Solutions of a Linear Equation

For a single-variable equation like $2x = 6$, "solving" means finding the value of $x$ , here, $3$. For a two-variable equation like $2x + 3y = 6$, the question changes shape. There are infinitely many pairs $(x, y)$ that satisfy it; the task is to describe them all. This lesson handles the bookkeeping.

Definitions

An ordered pair $(p, q)$ is a solution of the linear equation $ax + by + c = 0$ if substituting $x = p$ and $y = q$ makes the equation true: $a p + b q + c = 0.$

The solution set of the equation is the collection of all such pairs.

Concept and method

Infinitely many solutions. Pick any value $p$ for $x$. Substitute into $ax + by + c = 0$: $a p + by + c = 0 \implies y = -\frac{a p + c}{b} \quad (\text{provided } b \ne 0).$ For each choice of $p$ you get a $y$, so you get one solution. There are infinitely many possible $p$, hence infinitely many solutions.

When $b = 0$ (i.e. the equation is $ax + c = 0$, or $x = -\dfrac{c}{a}$), there is exactly one allowed $x$, and $y$ can be anything. Again infinitely many solutions, paired as $\left(-\dfrac{c}{a}, q\right)$ for any $q$.

Generating a table. To list a few solutions, pick three or four convenient values for $x$ (or $y$) and compute the other coordinate. Convenient values are usually $0, 1, 2, -1, -2$.

Example. Equation: $2x + 3y = 12$.

• $x = 0 \Rightarrow 3y = 12 \Rightarrow y = 4$. Solution: $(0, 4)$.
• $x = 3 \Rightarrow 6 + 3y = 12 \Rightarrow y = 2$. Solution: $(3, 2)$.
• $x = 6 \Rightarrow 12 + 3y = 12 \Rightarrow y = 0$. Solution: $(6, 0)$.
• $x = -3 \Rightarrow -6 + 3y = 12 \Rightarrow y = 6$. Solution: $(-3, 6)$.

A neat table:

This single pattern , plug in convenient $x$, solve for $y$ , will be the heart of every graphing task in the next lesson.

Checking a candidate. Given a specific ordered pair, substitute both coordinates and check the equation.

Example. Is $(3, 2)$ a solution of $2x + 3y = 12$? Substitute: $2(3) + 3(2) = 6 + 6 = 12$. Yes.

Example. Is $(1, 5)$ a solution? Substitute: $2(1) + 3(5) = 2 + 15 = 17 \ne 12$. No.

Building an equation from a solution. If you are told that $(2, 3)$ is a solution, you have a constraint relating $a, b, c$: $2a + 3b + c = 0$. With one solution you cannot pin down the equation , you would need at least two distinct solutions (which determine the line).

Two solutions determine the line. If $(p_1, q_1)$ and $(p_2, q_2)$ are two distinct solutions, then they lie on the line representing the equation; and (as we will see in the next lesson) two points determine a unique line. So two solutions are enough to identify the equation up to a non-zero scalar multiple.

Solutions of $x = 4$. All pairs $(4, q)$ for any real $q$. Examples: $(4, 0), (4, 5), (4, -3), (4, \sqrt{2})$. The solution set is the vertical line $x = 4$.

Solutions of $y = -2$. All pairs $(p, -2)$ for any real $p$. Examples: $(0, -2), (3, -2), (-1, -2)$. The solution set is the horizontal line $y = -2$.

Worked examples

Example 1. Find four solutions of the equation $3x + 4y = 12$.

• $x = 0 \Rightarrow y = 3$. $(0, 3)$.
• $y = 0 \Rightarrow x = 4$. $(4, 0)$.
• $x = 4 \Rightarrow 12 + 4y = 12 \Rightarrow y = 0$ , same as above. Try $x = -4$: $-12 + 4y = 12 \Rightarrow y = 6$. $(-4, 6)$.
• $x = 8 \Rightarrow 24 + 4y = 12 \Rightarrow y = -3$. $(8, -3)$.

Example 2. Is $(-3, 2)$ a solution of $5x + 2y + 11 = 0$?

Substitute: $5(-3) + 2(2) + 11 = -15 + 4 + 11 = 0$. Yes.

Example 3. Find three solutions of $x - 2y = 4$.

Solve for $y$: $y = \dfrac{x - 4}{2}$. Choices: $x = 0 \Rightarrow y = -2$, $(0, -2)$. $x = 4 \Rightarrow y = 0$, $(4, 0)$. $x = 2 \Rightarrow y = -1$, $(2, -1)$.

Example 4. Write three solutions of the equation $x = 5$.

$(5, 0), (5, 2), (5, -7)$. The first coordinate is always $5$; the second can be anything.

Example 5. If $(2, k)$ is a solution of $3x - 2y = 4$, find $k$.

Substitute: $6 - 2k = 4 \Rightarrow 2k = 2 \Rightarrow k = 1$. The solution is $(2, 1)$.

Try it yourself

1. Find four solutions of $2x + 3y = 6$.
2. Is $(1, 1)$ a solution of $x + y = 2$?
3. Find three solutions of $y = 2x - 1$.
4. If $(k, 3)$ is a solution of $4x - y = 5$, find $k$.
5. Find three solutions of $x + y = 0$ that include negative numbers.
6. Find a solution of $3x - y = 7$ with $x = 0$, and one with $y = 0$.
7. List three solutions of the equation $y = 3$.
8. Is $(0, 0)$ a solution of $5x - 3y = 0$?
9. Find $a$ so that $(a, 2)$ is a solution of $2x + 5y - 18 = 0$.
10. Show that $(0, 4), (3, 2)$ and $(6, 0)$ are all solutions of $2x + 3y = 12$.

Pitfalls / Insight

• Pick convenient values. Setting $x = 0$ gives $y$-intercept immediately; setting $y = 0$ gives $x$-intercept immediately.
• Always check by substitution. It is the only way to be certain a pair is a solution.
• Infinitely many means infinitely many , not "all". Not every ordered pair is a solution; only those that satisfy the specific equation.

Insight. The solution set of a linear equation is a line in the plane. The table method gives you a few sample points; the next lesson uses two of those points to draw the entire line. Algebra and geometry , equation and graph , are the same story told twice.