Systems of Linear Equations and Inequalities

A system combines conditions that must be true at the same time. Linear systems are useful whenever quantities are connected by fixed rates: prices and totals, mixtures, production limits, or travel plans. This lesson builds on algebraic rearranging and graphing lines; review the Algebra and Functions lessons if solving a one-variable equation or reading slope is unfamiliar.

What a Solution Means

A solution to a system of equations is an ordered pair, or ordered triple, that makes every equation true. For two linear equations in two variables, each equation is a line. A solution is therefore a point common to both lines.

One solution: distinct non-parallel lines intersect once.
No solution: distinct parallel lines never meet.
Infinitely many solutions: the equations describe the same line.

Worked example: check a proposed point

Does (2, 3) solve the system x + y = 5 and 2x − y = 1?

First equation:  2 + 3 = 5       yes
Second equation: 2(2) − 3 = 1    yes

Because both equations are true, (2, 3) is the solution.

Substitution

Use substitution when one equation already isolates a variable, or can do so easily. Replace that variable in the other equation with an equal expression. The result is one equation in one variable; then substitute back to find the other coordinate.

Worked example: solve by substitution

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

Replace y in the second equation with 2x + 1:
3x + (2x + 1) = 13
5x + 1 = 13
5x = 12
x = 12/5

y = 2(12/5) + 1 = 24/5 + 5/5 = 29/5

Solution: (12/5, 29/5)

A quick check is worthwhile: 3(12/5) + 29/5 = 65/5 = 13.

Elimination

Elimination uses addition or subtraction to make one variable disappear. You may multiply an entire equation by a nonzero number first. This operation preserves its solutions, provided every term on both sides is multiplied.

Worked example: align opposite coefficients

Solve 2x + 3y = 17 and 4x − 3y = 7.

  2x + 3y = 17
  4x − 3y =  7
  -------------
  6x      = 24
   x      = 4

Substitute x = 4 into 2x + 3y = 17:
8 + 3y = 17
3y = 9
y = 3

Solution: (4, 3)

Worked example: multiply before eliminating

Solve 3x + 2y = 16 and 5x − 3y = 7.

Multiply the first equation by 3 and the second by 2:
  9x +  6y = 48
 10x −  6y = 14
  --------------
 19x       = 62
  x        = 62/19

Use 3x + 2y = 16:
2y = 16 − 3(62/19) = 118/19
y = 59/19

Solution: (62/19, 59/19)

Special Outcomes

After elimination, a statement with no variable reveals the type of system. A false statement means no point can satisfy both equations. A true statement means the equations were dependent, so one free variable remains and there are infinitely many solutions.

Elimination result	Meaning	Example conclusion
0 = 5	False; inconsistent system	No solution
0 = 0	Always true; dependent system	Infinitely many solutions

Worked example: parallel or identical?

System A:  x + 2y = 6       System B:  x + 2y = 6
           2x + 4y = 15                3x + 6y = 18

Double the first equation:              Triple the first equation:
2x + 4y = 12                            3x + 6y = 18

System A gives 12 = 15: no solution.   System B gives 18 = 18:
                                        infinitely many solutions.

In System B, all solutions can be described by x + 2y = 6, or y = 3 − x/2.

Graphing a System

Graphing makes the geometry visible and gives an estimate of a solution. Put each line into y = mx + b when possible, then locate their intersection. Algebra is generally more accurate when the intersection has fractional coordinates.

Worked example: read and verify an intersection

For y = −x + 5 and y = 2x − 1, a graph shows the lines meeting near (2, 3). Verify it algebraically.

At an intersection, the two y-expressions are equal:
−x + 5 = 2x − 1
6 = 3x
x = 2

y = 2(2) − 1 = 3
The exact intersection is (2, 3).

Three Variables and 3 × 3 Systems

With three unknowns, a solution is an ordered triple (x, y, z). Eliminate the same variable from two different pairs of equations, producing two equations in y and z. Solve that smaller system, then substitute back.

Worked example: systematic elimination

Solve the system x + y + z = 6, 2x − y + z = 3, and x + 2y − z = 2.

(1) x + y + z = 6
(2) 2x − y + z = 3
(3) x + 2y − z = 2

Eliminate x using (2) − 2(1):  −3y − z = −9
Eliminate x using (3) − (1):     y − 2z = −4

From y − 2z = −4, y = 2z − 4.
Put this into −3y − z = −9:
−3(2z − 4) − z = −9
−7z = −21, so z = 3.
y = 2(3) − 4 = 2.
x = 6 − 2 − 3 = 1.

Solution: (1, 2, 3)

Systems of Linear Inequalities

An inequality describes a half-plane rather than a single line. Its boundary is dashed for < or >, because points on it are excluded; it is solid for ≤ or ≥, because they are included. The solution to a system is the overlap of all required half-planes, called the feasible region.

Worked example: describe a feasible region

A club sells x adult tickets and y student tickets. There are at most 120 seats, at least 30 student tickets must be sold, and ticket counts cannot be negative. Translate these constraints.

x + y ≤ 120    (capacity)
y ≥ 30         (minimum student tickets)
x ≥ 0, y ≥ 0   (counts cannot be negative)

Graph the solid boundary lines x + y = 120 and y = 30 in the first quadrant. Shade below the capacity line and above y = 30. The shared region, including its boundaries, is feasible. Its corner points are (0, 30), (0, 120), and (90, 30).

Modelling and Optimisation

A linear model begins by defining variables with units, translating each sentence into a constraint, and checking that the answer makes physical sense. For a continuous linear-programming problem with a bounded feasible region, a linear objective such as profit reaches a maximum or minimum at a corner point. Evaluate the objective at each corner.

Worked example: choose a production plan

A workshop makes desks (d) and stools (s). A desk uses 3 labour-hours and a stool uses 1; at most 60 labour-hours are available. At most 24 desks can be made. Profit is $80 per desk and $30 per stool. For this continuous linear-programming model, assume d and s are nonnegative real quantities.

Constraints: 3d + s ≤ 60, d ≤ 24, d ≥ 0, s ≥ 0
Profit: P = 80d + 30s

Corners of the feasible region:
(0, 0), (20, 0), and (0, 60)
The limit d ≤ 24 does not add a corner because labour already gives d ≤ 20.

P(0, 0)  = $0
P(20, 0) = 80(20) = $1600
P(0, 60) = 30(60) = $1800

The model recommends 0 desks and 60 stools for maximum profit of $1800.

A model can yield an impractical extreme answer. If the workshop must make at least one of each product, add d ≥ 1 and s ≥ 1 before optimizing. If desks and stools must instead be whole numbers, this is an integer-optimization problem: the corner theorem identifies the continuous optimum but does not by itself prove the best integer point, so feasible integer points near the relevant boundary must also be checked.

Common Mistakes

Changing only one term when multiplying an equation: multiply every term on both sides.
Stopping after finding one variable: substitute back to obtain all coordinates.
Calling 0 = 0 one solution: it signals a dependent equation and infinitely many solutions, unless another constraint changes the system.
Reversing an inequality incorrectly: reverse its direction only when multiplying or dividing that inequality by a negative number.
Shading a boundary by guesswork: test a point not on the line, often (0, 0), in the original inequality.
Ignoring units or whole-number requirements: a negative number of tickets or a fractional machine is not feasible.

Practice Set

Solve by substitution: y = 3x − 2 and x + y = 14.
Solve by elimination: 4x + 3y = 18 and 2x − 3y = 6.
Classify the system 2x − 4y = 8 and x − 2y = 1.
Classify the system 3x + y = 7 and 6x + 2y = 14.
Find the intersection of y = 4 − x and y = x − 2.
Solve: x + y + z = 9, x − y + z = 3, and x + y − z = 5.
State whether the boundary of 2x − y > 4 is solid or dashed, and rewrite it in terms of y.
List the constraints for nonnegative numbers x and y with x + y at most 40 and x at least twice y.
For x ≥ 0, y ≥ 0, x + y ≤ 10, and x ≤ 6, list all corner points.
A taxi charges a $4 base fee plus $2.50 per kilometre. A second service charges $1 plus $3 per kilometre. At what distance do they cost the same, and what is that cost?
Maximize R = 5x + 4y subject to x + y ≤ 8, x ≤ 5, x ≥ 0, and y ≥ 0.

Answer Checks

Substitute: x + 3x − 2 = 14, so x = 4 and y = 10. Solution: (4, 10).
Add the equations: 6x = 24, so x = 4. Then 2(4) − 3y = 6 gives y = 2/3. Solution: (4, 2/3).
Doubling the second equation gives 2x − 4y = 2, which contradicts 2x − 4y = 8. No solution.
The second equation is twice the first, so there are infinitely many solutions on 3x + y = 7.
Set 4 − x = x − 2. Then 6 = 2x, so x = 3 and y = 1. Intersection: (3, 1).
Adding the first two equations gives 2x + 2z = 12, so x + z = 6. Comparing with the first equation gives y = 3; then the third gives x − z = 2. Thus x = 4, z = 2: (4, 3, 2).
The boundary is dashed. From 2x − y > 4, subtract 2x and multiply by −1: y < 2x − 4.
x ≥ 0, y ≥ 0, x + y ≤ 40, and x ≥ 2y.
The corners are (0, 0), (6, 0), (6, 4), and (0, 10).
Set costs equal: 4 + 2.50k = 1 + 3k. Thus 3 = 0.50k, so k = 6 km. The cost is $19.
Corners are (0, 0), (5, 0), (5, 3), and (0, 8). Their revenues are 0, 25, 37, and 32. The maximum is 37 at (5, 3).