Math Detours · Grade 8 · Sharpen · 4

Two equations, two unknowns

Substitution, elimination, and the picture: two lines cross once, never, or everywhere.

Name ____________________Date ____________

Tonight's idea

An equation in \(x\) and \(y\) with no powers, like \(2x + 3y = 12\), is a line: the points \((x, y)\) that make it true. Two such equations together are a system. A solution makes both true, so it is on both lines: where they cross.

xy (3, 2) 32 2x + 3y = 12 x − y = 1
Every point on the blue line satisfies the first equation, every point on the orange line the second. Only \((3, 2)\) satisfies both.

Two lines can do only three things. Cross once: one solution. Run parallel: none. Lie on top of each other: every point is a solution. This is the picture behind One, none, or infinitely many.

Two ways to find it without drawing.

Substitution. If one equation gives a letter on its own, such as \(x = y + 1\), write \(y + 1\) wherever \(x\) appears in the other. At the crossing both equations hold, so swapping equal expressions changes nothing. One equation in one letter remains.

Elimination. Add equal things to equal things and the results are equal. If one equation has \(+3y\) and the other \(-3y\), adding them removes \(y\). If they do not match, multiply one whole equation by a number first.

Both methods end with one letter. Find it, go back for the other, check both.

Watch one: \(2x + 3y = 12\) and \(x - y = 1\)

Substitution: the second equation gives \(x = y + 1\). Put that into the first: \(2(y + 1) + 3y = 12\), so \(5y + 2 = 12\) and \(y = 2\). Then \(x = 3\).

Elimination: multiply the second equation by 3 to get \(3x - 3y = 3\). Add it to the first: \(5x = 15\), so \(x = 3\). Then \(3 - y = 1\) gives \(y = 2\).

Same point, \((3, 2)\). Check in both: \(6 + 6 = 12\) and \(3 - 2 = 1\). That is where the lines cross in the figure.

Move the two lines and watch the crossing point, or make them parallel: mathdetours.com/demos/systems.html.

Your turn

  1. Solve by substitution: \(y = 2x - 1\) and \(3x + y = 14\). Check your point in both equations.
  2. Solve by elimination: (a) \(3x + 2y = 16\) and \(5x - 2y = 8\); (b) \(4x + 3y = 1\) and \(2x + 5y = -3\).
  3. For each pair, say how many solutions there are, then find them if there are any. (a) \(2x - y = 3\) and \(4x - 2y = 6\); (b) \(2x - y = 3\) and \(4x - 2y = 10\); (c) \(2x - y = 3\) and \(4x + 2y = 10\). Describe what elimination does in (a) and (b).
  4. A show sells 250 tickets. Adults pay £8, children pay £5, and the takings are £1565. (a) How many adults and how many children? (b) The next night's report says 250 tickets and £2010. Set up the same system and solve it. Explain why the lines still cross but the report must be wrong.

Stretch: when do the lines go parallel?

The system \(x + ky = 3\) and \(kx + 4y = 6\) depends on a number \(k\). For which values of \(k\) does it have no solution? For which does it have infinitely many? Think about when the two lines have the same slope.

Answers

  1. \(3x + (2x - 1) = 14\), so \(5x = 15\), \(x = 3\), \(y = 5\). Check: \(2(3) - 1 = 5\) and \(9 + 5 = 14\).
  2. (a) Add: \(8x = 24\), \(x = 3\); then \(9 + 2y = 16\), \(y = \tfrac72\). (b) Double the second: \(4x + 10y = -6\). Subtract the first: \(7y = -7\), \(y = -1\); then \(4x - 3 = 1\), \(x = 1\).
  3. (a) Doubling the first gives the second exactly. Subtracting leaves \(0 = 0\): the same line, infinitely many solutions. (b) Doubling the first gives \(4x - 2y = 6\), but the second says \(4x - 2y = 10\). Subtracting leaves \(0 = 4\), which is false: parallel lines, no solution. (c) Double the first and add: \(8x = 16\), \(x = 2\), \(y = 1\). One solution, \((2, 1)\).
  4. (a) \(a + c = 250\) and \(8a + 5c = 1565\). Substitute \(c = 250 - a\): \(8a + 1250 - 5a = 1565\), so \(3a = 315\), \(a = 105\) adults and \(c = 145\) children. (b) Now \(3a = 760\), so \(a = 253\tfrac13\) and \(c = -3\tfrac13\). The lines cross, but at a point with a negative number of children and a fraction of an adult. With 250 tickets the most you can take is \(250 \times 8 = £2000\), so £2010 cannot have happened.

Stretch. The slopes are \(-\dfrac{1}{k}\) and \(-\dfrac{k}{4}\). They are equal when \(k^2 = 4\), so \(k = 2\) or \(k = -2\). With \(k = 2\) the second equation is exactly twice the first: the same line, infinitely many solutions. With \(k = -2\) the equations are \(x - 2y = 3\) and \(-2x + 4y = 6\); doubling the first and adding gives \(0 = 12\): parallel, no solution. Every other \(k\), including \(k = 0\), gives exactly one crossing.

← PreviousNext: Why a⁰ = 1 and a⁻³ = 1/a³ →