Systems of two linear equations
Learning Objectives
- Recognise a linear equation in two unknowns and its solutions as ordered pairs.
- State what it means for a pair to solve a system of two equations.
- Solve a system by substitution.
- Solve a system by elimination (reduction).
- Interpret the solution as the intersection point of two lines.
Two unknowns, two conditions
Every equation so far had one unknown. But many questions involve two quantities tied together by two pieces of information. A classic: two numbers have sum \(7\) and difference \(3\) — what are they? Writing \(x\) and \(y\) for the numbers, the two conditions are
\[x + y = 7 \qquad\text{and}\qquad x - y = 3.\]
Neither equation alone pins the numbers down; the pair \((6, 1)\) satisfies the first but not the second, and \((5, 2)\) satisfies both. Asking for all pairs that satisfy both conditions at once is what a system of equations is.
Linear equation in two unknowns
A linear equation in two unknowns \(x\) and \(y\) is an equation that can be written
\[ax + by = c,\]
where \(a, b, c \in \mathbb{R}\) and \(a, b\) are not both zero. A solution is an ordered pair \((x_0, y_0)\) of real numbers that makes the equality true.
One such equation has infinitely many solutions. In \(x + y = 7\), choose any value for \(x\) and \(y = 7 - x\) completes a solution: \((0, 7),\ (1, 6),\ (2.5, 4.5),\ \ldots\) Indeed, when \(b \neq 0\) we can solve for \(y\),
\[y = -\frac{a}{b}\,x + \frac{c}{b},\]
and recognise the graph of a linear (or constant) function: the solutions of the equation are exactly the points of a line in the Cartesian plane. (When \(b = 0\) the equation reads \(ax = c\), a vertical line — also a line, though not a function's graph.)
System of two linear equations
A system of two linear equations in two unknowns is a pair of such equations considered together:
\[\begin{cases} a_1 x + b_1 y = c_1 \\ a_2 x + b_2 y = c_2 \end{cases}\]
A solution of the system is a pair \((x_0, y_0)\) that satisfies both equations simultaneously. To solve the system is to find all such pairs.
Two systems are equivalent if they have the same solution set — the same notion we used for single equations, and the same strategy applies: transform the system step by step into a simpler equivalent one.
Method 1: substitution
Solve one equation for one unknown, then substitute into the other.
\[\begin{cases} y = 2x - 1 \\ 3x + y = 9 \end{cases}\]
The first equation already gives \(y\) in terms of \(x\). Substitute it into the second:
\[3x + (2x - 1) = 9 \;\Rightarrow\; 5x - 1 = 9 \;\Rightarrow\; x = 2.\]
Then back-substitute: \(y = 2(2) - 1 = 3\). The solution is the pair \((2, 3)\) — and a check in both original equations confirms it: \(3 = 2(2) - 1\) ✓ and \(3(2) + 3 = 9\) ✓.
Substitution turns a two-unknown problem into the one-unknown equation we already know how to solve. That is the whole idea.
Method 2: elimination (reduction)
Add or subtract a multiple of one equation to the other so that one unknown cancels.
Adding to cancel. In
\[\begin{cases} x + y = 7 \\ x - y = 3 \end{cases}\]
the \(y\)-coefficients are opposite, so adding the two equations eliminates \(y\): \(2x = 10\), so \(x = 5\); then from the first equation \(y = 2\). Solution: \((5, 2)\) — our two numbers from the opening.
Subtracting when coefficients match. In
\[\begin{cases} 2x + 3y = 12 \\ 2x - y = 4 \end{cases}\]
the \(x\)-coefficients are equal, so subtract the second equation from the first: \(4y = 8\), hence \(y = 2\), and then \(2x - 2 = 4\) gives \(x = 3\). Solution: \((3, 2)\).
Scaling first. In
\[\begin{cases} 3x + 2y = 7 \\ x + y = 3 \end{cases}\]
no coefficients match yet. Multiply the second equation by \(2\) — a non-zero number, so the system stays equivalent — to get \(2x + 2y = 6\). Now the \(y\)-coefficients match, and subtracting this from the first equation eliminates \(y\): \(x = 1\). Then \(1 + y = 3\) gives \(y = 2\). Solution: \((1, 2)\); check: \(3(1) + 2(2) = 7\) ✓ and \(1 + 2 = 3\) ✓.
The geometric reading
Each equation's solutions form a line; the system asks for the pairs on both lines — that is, the points where the lines meet. In the substitution example the lines \(y = 2x - 1\) and \(y = 9 - 3x\) have different slopes (\(2\) and \(-3\)), so they cross in exactly one point, the solution \((2, 3)\).
Two distinct slopes always force exactly one crossing — which is why each of our examples had exactly one solution. But two lines in the plane can also be parallel, or even the same line. What happens to the system then? That is the subject of the next lesson.
Connection to Computer Science
Solving two equations in two unknowns by hand scales up: weather forecasts, engineering simulations, and the graphics in a video game all come down to solving systems like ours, but with thousands or millions of unknowns. Computers solve them with the same elimination idea you just used — systematically cancelling one unknown at a time — repeated at a scale no human could manage.
Could a step lose — or invent — a solution?
Substitution and elimination feel like tricks. Why are we sure the pair they produce is the answer — that no solution was lost on the way, and none invented?
Because every step replaces the system with an equivalent one. Substituting \(y = 2x - 1\) into the other equation only rewrites the condition "both equations hold" in different words. Adding one equation to the other is reversible — subtract it back and you recover the original system — and multiplying an equation by a non-zero number is reversible too. Reversible steps cannot change the solution set.
The word non-zero matters once more: multiplying an equation by \(0\) erases it (it becomes \(0 = 0\)), and that step is not reversible — information is destroyed, and invented "solutions" can creep in. The safe moves are exactly the reversible ones.
The final safeguard is the humble check: substituting the answer into both original equations. It costs one line and catches any slip in the algebra.
Exercises
Exercise 1
Solve the system \(\begin{cases} x + y = 10 \\ x - y = 4 \end{cases}\)
Exercise 2
Consider the system \(\begin{cases} y = 3x - 2 \\ 2x + y = 8 \end{cases}\)
Exercise 3
Decide whether the pair \((2, 1)\) is a solution of each system (it must satisfy both equations).
Exercise 4
Decide whether each statement about systems is true or false.
Exercise 5
Solve the system \(\begin{cases} 2x + 3y = 12 \\ 2x - y = 4 \end{cases}\)
Summary
A linear equation in two unknowns, \(ax + by = c\), has ordered pairs as solutions — infinitely many, forming a line in the plane.
A system asks for the pairs satisfying both equations at once. Substitution expresses one unknown from one equation and replaces it in the other; elimination adds a multiple of one equation to the other to cancel an unknown. Both reduce the problem to a one-unknown equation.
Each step is reversible, so the transformed system is equivalent — no solutions are lost or invented. Checking the answer in both original equations is the final safeguard.
Geometrically, solving the system means intersecting two lines; different slopes give exactly one crossing. Parallel or coincident lines are the story of the next lesson.