This note covers solving systems of up to three linear equations in three unknowns. You will see how to use elimination and row operations, how to use your GDC, and how to recognise systems with one solution, infinitely many solutions or no solution. The geometry of planes explains why these three outcomes are the only possibilities.
Key ideas
A system of linear equations is a set of equations where each unknown appears only to the first power. For three unknowns \(x\), \(y\) and \(z\), each equation describes a plane in three-dimensional space. The solutions are the points that lie on all the planes at once.
There are three possible outcomes.
- A unique solution. The three planes meet at a single point.
- Infinitely many solutions. The planes share a whole line (or are the same plane). The solution is written using a parameter.
- No solution. The planes have no point in common. The system is called inconsistent.
Elimination. Add multiples of one equation to another to remove one unknown at a time, until you have an equation in a single unknown. Then substitute back. This is the same method as for two equations, repeated.
Row reduction. You can write the system as an array of numbers, with one row per equation, and apply the same operations to the rows. The allowed operations are swapping two rows, multiplying a row by a non-zero number, and adding a multiple of one row to another. Aim for a form where the left-hand part looks like a staircase.
Read the final rows carefully.
- A row such as \(0=0\) means one equation is redundant. If this leaves fewer independent equations than unknowns, there are infinitely many solutions.
- A row such as \(0=5\) is impossible. The system is inconsistent and has no solution.
Using your GDC. Your GDC has a simultaneous equation solver. Enter the number of equations and unknowns, then the coefficients and constants. The solver gives a unique solution when there is one. If the system has no solution or infinitely many, the solver reports an error or gives no useful answer. In that case you must do the elimination by hand.
Parameters. When there are infinitely many solutions, let one unknown equal a parameter, such as \(z=t\), and write the others in terms of \(t\). The result describes a line in space.
Definitions and formulae
| Term or formula | Meaning / when to use |
|---|---|
| \(a_1x+b_1y+c_1z=d_1\) | A linear equation in three unknowns. It represents a plane. |
| Consistent system | A system with at least one solution. |
| Inconsistent system | A system with no solution. Row reduction produces something like \(0=5\). |
| Row operations | Swap rows, multiply a row by a non-zero constant, add a multiple of one row to another. |
| Parameter | A letter such as \(t\) used to describe an infinite set of solutions. |
Worked examples
Example 1 Solve \(x+y+z=6\), \(2x-y+z=3\), \(x+2y-z=2\).
Label the equations (1), (2) and (3).
Add (1) and (3) to remove \(z\): \(2x+3y=8\). Call this (4).
Subtract (1) from (2): \(x-2y=-3\). Call this (5).
From (5), \(x=2y-3\). Substitute into (4): \(2(2y-3)+3y=8\), so \(7y=14\) and \(y=2\).
Then \(x=2(2)-3=1\). From (1), \(z=6-1-2=3\).
Check in (2): \(2(1)-2+3=3\). Correct. Check in (3): \(1+4-3=2\). Correct.
Answer: \(x=1,\ y=2,\ z=3\)
Example 2 Show that the system \(x+2y-z=4\), \(2x+y+z=5\), \(x-y+2z=1\) has infinitely many solutions, and describe them.
Label the equations (1), (2) and (3).
Subtract (1) from (2): \(x-y+2z=1\). This is exactly equation (3), so (3) gives no new information.
So there are only two independent equations for three unknowns, and there are infinitely many solutions.
Let \(z=t\). From (3), \(x=1+y-2t\). Substitute into (1): \((1+y-2t)+2y-t=4\), so \(3y=3+3t\) and \(y=1+t\).
Then \(x=1+(1+t)-2t=2-t\).
Check in (2): \(2(2-t)+(1+t)+t=4-2t+1+t+t=5\). Correct for every \(t\).
Answer: \(x=2-t,\ y=1+t,\ z=t\) for any real \(t\) (a line of solutions)
Example 3 Show that the system \(x+2y-z=4\), \(2x+y+z=5\), \(x-y+2z=2\) has no solution.
Label the equations (1), (2) and (3).
Subtract (1) from (2): \(x-y+2z=1\). Call this (4).
Equation (3) says \(x-y+2z=2\), but equation (4) says \(x-y+2z=1\).
The same expression cannot equal both 1 and 2, so the equations contradict each other.
Answer: no solution, because the system is inconsistent
Common errors
- Applying an operation to only one side of an equation. When you multiply or add, do it to every term, including the right-hand side.
- Eliminating different unknowns in different pairs of equations, which leaves you with equations that still have three unknowns. Choose one unknown and remove it from two pairs.
- Concluding "no solution" when the working gives \(0=0\). A row of \(0=0\) means infinitely many solutions, not none.
- Giving only values of \(x\) and \(y\) when there is a parameter. State all three unknowns in terms of \(t\).
- Not checking the answer in an equation that was not used to find it.
Exam tip: If your GDC gives an error for a system, do not assume you made a typing mistake. It is often a signal that the system has no unique solution, so look for a dependent or contradictory equation.
Check yourself
- Solve \(x+y+z=3\), \(x-y+z=1\), \(2x+y-z=2\).
Answer
Subtract the second from the first: \(2y=2\), so \(y=1\). Then \(x+z=2\) and \(2x-z=1\). Adding gives \(3x=3\), so \(x=1\) and \(z=1\). The solution is \(x=y=z=1\). - Explain why \(3x+2y=4\) and \(6x+4y=9\) have no solution.
Answer
Doubling the first gives \(6x+4y=8\), which contradicts \(6x+4y=9\). The lines are parallel and never meet. - For the system \(x+y=2\), \(2x+2y=k\), find the value of \(k\) for which there are infinitely many solutions.
Answer
Doubling the first equation gives \(2x+2y=4\). The second equation matches it only when \(k=4\). Then every point on the line \(x+y=2\) is a solution. - For which value of \(k\) does the system \(x+2y-z=4\), \(2x+y+z=5\), \(x-y+2z=k\) have a solution?
Answer
Subtracting the first from the second gives \(x-y+2z=1\). So the third equation is consistent only if \(k=1\). For any other \(k\) the system has no solution.