Solutions of linear equations

What counts as a solution. For the equation $ax + by + c = 0$, a solution is an ordered pair $(x, y)$ of real numbers which, when substituted in for $x$ and $y$ respectively, makes the left-hand side equal to the right-hand side (here, zero).

Why "ordered" matters. The pair $(x, y)$ is written in a fixed order: the first number always goes to $x$, the second always to $y$. Swapping them, $(y, x)$, generally gives a different pair altogether and can fail the equation even when $(x, y)$ succeeds.

Why there are infinitely many. Pick any real number for $x$ — there is nothing stopping you from choosing $0$, $5$, $-3$, or a fraction like $\tfrac{1}{2}$. Once $x$ is fixed, the equation becomes a one-variable equation in $y$ alone, which has exactly one value of $y$. So every choice of $x$ produces exactly one matching $y$, and hence one solution $(x, y)$. Since there are infinitely many choices for $x$, there are infinitely many solutions.

where: $a$, $b$, $c$ are the fixed real-number coefficients of the equation ($a$ and $b$ not both zero), $x$ is the value freely chosen, and $y$ is the value that follows by solving the equation once $x$ is fixed.

Checking a candidate pair. To test whether a given ordered pair $(p, q)$ solves $ax + by + c = 0$, substitute $x = p$ and $y = q$ into the left-hand side and simplify. If the result is $0$, the pair is a solution; if not, it isn't.

where: $p$ is the candidate value for $x$, $q$ is the candidate value for $y$, and the check succeeds exactly when the arithmetic balances to zero.

Problem. Find four different solutions of the equation $x + 2y = 6$, and then check whether the pair $(1, 1)$ is also a solution.

Step 1. Rewrite the equation with $y$ as the subject: $y = \dfrac{6-x}{2}$.

Step 2. Choose four convenient values of $x$ and compute the matching $y$ for each. For $x=0$: $y=\dfrac{6-0}{2}=3$, giving $(0,3)$. For $x=2$: $y=\dfrac{6-2}{2}=2$, giving $(2,2)$. For $x=4$: $y=\dfrac{6-4}{2}=1$, giving $(4,1)$. For $x=6$: $y=\dfrac{6-6}{2}=0$, giving $(6,0)$.

Step 3. Note that other choices work too — for instance $x=-2$ gives $y=\dfrac{6-(-2)}{2}=4$, so $(-2,4)$ is a fifth valid solution, confirming there is no limit to how many can be found.

Step 4. Now check $(1, 1)$ by substitution: $x + 2y = 1 + 2(1) = 1 + 2 = 3$. Since $3 \neq 6$, the pair does not satisfy the equation.

Final answer: $(0,3)$, $(2,2)$, $(4,1)$ and $(6,0)$ are four solutions of $x + 2y = 6$, while $(1,1)$ is not a solution.

Competitive stretch

Problem. Check whether $(3, 4)$ and $(1, 1)$ are solutions of $2x - y = 2$, and state what this shows about how many pairs actually satisfy a given equation.

Step 1. Test $(3, 4)$: substitute $x = 3$, $y = 4$ into $2x - y$, giving $2(3) - 4 = 6 - 4 = 2$. This equals the right-hand side, so $(3, 4)$ is a solution.

Step 2. Test $(1, 1)$: substitute $x = 1$, $y = 1$, giving $2(1) - 1 = 2 - 1 = 1$. This does not equal $2$, so $(1, 1)$ is not a solution.

Final answer: $(3,4)$ satisfies $2x-y=2$ but $(1,1)$ does not — a reminder that although infinitely many ordered pairs solve any given linear equation, not every ordered pair you can write down is one of them; the equation still acts as a filter.

• Writing the two numbers of a solution in the wrong order — $(x, y)$ and $(y, x)$ are not interchangeable, and checking with the values swapped gives a wrong verdict.
• Believing a two-variable linear equation has "the" solution, the way a one-variable equation does — there are always infinitely many, and no single correct pair.
• Restricting the search for solutions to positive whole numbers only — zero, negative numbers and fractions are equally valid choices for $x$ or $y$.
• Sign or division slips while isolating a variable, such as writing $y = 6 - x$ instead of $y = \dfrac{6-x}{2}$ when the coefficient of $y$ is not $1$.

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