Year 10 Extended · Unit 2 Generalizing, coordinate geometry and linear programming · Lesson 4 of 6

How many solutions can two linear equations have?

Two equations in x and y are not always a puzzle with one answer. The coefficients say what kind of answer to expect before anything is solved, and a letter in an equation shows where one kind turns into another.

Number and AlgebraGDC optional

Warm-up

Three quick ones from earlier work. Try each in your head or on paper before you open the answer.

  1. From Year 9: solve x + y = 7 and x − y = 3.
    Answerx = 5 and y = 2. Add the equations to get 2x = 10.
  2. From Year 8: what does multiplying both sides of an equation by −1 do to the signs?
    AnswerIt changes every sign, on both sides.
  3. From lesson 3: two lines with gradients 2/3 and 2/3 are what?
    AnswerParallel, because their gradients are equal.
Words used in this lesson
system of equations
two or more equations that must be true at the same time. For example: x + y = 9 and x − y = 1.
elimination
adding or subtracting equations to remove one letter. For example: adding 2y and −2y gives 0y.
parallel
lines with the same gradient that never meet. For example: y = 4x + 1 and y = 4x − 7.
coincident
the same line, so every point is a solution. For example: x + y = 2 and 3x + 3y = 6.
break-even
the amount where two costs are equal. For example: where two phone plans cost the same.
equivalence transformation
a step that keeps the same solutions. For example: adding 5 to both sides.

Two lines, three outcomes

Two friends in Nairobi compare phone offers. The plan from the first shop costs a fixed fee plus a price per gigabyte, and the second shop's plan is built the same way with a different fee and a different price. The cost of each plan is a straight line against gigabytes, and the question is where the lines meet.

Predict first: the first figure fixes the line 2x + 3y = 6 and lets you choose a second line ax + by = c. Write down one equation that you think gives a line that never meets the fixed one, and one that you think is the same line in disguise. Then set a, b and c to test them.

3
2
5
one solutionnumber of solutions
(3/5, 8/5)where the lines meet

3x + 2y = 5 against 2x + 3y = 6: 2 × 2 − 3 × 3 = −5, which is not 0, so the gradients differ and the lines cross once, at (3/5, 8/5).

Show me the steps
  1. a = 3, b = 2, c = 5: 2 × 2 − 3 × 3 = −5, not zero, so the lines cross once, at (3/5, 8/5).
  2. a = 4, b = 6, c = 5: 2 × 6 − 3 × 4 = 0, so the gradients are equal. 4x + 6y = 5 is not twice 2x + 3y = 6, so the lines are parallel: no solution.
  3. a = 4, b = 6, c = 12: 4x + 6y = 12 is twice 2x + 3y = 6. It is the same line: infinitely many solutions.
  4. a = 2, b = 3: with the gradients equal, c alone decides. c = 6 gives the same line; every other c gives a parallel line and no solution.

Two lines can meet once, never, or everywhere. For ax + by = c and px + qy = r, the lines have different gradients, and meet once, when aq − bp is not 0. When aq − bp = 0 the gradients are equal: the lines are parallel and never meet, unless the second equation is a multiple of the first, when it is the same line and every point is a solution.

Solving keeps the solutions with equivalence transformations: add the same thing to both sides, multiply both sides by a number that is not zero, or substitute equals for equals. Multiplying by zero, or by an expression that can be zero, can add false solutions, so it is not an equivalence.

Predict first: in the second figure Plan A is fixed. With Plan B's rate set to 4, which fee puts the break-even at exactly 30 gigabytes? Work it out, then slide the fee to test it.

100
5
25gigabytes at break-even
225cost at break-even

Plan A: 150 + 3g. Plan B: 100 + 5g. They cost the same when 150 + 3g = 100 + 5g, so 2g = 50 and g = 25, where both cost 225.

Show me the steps
  1. Fee 100, rate 5: 150 + 3g = 100 + 5g gives g = 25, where both cost 225. Below 25 gigabytes B is cheaper; above, A is.
  2. Fee 60, rate 4: 150 + 3g = 60 + 4g gives g = 90, where both cost 420. The lower fee wins for longer.
  3. Fee 140, rate 8: 150 + 3g = 140 + 8g gives g = 2, where both cost 156. The high rate catches up quickly.
  4. If Plan B's rate were also 3 the lines would be parallel: the gap stays the same size and there is no break-even.

Two common mistakes. Reading 0 = 0 as “no solution”: when elimination leaves 0 = 0 the equations are the same line and every point works; when it leaves 0 = 5 there is no solution. And multiplying only one side: 2(x + y = 3) is 2x + 2y = 6, not 2x + 2y = 3.

Worked examples

1. Solve 5x − 3y = 1 and 2x + 3y = 13.

  1. The y terms are −3y and +3y, so add the equations: 7x = 14, so x = 2.
  2. Put x = 2 into 2x + 3y = 13: 4 + 3y = 13, so y = 3. Check in the first: 10 − 9 = 1. The solution is (2, 3).

2. How many solutions do x + 2y = 4 and 3x + 6y = 5 have?

  1. Multiply the first equation by 3: 3x + 6y = 12. The left side matches the second equation.
  2. The right sides are 12 and 5, which differ, so subtracting leaves 0 = 7: no solution. The lines are parallel.

3. Find k so that 2x + ky = 8 and 3x + 6y = 12 have infinitely many solutions.

  1. For the same line the second equation is a multiple of the first. Multiplying 2x + ky = 8 by 1.5 gives 3x + 1.5ky = 12.
  2. That matches 3x + 6y = 12 when 1.5k = 6, so k = 4. The right sides agree too, because 1.5 × 8 = 12, so there are infinitely many solutions.

Your go. The first step is given. Find k so that kx + 4y = 7 and 3x + 2y = 5 have no solution. For no solution the gradients must be equal, so k ÷ 3 = 4 ÷ 2.

  1. k ÷ 3 = 2, so k = 6.
  2. Now check that the right-hand sides do not match, so the lines are parallel and not the same line.
Check the answer

With k = 6 the first equation is 6x + 4y = 7. Twice the second equation is 6x + 4y = 10. The left sides match and the right sides do not (7 and 10), so the lines are parallel and there is no solution when k = 6.

With a GDC (optional)

A GDC can graph both lines: rewrite each equation as y = … first, enter both, and use the intersect tool to read the crossing point. Parallel lines show no intersection, and two equations of the same line show as one line.

Take care: two lines with nearly equal gradients can look parallel on a small screen, and the coefficients are what decide. Everything on this page can be done by hand.

Check before you go on

1. Which of these systems has no solution?

Your turn

Start at Level 1 and go as far as you can. Type negative numbers with a minus sign, for example -7. A wrong answer gets a reason, not just a cross: read it, then try again.

Level 1 · Getting started

2. Solve 2x + 3y = 12 and 5x − 2y = 11. What is x?

3. The equations 2x + 5y = 10 and 4x + 10y = c describe the same line. What is c?

Level 2 · Building

4. For which value of k do 4x + ky = 9 and 6x + 15y = 5 have no solution?

5. Plan A costs 180 rand a month plus 2 rand for each gigabyte. Plan B costs a monthly fee f plus 5 rand for each gigabyte. The two plans cost the same at exactly 25 gigabytes. What is f?

Level 3 · Stretch

6. The equations ax + 9y = 21 and 2x + by = 7 describe the same line. What is a + b?

7. The lines x + y = 10 and x − y = 2 meet the y-axis at two points and meet each other at a third. What is the area of the triangle they make, in square units?

Something to think about

A system of two equations can have no solution. Can a system of three straight-line equations in x and y have exactly two solutions? Think about where straight lines can meet.

One way to think about it

No. Two different straight lines meet in at most one point, so a common solution of all three lines is at most one point. If two of the equations describe the same line there is still a single line, so the system has no solution, one solution or infinitely many, never exactly two.

Exit ticket

8. The equations 3x − 6y = 9 and −x + 2y = k have infinitely many solutions. What is k?