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.