Solutions to Linear Equations

Q: How many solutions does a two-variable equation have?

Infinitely many. Every $x$-value produces a matching $y$-value, and each pair is one solution. That is why the graph is an unbroken line rather than a single dot.

Q: Can the coordinates be negative or fractions?

Yes. For $y = 2x + 1$, the pairs $(-3, -5)$ and $\left(\dfrac{1}{2}, 2\right)$ are both solutions. Run the same substitution test regardless of what the numbers look like.

An equation with two variables, like $y = 2x + 1$ , is not asking for one answer. Its solutions are ordered pairs — an $x$ and a $y$ together — and a pair is a solution exactly when plugging both numbers in makes the equation true.

That is a real shift from one-variable equations. $3x + 5 = 20$ has a single solution, but $y = 2x + 1$ has infinitely many pairs that work: $(0, 1)$ , $(1, 3)$ , $(2, 5)$ , and on forever. The skill here is testing a given pair quickly — and finding the partner coordinate when only half a pair is given.

How to test an ordered pair

Substitute the $x$ -value and the $y$ -value into the equation at the same time, then simplify. If both sides come out equal, the pair is a solution. If they disagree — even by a little — it is not.

To test $(2, 5)$ in $y = 2x + 1$ : the right side becomes $2(2) + 1 = 5$ , which matches the $y$ -value of $5$ . True equation, so $(2, 5)$ is a solution. Testing $(3, 4)$ : the right side is $2(3) + 1 = 7$ , but the pair claims $y = 4$ . Since $4 \neq 7$ , it is not a solution.

Solutions live on the line

Graph the equation and something clean happens: every solution is a point on the line, and every point on the line is a solution. The line is a picture of the entire solution set at once.

The line below is $y = x + 2$ . The point $(1, 3)$ sits on it, and sure enough $1 + 2 = 3$ . The point $(3, 4)$ sits just below it, and sure enough $3 + 2 = 5$ , not $4$ . On the line means solution; off the line means not.

Finding a missing coordinate

Sometimes you get half a pair, like $(4, \;?\;)$ for the equation $y = 3x - 5$ . Substitute the coordinate you have and compute the one you need: $y = 3(4) - 5 = 7$ , so the pair is $(4, 7)$ .

If the missing coordinate is $x$ , the substitution leaves you a one-step or two-step equation to solve. Given $(\;?\;, 9)$ for $y = 2x + 3$ : substitute $9$ for $y$ to get $9 = 2x + 3$ , subtract $3$ , divide by $2$ , and $x = 3$ .

Worked examples

Example 1: the pair works

Is $(2, 5)$ a solution of $y = 2x + 1$ ?

Substitute

x = 2

into the right side

2(2) + 1

Simplify

4 + 1 = 5

Compare with the pair's

y

-value

5 = 5

Answer: Yes — $(2, 5)$ is a solution

Example 2: the pair fails

Is $(3, 4)$ a solution of $y = 2x + 1$ ?

Substitute

x = 3

into the right side

2(3) + 1

Simplify

6 + 1 = 7

Compare with the pair's

y

-value

4 \neq 7

Answer: No — the equation says $y$ should be $7$ when $x = 3$

Example 3: find the missing coordinate

Find the missing coordinate so that $(4, \;?\;)$ is a solution of $y = 3x - 5$ .

Substitute the known coordinate

y = 3(4) - 5

Multiply first

y = 12 - 5

Subtract

y = 7

Answer: $(4, 7)$

Try one yourself

Common questions

How many solutions does a two-variable equation have?

Infinitely many. Every $x$ -value produces a matching $y$ -value, and each pair is one solution. That is why the graph is an unbroken line rather than a single dot.

What does a solution look like on the graph?

It is a point on the line — no exceptions in either direction. If a pair checks out algebraically, its point lands on the line; if a point is on the line, its coordinates make the equation true.

Can the coordinates be negative or fractions?

Yes. For $y = 2x + 1$ , the pairs $(-3, -5)$ and $\left(\dfrac{1}{2}, 2\right)$ are both solutions. Run the same substitution test regardless of what the numbers look like.

What if the y-value is given instead of the x-value?

Substitute it for $y$ and solve the equation that remains. Given $y = 11$ for $y = 2x + 3$ : from $11 = 2x + 3$ , subtract $3$ to get $8 = 2x$ , then divide by $2$ to get $x = 4$ .

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