Equations with Two Variables

Intuition

Every equation you have solved so far had one unknown and, usually, one answer: $2x + 3 = 11$ forces $x = 4$ and nothing else. But most real situations tie two unknowns together at once. “Two adult tickets and three child tickets cost 26 dollars” is a single fact linking two prices, and it does not pin down either price by itself. Cheap adult tickets with pricey child tickets can hit the same total as the reverse. An equation with two variables captures exactly this kind of statement. Instead of one solution it has a whole family of them, one for each way the total can be split. The striking part is that plotting those solutions on the coordinate plane lines them up into a perfectly straight line. That line is the picture the rest of this chapter is built on, because solving a system means finding where two such lines meet.

Learning goals

Write a linear equation in two variables as $ax + by = c$
Test an ordered pair by substituting in the right order
Generate solutions by solving for one variable and choosing values
Recognize the graph as the whole solution set, a line
Explain why a second equation is needed to fix one pair

What a two-variable equation is

A linear equation in two variables is an equation that can be arranged into the standard form

ax + by = c,

where $a$ , $b$ , and $c$ are numbers and $a$ and $b$ are not both zero. The two variables are usually called $x$ and $y$ . Each one appears only to the first power: never squared, never multiplied by the other, never under a root, and never in a denominator. That restriction is what keeps the equation linear, exactly as it did for the one-variable equations of the last chapter.

These are linear equations in two variables:

2x + 3y = 12, \qquad x - y = 4, \qquad y = 3x - 1.

The last one does not look like the standard form at first, but subtracting $3x$ from both sides turns it into $-3x + y = -1$ , which fits the pattern. These are not linear:

xy = 6 \quad (\text{the variables are multiplied}), \qquad x^2 + y = 5 \quad (x \text{ is squared}).

A one-variable linear equation such as $2x = 6$ locks its variable to a single value. With two variables in a single equation, one number can no longer settle both. So instead of asking for the value of $x$ we look for pairs of values, one for $x$ and one for $y$ , that fit together.

When is a pair a solution?

A solution of a two-variable equation is an ordered pair $(x, y)$ . It counts as a solution when putting the first number in for $x$ and the second for $y$ makes the equation true. You already have the tool to test one: substitution, the same check you used for one-variable equations.

Take $2x + 3y = 12$ and test the pair $(3, 2)$ . Substitute $x = 3$ and $y = 2$ into the left side:

2(3) + 3(2) = 6 + 6 = 12.

The left side equals $12$ , matching the right, so $(3, 2)$ is a solution. Now test $(1, 1)$ :

2(1) + 3(1) = 2 + 3 = 5,

which is not $12$ , so $(1, 1)$ is not a solution.

The word ordered matters as much here as it did when you plotted points. The pair $(2, 3)$ is a different claim from $(3, 2)$ : it puts $x = 2$ and $y = 3$ , giving $2(2) + 3(3) = 4 + 9 = 13$ , not $12$ . So $(3, 2)$ solves the equation while $(2, 3)$ does not, even though they use the same two numbers. Always feed the first coordinate to $x$ and the second to $y$ .

Worked example 1 Test whether pairs solve $3x + 2y = 12$

Check each pair by substituting the first number for $x$ and the second for $y$ , then compare the left side with $12$ .

For $(2, 3)$ :

3(2) + 2(3) = 6 + 6 = 12. \quad \text{True, so } (2, 3) \text{ is a solution.}

For $(4, 1)$ :

3(4) + 2(1) = 12 + 2 = 14. \quad \text{Not } 12, \text{ so } (4, 1) \text{ is not a solution.}

For $(0, 6)$ :

3(0) + 2(6) = 0 + 12 = 12. \quad \text{True, so } (0, 6) \text{ is a solution.}

Two of the three pairs check out. Notice that $(0, 6)$ works even though $x = 0$ : the term $3x$ becomes $0$ , but it is still there and still counted. A zero coordinate never lets you drop a term.

Check your understanding

Which ordered pair is a solution of $2x + y = 10$ ?

(4, 3)

(3, 4)

(2, 5)

(1, 9)

Why there are infinitely many solutions

A single two-variable equation does not have one solution or a handful. It has infinitely many, and there is a simple procedure that produces them on demand. Solve the equation for one variable, then feed in any value you like for the other.

Take $2x + y = 6$ . Solving for $y$ takes one move, subtracting $2x$ from both sides:

y = 6 - 2x.

Now $y$ is written directly in terms of $x$ . Choose any number for $x$ , and this formula hands back the matching $y$ , so the pair $(x, y)$ solves the equation. Choose $x = 1$ and you get $y = 6 - 2 = 4$ , the pair $(1, 4)$ . Because you can pick $x$ to be any number at all, and every pick produces a solution, there is no end to them.

Here is the argument in full, for any linear equation in two variables.

Why $ax + by = c$ has infinitely many solutions#

Start with $ax + by = c$ , and suppose $b \neq 0$ (if instead $a \neq 0$ , the same reasoning works with the roles of $x$ and $y$ swapped). Solve for $y$ : subtract $ax$ from both sides and then divide by $b$ , which is allowed precisely because $b$ is not zero,

y = \frac{c - ax}{b}.

Now pick any number whatsoever for $x$ . The right-hand side is then a definite arithmetic expression, so it evaluates to exactly one number, and that number is the matching $y$ . By construction the pair $\left(x, \frac{c - ax}{b}\right)$ satisfies the original equation, because we obtained $y$ from it.

There are infinitely many numbers to choose from for $x$ : the whole numbers $0, 1, 2, \ldots$ alone are already endless, and each different choice of $x$ produces a different pair. So the equation has infinitely many solutions. One equation, then, is nowhere near enough to pin down two unknowns. A single equation narrows the possibilities from every pair in the plane down to a single line’s worth of them, but no further. That is exactly why the next lessons bring in a second equation to isolate one pair.

A neat way to keep track of these solutions is a table. Solve for one variable, list a few values of the other, and compute. Using $y = 6 - 2x$ :

$x$	$y = 6 - 2x$	Solution $(x, y)$
$-1$	$8$	$(-1, 8)$
$0$	$6$	$(0, 6)$
$1$	$4$	$(1, 4)$
$2$	$2$	$(2, 2)$
$3$	$0$	$(3, 0)$

Every row is a genuine solution, and you could extend the table forever in either direction. Notice a pattern already: each time $x$ goes up by $1$ , $y$ drops by $2$ . Hold on to that, because it is what makes the graph a straight line.

Worked example 2 Fill in the missing coordinate for $3x + 2y = 18$

When one coordinate of a solution is known, substitute it and solve the one-variable equation that remains for the other coordinate.

Find the solution with $x = 4$ . Put $x = 4$ into $3x + 2y = 18$ :

3(4) + 2y = 18, \qquad 12 + 2y = 18, \qquad 2y = 6, \qquad y = 3.

So $(4, 3)$ is a solution. Find the solution with $y = 0$ . Put $y = 0$ :

3x + 2(0) = 18, \qquad 3x = 18, \qquad x = 6.

So $(6, 0)$ is a solution. Each known coordinate turns the two-variable equation into an ordinary one-variable equation, which you already know how to solve.

Check your understanding

Rewriting $4x + y = 9$ to give $y$ in terms of $x$ , what do you get?

y = 9 - 4x

y = 4x - 9

y = 9 + 4x

y = 4x + 9

The solutions form a straight line

Solutions are pairs $(x, y)$ , and pairs are points on the coordinate plane, so a natural question is what the whole collection of solutions looks like once plotted. Take the solutions of $2x + y = 6$ from the table and plot the four that fit on the grid.

The four plotted solutions (0, 6), (1, 4), (2, 2), (3, 0) of 2x + y = 6 all fall on one straight line. Every point on that line is a solution, and every solution is a point on the line.

The points do not scatter. They fall on one straight line, and this happens for every linear equation, which is where the name linear comes from. The reason is the pattern you spotted in the table: each step of $1$ to the right in $x$ drops $y$ by the same $2$ . So from any plotted point the next one sits one across and two down, every time. Equal steps across paired with equal steps down trace a straight path and never bend. (That steady rate of change is what the graphing chapter will call the line’s slope; here we need only the fact that the solutions line up.)

The line is the complete picture of the solution set, and this goes both ways. Every solution of the equation is one of the points on the line. Every point on the line, including the ones between the whole-number dots, is a solution. So “the graph of $2x + y = 6$ ” and “the set of all pairs that solve $2x + y = 6$ ” name the very same thing. To draw the line you only need two solutions, since two points determine a line. But plotting a third is a good check: if it does not land on the same line, one of your pairs is wrong.

Worked example 3 Solutions of $2x + 3y = 12$ for chosen values of $x$

Solve for $y$ first, so each value of $x$ is quick to evaluate. Subtract $2x$ and divide by $3$ :

y = \frac{12 - 2x}{3}.

At $x = 0$ :

y = \frac{12 - 0}{3} = 4, \qquad \text{the pair } (0, 4).

At $x = 3$ :

y = \frac{12 - 6}{3} = 2, \qquad \text{the pair } (3, 2).

At $x = 1$ :

y = \frac{12 - 2}{3} = \frac{10}{3}, \qquad \text{the pair } \left(1, \tfrac{10}{3}\right).

The first two pairs land on whole-number grid points, but the third does not, and it is every bit as valid a solution. Solutions of a linear equation are not required to be whole numbers. Between any two of the tidy grid points, the line is packed with fractional solutions like this one.

Check your understanding

Which of these points lies on the graph of $x + y = 5$ ?

(1, 3)

(4, 2)

(2, 3)

(3, 3)

The figure below is that checkpoint, left open. The line drawn on it is $x + y = 5$ , which is the same rule written as $y = -x + 5$ . The hollow marker is a pair you can move anywhere on the grid. It starts at $(1, 3)$ , the first option, and the sentence under the figure does the substitution out loud. Walk the marker through $(4, 2)$ , $(2, 3)$ and $(3, 3)$ and watch which one lands.

What is worth noticing is not that one of the four works. It is that the marker landing on the line and the substitution checking out are the same event, every single time. You can also see them happen together rather than being told they agree. Send the marker somewhere the checkpoint never mentioned, $(5, 0)$ or $(-1, 6)$ , and the rule holds there too. The line is not a summary of the four tidy pairs, it is the complete list of them.

Then change the line itself. Move the rise and the run and the marker stays exactly where you put it, now testing the same pair against a different equation. That is the other half of the same idea: a pair is not a solution on its own, it is a solution of something.

Is the pair a solution of $x + y = 5$ ?

Rise Run Crosses at Point x Point y

y = -x + 5. Rise -1 over run 1 is a slope of -1, so from any point on the line, 1 to the right and 1 down lands back on it. And it crosses the vertical axis at 5. The point (1, 3) is not a solution: putting x = 1 into the equation gives y = 4, and the point sits at 3 instead.

A coordinate plane carrying a straight line and a separate hollow marker. The marker can be moved to any grid point, and the sentence below the figure substitutes its first coordinate into the equation and compares the result with its second.

Looking ahead: two equations, one point

A single two-variable equation leaves a whole line of possibilities open. To single out one specific pair, you need a second equation, a second condition the pair must also satisfy at the same time. Two such equations considered together are called a system of equations, and that is the subject of the rest of this chapter.

The graph makes the goal vivid. Each equation is a line, so a pair that solves both must lie on both lines at once, which means it sits exactly where the two lines cross. Consider the pairs solving $x + y = 6$ and, separately, the pairs solving $x - y = 2$ . The single pair $(4, 2)$ lies on both lines, because $4 + 2 = 6$ and $4 - 2 = 2$ are both true, so it is the one pair that solves the system.

Two equations, two lines. The pairs solving x + y = 6 form one line and the pairs solving x - y = 2 form another. The single crossing point (4, 2) is the one pair that solves both, the solution of the system.

Reading a crossing point off a graph is quick to picture but hard to do exactly, especially when the answer is a fraction. So the coming lessons develop algebra that finds the crossing point without drawing anything: substitution and elimination. Both rest on the idea you built here, that a two-variable equation stands for a line’s worth of solution pairs. Both also rest on the second part of that idea, that where a system’s two lines cross, the shared pair is its answer.

Common mistakes

Writing a pair in the wrong order. A solution is an ordered pair $(x, y)$ . For $2x + 3y = 12$ the pair $(3, 2)$ works but $(2, 3)$ does not, because $2(2) + 3(3) = 13$ . Always put the first number in for $x$ and the second for $y$ .
Expecting a single answer. A single two-variable equation has infinitely many solutions, not one. It stands for an entire line of pairs, so “solve it” means describe that line, not find one point.
Thinking solutions must be whole numbers. Any value of $x$ gives a solution, so between the tidy grid points sit fractional pairs like $\left(1, \tfrac{10}{3}\right)$ . They count too.
Dropping a term when a coordinate is zero. In $2x + y = 6$ the pair $(0, 6)$ is a solution because $2(0) + 6 = 6$ . The $2x$ term does not disappear just because $x = 0$ ; it becomes $0$ and is still added in.
Sign slips when solving for a variable. Solving $2x + y = 6$ for $y$ gives $y = 6 - 2x$ , not $y = 6 + 2x$ . The $2x$ crosses the equals sign with the opposite sign.
Drawing a line through a single point. One point does not fix a line. Plot at least two solutions, and use a third as a check that they truly line up.

Key takeaways

A linear equation in two variables can be written $ax + by = c$ , with $x$ and $y$ each only to the first power. A solution is an ordered pair $(x, y)$ that makes it true.
Test a pair by substitution, keeping the order: the first number replaces $x$ , the second replaces $y$ .
One such equation has infinitely many solutions. Solve for one variable, for example $y = 6 - 2x$ , then choose any value of the other to generate solution pairs, and collect them in a table.
Plotting all the solutions traces a straight line. The graph of the equation is its solution set: every point on the line is a solution and every solution is on the line. Two points are enough to draw it.
One equation fixes a whole line of pairs. A second equation is needed to single out one pair, which sits where the two lines cross. That pair is the solution of a system of equations, the focus of the coming lessons.

Equations with Two Variables

Learning goals

What a two-variable equation is

When is a pair a solution?

Why there are infinitely many solutions

The solutions form a straight line

Looking ahead: two equations, one point

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Free Response Questions (FRQ)