Three equations, three unknowns
Tonight's idea
You can already solve two equations in two unknowns. Three unknowns need the same idea, kept in order.
Three moves never change the solutions of a system: swap two equations, multiply an equation by a nonzero number, or add a multiple of one equation to another. Each move can be undone, so no solution is lost or gained.
Use the first equation to remove \(x\) from the other two. Then use the new second equation to remove \(y\) from the third. This elimination leaves a staircase:
The last equation now contains only \(z\). Solve it, put \(z\) into the equation above to get \(y\), then both into the first to get \(x\): back-substitution.
Eliminate downward until the system is a staircase. Then substitute upward.
Sometimes the last row collapses. If it reads \(0 = 5\), there is no solution. If it reads \(0 = 0\), one equation repeated the others, and there are infinitely many (the cases from grade 8). This routine, Gaussian elimination, is the same for any number of unknowns.
Watch one: label the rows and keep them in order
\[\begin{aligned} x + y + z &= 6 && (1)\\ 2x + 3y + z &= 11 && (2)\\ x - y + 2z &= 5 && (3)\end{aligned}\]
Remove \(x\): \((2) - 2\times(1)\) gives \(y - z = -1\), call it \((4)\). And \((3) - (1)\) gives \(-2y + z = -1\), call it \((5)\).
Remove \(y\): \((5) + 2\times(4)\) gives \(-z = -3\).
Climb back up. \(z = 3\). Then \((4)\): \(y - 3 = -1\), so \(y = 2\). Then \((1)\): \(x + 2 + 3 = 6\), so \(x = 1\). Check in \((3)\): \(1 - 2 + 6 = 5\). Correct.
Enter any system and watch the staircase form, one row operation at a time: mathdetours.com/demos/rowreduce.html.
Your turn
- This system is already a staircase. Solve it by back-substitution: \(x + 2y - z = 3\), \(\;y + 3z = 7\), \(\;2z = 4\).
- Solve by elimination, then check your answer in all three equations: \(x + y + z = 2\), \(\;x + 2y + 3z = 2\), \(\;2x + 3y + z = 1\).
- Solve: \(2x + y - z = 1\), \(\;4x - y + 3z = 15\), \(\;-2x + 3y + 2z = -3\).
- Run elimination on \(x + y + z = 3\), \(\;x + 2y + 3z = 6\), \(\;2x + 3y + 4z = 10\). What does the last row say, and what does that mean? Now change the 10 to 9 and run it again. Describe all the solutions.
Stretch: three points fix a parabola
The parabola \(y = ax^2 + bx + c\) passes through \((1, 2)\), \((2, 3)\) and \((3, 8)\). Each point gives one equation in the unknowns \(a\), \(b\) and \(c\). Solve the system and write down the parabola.
Answers
- \(z = 2\). Then \(y + 6 = 7\), so \(y = 1\). Then \(x + 2 - 2 = 3\), so \(x = 3\). Solution \((3, 1, 2)\).
- \((2) - (1)\): \(y + 2z = 0\). \((3) - 2\times(1)\): \(y - z = -3\). Subtract these: \(3z = 3\), so \(z = 1\), \(y = -2\), \(x = 3\). Check: \(3 - 2 + 1 = 2\), \(3 - 4 + 3 = 2\), \(6 - 6 + 1 = 1\).
- \((2) - 2\times(1)\): \(-3y + 5z = 13\). \((3) + (1)\): \(4y + z = -2\). Add 4 times the first to 3 times the second: \(23z = 46\), so \(z = 2\). Then \(4y = -4\), \(y = -1\), and \(2x - 1 - 2 = 1\), \(x = 2\). Solution \((2, -1, 2)\).
- \((2) - (1)\): \(y + 2z = 3\). \((3) - 2\times(1)\): \(y + 2z = 4\). Subtracting gives \(0 = 1\), which is impossible: no solution. With 9 in place of 10, the last row is \(0 = 0\): the third equation is just the sum of the first two, so it adds no information. Choose any value \(z = t\); then \(y = 3 - 2t\) and \(x = 3 - y - z = t\). The solutions are \((t,\, 3 - 2t,\, t)\) for every \(t\): infinitely many, for example \((0, 3, 0)\) and \((1, 1, 1)\).
Stretch. The equations are \(a + b + c = 2\), \(4a + 2b + c = 3\), \(9a + 3b + c = 8\). Subtracting the first from the second, and the second from the third, removes \(c\): \(3a + b = 1\) and \(5a + b = 5\). So \(2a = 4\), \(a = 2\), \(b = -5\), \(c = 5\). The parabola is \(y = 2x^2 - 5x + 5\). Check at \(x = 3\): \(18 - 15 + 5 = 8\).