From Conditions to a System
A linear system is a collection of linear equations that must hold at the same time. Solving the system means finding every assignment of values to its variables that satisfies every equation. The equations may come already written symbolically, or they may need to be built from a description. In either case, the main discipline is the same: decide what each variable represents, keep that order consistent, and check any proposed answer in the original equations.
In this tutorial all variables take values in \(\mathbb{R}\). A solution of a system in \(n\) variables is an ordered \(n\)-tuple of real numbers that makes every equation true. The system has no solution if no such tuple exists, a unique solution if exactly one tuple exists, and infinitely many solutions if there are infinitely many such tuples. These possibilities are different: satisfying one equation is not enough if another equation in the system fails.
When modeling a word problem, define the variables before writing equations. Then translate each independent condition into its own equation. A total count usually gives an equation in which the coefficients are counts; a total cost or weight uses the relevant per-item amounts as coefficients. Units provide a useful check: the two sides of an equation should measure the same kind of quantity.
A Reliable Workflow
State exactly what each variable counts or measures, and keep that order consistent throughout the work.
Write one equation for each stated condition, preserving signs, totals, and units.
Use substitution or elimination to combine the equations. Keep track of which equations are being combined.
Determine whether the work gives a contradiction, one assignment, or a parameterized family, and test the result in the original equations.
Elimination is valid because replacing one equation by its difference with a multiple of another does not change which assignments satisfy the system. This fact justifies a common solving step, rather than merely suggesting a convenient calculation.
Proof. If an assignment satisfies \(E_1\) and \(E_2\), then it satisfies \(E_2-cE_1\), because subtracting \(c\) times a true equation from another true equation gives a true equation. Conversely, if an assignment satisfies \(E_1\) and \(E_2-cE_1\), then it satisfies \(E_2=(E_2-cE_1)+cE_1\). Every other equation in the system is unchanged, so the two systems have exactly the same solutions. \(\square\)
Modeling and Solving a Two-Variable Problem
Worked Example: Tickets and Total Cost
A theatre sells adult tickets for \(14\) dollars and student tickets for \(8\) dollars. A group buys \(17\) tickets for a total of \(190\) dollars. How many of each type did it buy?
Let \(a\) be the number of adult tickets and \(s\) the number of student tickets. The ticket count and the cost give two separate conditions:
From the first equation, \(s=17-a\). Substitute this into the cost equation and solve:
Then \(s=17-9=8\). The group bought \(9\) adult tickets and \(8\) student tickets. Check both original conditions: \(9+8=17\), and \(14(9)+8(8)=126+64=190\). Both conditions hold.
Solving the equations produces a candidate answer to the original problem. The check matters because the equations encode the assumptions: it confirms that the answer meets every condition that was modeled. It is also a useful way to catch arithmetic errors even when the model itself is correct.
Classifying a System in Two Variables
For two equations in two variables, elimination can expose the three possible outcomes. If elimination produces an equation such as \(0=5\), there is no solution. If it gives a definite value for one variable and then for the other, there is one solution. If the equations reduce to an identity such as \(0=0\), one equation may be redundant, leaving a free choice and infinitely many solutions.
Proof. Write the equations as \(a_1x+b_1y=c_1\) and \(a_2x+b_2y=c_2\). If both equations have zero coefficients on \(x\) and \(y\), each is either \(0=0\) or a contradiction \(0=c\) with \(c\ne0\). A contradiction makes the system unsolvable; if both equations are identities, every pair \((x,y)\in\mathbb{R}^2\) is a solution. If exactly one equation has nonzero variable coefficients, any other equation with zero variable coefficients either contradicts the system or adds no restriction. The nonzero-coefficient equation describes infinitely many pairs: if its \(x\)-coefficient is nonzero, choose any real \(y\) and solve uniquely for \(x\); if its \(x\)-coefficient is zero, its \(y\)-coefficient is nonzero, so choose any real \(x\) and solve for \(y\). Thus these cases give no solutions or infinitely many, not a finite positive number.
It remains to consider the case where both equations have at least one nonzero variable coefficient. If either equation has a nonzero \(x\)-coefficient, use that equation and, if needed, exchange the names \(x\) and \(y\) so that the first equation has a nonzero \(x\)-coefficient. If neither has a nonzero \(x\)-coefficient, both have nonzero \(y\)-coefficients; exchange the variable names and reach the same setup. The first equation can now be solved as \(x=(c_1-b_1y)/a_1\), where \(a_1\ne0\). Substitution into the second equation leaves one equation in \(y\), of the form \(dy=e\), for real \(d,e\). If \(d\ne0\), there is exactly one \(y=e/d\), and then exactly one \(x\). If \(d=0\) and \(e\ne0\), the equation is contradictory and there is no solution. If \(d=0\) and \(e=0\), every real \(y\) works in the substituted equation, and each choice determines an \(x\), giving infinitely many solutions. These cases exhaust the possibilities, proving the theorem. \(\square\)
Worked Example: A Three-Variable System with One Solution
Solve and classify the system
Subtract the first equation from the second, and also subtract the first from the third. These operations preserve the solutions by the proposition above:
The second derived equation gives \(y=2z-7\). The first gives \(x=4+2y\). Substitute both expressions into the first original equation:
Using \(y=2z-7\), we obtain \(3(2z-7)+z=0\), so \(7z-21=0\), hence \(z=3\). Then \(y=2(3)-7=-1\) and \(x=4+2(-1)=2\). This gives the candidate \((x,y,z)=(2,-1,3)\).
Check: \(2+(-1)+3=4\); \(2(2)-(-1)+3=8\); and \(2+2(-1)-3=-3\). All three original equations hold. The substitutions determine each variable to have only one value, so the solution is unique.
Contradictions and Free Choices in Three Variables
With three variables, a contradiction can occur even when there are fewer than three equations. An equation that is a multiple of another may instead add no new restriction. When fewer independent conditions remain than variables, a free choice can describe an infinite family. The following examples show how to identify these outcomes directly, without needing to use a general classification rule for larger systems.
Worked Example: A Contradiction Means No Solution
Classify the system
Doubling the first equation gives \(2x+2y+2z=4\). The second equation requires the same left-hand side to equal \(5\), so together they imply \(4=5\), which is false. Therefore no triple can satisfy both equations, and the system has no solution. In particular, the failure is not caused by a value missed during solving; the two conditions themselves are incompatible.
Worked Example: A Parameter Describes Infinitely Many Solutions
Solve and classify
Subtract the first equation from the second to get \(x-2y=1\), so \(x=1+2y\). Substituting this into the first equation gives \(1+2y+y+z=4\), or \(z=3-3y\). The variable \(y\) is unrestricted, so write \(y=t\), where \(t\in\mathbb{R}\). The full solution set is
Check: For the first equation, \((1+2t)+t+(3-3t)=4\). For the second, \(2(1+2t)-t+(3-3t)=5\). Both equations hold for every real \(t\), and different values of \(t\) give different \(y\)-coordinates. The system therefore has infinitely many solutions.
Common Pitfall: Losing a Sign or Changing Variable Order
Suppose \(x\) is the number of large containers and \(y\) is the number of small containers. If there are \(10\) containers altogether, the count equation is \(x+y=10\). A common mistake is to solve it as \(x=10+y\). That rearrangement has the wrong sign: subtracting \(y\) from both sides gives \(x=10-y\). For example, if \(y=4\), the correct equation gives \(x=6\); the incorrect expression gives \(x=14\), and \(14+4\ne10\).
A separate but related error occurs when an ordered answer is reported in the opposite variable order from the one defined. If the variables were defined as \(x=\) large containers and \(y=\) small containers, then \((6,4)\) means six large and four small. Reporting \((4,6)\) silently swaps the meanings. Write the variable definitions beside the result, and substitute the values into the equations in that same order.
| What the work produces | Classification | What to verify |
|---|---|---|
| A false statement such as \(0=5\) | No solution | The equations really do imply incompatible conditions. |
| A determined value for each variable | One solution | Every original equation holds for the resulting tuple. |
| A free parameter with valid solutions | Infinitely many solutions | The parameter can vary and each resulting tuple satisfies every equation. |
| A proposed word-problem answer | Check the equations and context | Counts, units, variable meanings, and original conditions agree. |
Classification is part of solving, not an extra label added afterward. A contradiction explains why no answer exists; a determined tuple must pass every equation; and a parameterized family must satisfy all equations for every permitted parameter value. In a modeled problem, a final check should also confirm that the values make sense in context—for example, a number of tickets must be a whole number and cannot be negative.
Check Your Understanding
For each system, show enough work to justify the classification, and substitute any proposed solution into the original equations.
- A shop sells notebooks for \(5\) dollars and pens for \(2\) dollars. It sells \(12\) items for \(42\) dollars. Define variables and find the number of each item.
- Classify \(x+y=6\) and \(2x+2y=12\). Describe the solution set if it is infinite.
- Classify \(x+y+z=3\) and \(2x+2y+2z=8\). Explain the contradiction or give a solution.
- Solve and classify \(x+y+z=5\), \(x-y+z=1\), and \(2x+z=6\).
- If \(x\) is the number of red objects and \(y\) the number of blue objects, explain what ordered pair \((3,7)\) means and how to check it in a total-count equation.
