever and never meet. The system has no solution. Such a system is called inconsistent.
  3. They are the same line. Same slope and same intercept: the two equations are really one equation in disguise (for example, one is double the other). Every point on the line is a solution, so there are infinitely many.

Why there is no fourth picture

Could two straight lines cross at exactly two points? No. Through any two points there is only one straight line. So if two lines shared two points, they would be the same line, and then they would share all their points. That is why the number of solutions of a linear system is always 1, 0 or infinitely many, never 2 or 3. (Curves are different: a line and a parabola can meet twice, as you will see in level 7.)

Telling the case without drawing

Rewrite both equations in the form \(y = mx + b\) and compare:

Slopes Intercepts Lines Solutions
different anything cross exactly one
equal different parallel none
equal equal same line infinitely many

The limits of graphing

Graphing shows you the whole situation at a glance, and it is the best way to understand a system. But reading a crossing point off a graph is only as accurate as the drawing. If the lines meet at \(\left(\tfrac{12}{7}, \tfrac{31}{7}\right)\), no hand-drawn graph will tell you that exactly. The next lesson gives two algebraic methods that find the exact answer every time. Graphing remains the way to check that the answer is sensible.

Worked examples

1. The phone plans (shown above): the lines \(y = 2g + 30\) and \(y = 4g + 10\) cross at \((10, 50)\).

2. Slope and intercept. Solve by graphing: \(y = x + 1\) and \(y = -2x + 7\).

  • First line: start at \((0, 1)\), step 1 across and 1 up: \((1, 2)\), \((2, 3)\), \((3, 4)\).
  • Second line: start at \((0, 7)\), step 1 across and 2 down: \((1, 5)\), \((2, 3)\), \((3, 1)\).
  • Both lists contain \((2, 3)\), so the lines cross there. Check: \(2 + 1 = 3\) ✓ and \(-4 + 7 = 3\) ✓.

3. Intercepts. Solve by graphing: \(x + y = 6\) and \(x - y = 2\).

  • First line: \(x = 0\) gives \(y = 6\); \(y = 0\) gives \(x = 6\). Points \((0, 6)\) and \((6, 0)\).
  • Second line: \(x = 0\) gives \(y = -2\); \(y = 0\) gives \(x = 2\). Points \((0, -2)\) and \((2, 0)\).
246 246 (4, 2) x + y = 6x − y = 2
  • The lines cross at \((4, 2)\). Check: \(4 + 2 = 6\) ✓ and \(4 - 2 = 2\) ✓.

4. Parallel lines. How many solutions has the system \(y = 3x + 2\) and \(6x - 2y = 8\)?

  • Rewrite the second: \(-2y = -6x + 8\), so \(y = 3x - 4\).
  • Both slopes are 3; the intercepts are 2 and \(-4\). The lines are parallel: no solution.
  • In words: for any \(x\), the first line is always 6 higher than the second.

5. The same line. How many solutions has the system \(2x + 4y = 8\) and \(x + 2y = 4\)?

  • The first equation is exactly double the second. Both rewrite as \(y = -\tfrac12 x + 2\).
  • Same line: infinitely many solutions, for example \((0, 2)\), \((4, 0)\) and \((2, 1)\). Any point with \(x + 2y = 4\) works.

Try it

  1. Is \((3, 1)\) a solution of the system \(x + 2y = 5\) and \(2x - y = 5\)? Is \((1, 2)\)?
  2. Solve by graphing: \(y = 2x - 1\) and \(y = -x + 5\).
  3. Solve by graphing: \(x + y = 4\) and \(y = x + 2\).
  4. Without graphing, say how many solutions: \(y = -2x + 3\) and \(4x + 2y = 6\).
  5. Without graphing, say how many solutions: \(y = \tfrac12 x + 1\) and \(x - 2y = 6\).
  6. Gym A charges a 50 dollar joining fee plus 20 dollars a month. Gym B has no joining fee but charges 30 dollars a month. After how many months do they cost the same? Which is cheaper for a full year?
  7. Explain why a system of two linear equations cannot have exactly two solutions.
Answers
  1. \((3, 1)\): \(3 + 2 = 5\) ✓ and \(6 - 1 = 5\) ✓, so yes. \((1, 2)\): \(1 + 4 = 5\) ✓ but \(2 - 2 = 0 \ne 5\), so no: it is on the first line only.
  2. First line: \((0, -1)\), \((1, 1)\), \((2, 3)\). Second: \((0, 5)\), \((1, 4)\), \((2, 3)\). They cross at \((2, 3)\). Check: \(4 - 1 = 3\) ✓ and \(-2 + 5 = 3\) ✓.
  3. First line through \((0, 4)\) and \((4, 0)\); second through \((0, 2)\) with slope 1: \((1, 3)\). Cross at \((1, 3)\). Check: \(1 + 3 = 4\) ✓ and \(1 + 2 = 3\) ✓.
  4. \(4x + 2y = 6\) gives \(2y = -4x + 6\), so \(y = -2x + 3\): the same line. Infinitely many solutions.
  5. \(x - 2y = 6\) gives \(-2y = -x + 6\), so \(y = \tfrac12 x - 3\). Same slope \(\tfrac12\), intercepts 1 and \(-3\): parallel, no solution.
  6. Costs \(y = 20m + 50\) and \(y = 30m\). Equal when \(20m + 50 = 30m\), so \(m = 5\) months, at 150 dollars. For 12 months: A costs \(240 + 50 = 290\) dollars and B costs 360 dollars, so A is cheaper. (After month 5, A's line is below B's because its slope is smaller.)
  7. Through two points there is exactly one straight line. If two lines shared two points they would both be that line, so they would share every point: infinitely many solutions, not two.

Common mistake

Checking in only one equation. A point on one line is not a solution of the system. In question 1, \((1, 2)\) fits the first equation perfectly but not the second. Always check a solution in both equations.

Trusting a rough graph too much. If your lines seem to cross "at about \((2.4, 3.8)\)", that is an estimate, not an answer. Use the methods of the next lesson to find the exact point, then use the graph to see that it is reasonable.

Where it leads