A system of linear equations in two unknowns is a pair of equations in variables \(x\) and \(y\), where each equation is linear (the highest power of each variable is 1). A solution is an ordered pair \((x, y)\) that satisfies both equations at the same time.
Assumed knowledge
A convenient form is \[ ax + by = c, \qquad dx + ey = f, \] where \(a, b, c, d, e, f\) are constants. The same system can be written in other equivalent forms — for example \(y = 2x + 1\) is the same as \(-2x + y = 1\).
Substitution means isolate one unknown in one equation, then replace that expression in the other equation.
Example. Solve
\[ \begin{cases} y = 2x + 1 \\ x + y = 7 \end{cases} \]
From the first equation, \(y = 2x + 1\). Substitute into the second: \[ x + (2x + 1) = 7 \Rightarrow 3x + 1 = 7 \Rightarrow 3x = 6 \Rightarrow x = 2. \] Then \(y = 2(2) + 1 = 5\). The solution is \((2, 5)\).
Check: \(2(2)+1 = 5\) and \(2 + 5 = 7\).
Elimination adds or subtracts the equations so that one unknown cancels. You may need to multiply one or both equations first so that the coefficients of \(x\) or \(y\) match in size (one positive, one negative).
Example. Solve
\[ \begin{cases} x + y = 5 \\ 2x - y = 1 \end{cases} \]
The coefficients of \(y\) are \(1\) and \(-1\), so add the equations: \[ (x+y) + (2x-y) = 5 + 1 \Rightarrow 3x = 6 \Rightarrow x = 2. \] Substitute into \(x + y = 5\): \(2 + y = 5\), so \(y = 3\). Solution: \((2, 3)\).
Each equation \(ax + by = c\) (with \(b \neq 0\)) can be rearranged to \(y = -\frac{a}{b}x + \frac{c}{b}\), which is a straight line. The solution of the system is the point where the two lines meet — their intersection.
| Lines | Algebraic picture | Solutions |
|---|---|---|
| Intersect at one point | Different slopes | Exactly one \((x, y)\) |
| Parallel, distinct | Same slope, different intercepts | No solution |
| Coincident (same line) | One equation is a multiple of the other | Infinitely many |
Adjust the two lines or use presets. Read off the intersection and how the classification changes when the lines are parallel or coincident.
Pick a preset system and watch substitution or elimination steps. Systems with no unique solution show an explanation instead of incorrect steps.
For story problems, define unknowns clearly, write two independent equations, then solve by substitution, elimination, or a graph.
Example (tickets). Adult tickets cost \$12 and child tickets \$7. A group buys 10 tickets for \$95. How many of each type?
Let \(a\) be the number of adult tickets and \(c\) the number of child tickets. Then \(a + c = 10\) and \(12a + 7c = 95\). From the first, \(c = 10 - a\). Substitute: \[ 12a + 7(10 - a) = 95 \Rightarrow 12a + 70 - 7a = 95 \Rightarrow 5a = 25 \Rightarrow a = 5, \] so \(c = 5\). Five adult and five child tickets.
Problems that need two unknowns at once are ancient. The Chinese text Nine Chapters on the Mathematical Art (around 200 BCE, with later commentary) includes fangcheng methods — arranging coefficients in a table and eliminating unknowns, much like modern elimination.
In the 9th century, al-Khwarizmi organised linear and quadratic problems in systematic rules that spread to Europe. By the 17th century, Descartes linked algebra with geometry: an equation in \(x\) and \(y\) describes a curve, so solving a system is finding where two curves meet. Today, computers solve huge systems, but the same ideas — replace, combine, or intersect — remain the core of school algebra.
If \((x_0, y_0)\) satisfies both \(ax + by = c\) and \(dx + ey = f\), then any linear combination of the two equations is still true at \((x_0, y_0)\). For example, adding the left sides and adding the right sides gives \[ (a+d)x + (b+e)y = c + f, \] which \((x_0, y_0)\) also satisfies. Elimination chooses combinations so one unknown disappears, leaving a single equation in one unknown. Multiplying an equation by a non-zero constant before adding does not change the solution set either.
The graph of \(ax + by = c\) is all points \((x, y)\) that make the equation true. A solution of the system must lie on both graphs, so it must be a point common to both lines. If the lines cross once, that point is the unique solution; if they never meet, there is no solution; if they are the same line, every point on the line works.
Pause and reason before continuing. Discuss with a partner or write your work in a notebook.
Checkpoint
In the graph explorer, use the “Parallel” preset. What does the status panel say about solutions? Does the canvas show an intersection point?
Checkpoint
Solve \(y = 3x - 2\) and \(2x + y = 8\) by substitution without a calculator. Does your pair \((x, y)\) satisfy both originals?
Checkpoint
For \(x + 2y = 8\) and \(2x + 4y = 16\), are the two equations independent, or is one a multiple of the other? How many solutions do you expect?
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When eliminating, why do we sometimes subtract equations instead of add them? Give an example where subtracting cancels \(x\).
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A shop sells notebooks for \$4 and pens for \$2. You buy 6 items for \$18. What two equations describe this? Which method would you use first?
Checkpoint
In the step-by-step workbench, run elimination on the ticket-style preset. At which step does \(y\) (or your chosen variable) disappear?
Cinemas and transport systems often use different prices for adults and children. If you know the total number of tickets and the total cost, you have two equations in two unknowns — the counts of each type. The same structure appears in fundraising dinners (plate vs concession) and sports seating.
A chemist may mix a \(20\%\) solution with a \(50\%\) solution to obtain 2 L of \(35\%\) solution. Let \(x\) and \(y\) be the volumes (in litres) of the weaker and stronger solutions. Then \(x + y = 2\) (total volume) and \(0.2x + 0.5y = 0.35 \times 2\) (total amount of pure solute). Solving the system gives the recipe.
A small business might need at least 100 units to cover fixed costs and also stay within a production cap of 150 units while hitting a revenue target. Each constraint is linear; the feasible plan is where the constraints overlap — a preview of ideas used later in linear programming.
Work each question, then open the solution to check your reasoning. Difficulty increases through the sets.
Answer: \((3, 5)\).
Steps: Substitute \(y = x + 2\) into \(x + y = 8\): \(x + (x+2) = 8\), so \(2x = 6\), \(x = 3\), \(y = 5\).
Answer: \((4, 2)\).
Steps: Add: \(2x = 8\), \(x = 4\). Then \(y = 6 - 4 = 2\).
Answer: Yes.
Check: \(2(1)+2 = 4\) and \(1 - 2 = -1\).
Answer: \((3, 1)\).
Steps: Add equations: \(3x = 9\), \(x = 3\). Then \(3 - y = 2\), \(y = 1\).
Answer: \((2, 3)\).
Steps: From \(x + y = 5\), \(y = 5 - x\). Substitute: \(3x + 2(5-x) = 12\), \(3x + 10 - 2x = 12\), \(x = 2\), \(y = 3\).
Answer: No solution (parallel lines).
Explanation: Same slope \(2\), different intercepts — the lines never meet.
Answer: \((2, 3)\).
Steps: From \(4x - y = 5\), \(y = 4x - 5\). Substitute into \(2x + 3y = 13\): \(2x + 3(4x-5) = 13\), \(2x + 12x - 15 = 13\), \(14x = 28\), \(x = 2\), \(y = 3\).
Answer: \((3, 1)\).
Steps: Add equations: \(8x = 24\), \(x = 3\). Then \(3(3) + 2y = 11\), \(2y = 2\), \(y = 1\).
Answer: 5 coins at 5¢ and 7 coins at 10¢.
Steps: Let \(n\) be 5¢ coins and \(m\) be 10¢ coins. Then \(n + m = 12\) and \(5n + 10m = 95\). From the first, \(n = 12 - m\). Substitute: \(5(12-m) + 10m = 95\), \(60 - 5m + 10m = 95\), \(5m = 35\), \(m = 7\), \(n = 5\).
Answer: Every \(k\) (the second equation is \(2\times\) the first).
Explanation: The lines are coincident for all \(k\) where the second is a multiple of the first; here they are always multiples, so infinitely many solutions.
Answer: \((4, 3)\).
Steps: Multiply the first equation by 2: \(x + 2y = 10\). With \(x - y = 1\), substitute \(x = 1 + y\) into \(x + 2y = 10\): \(1 + y + 2y = 10\), \(3y = 9\), \(y = 3\), \(x = 4\).
Answer: width \(4.5\) cm, length \(8.5\) cm.
Steps: \(2\ell + 2w = 26\), \(\ell = w + 4\). Substitute: \(2(w+4)+2w=26\), \(4w+8=26\), \(w=4.5\), \(\ell=8.5\).