Linear Equations In One Variable: Rules & Examples

What Is a Linear Equation in One Variable?

A linear equation in one variable is an equation with a single unknown that appears only to the first power: no squares, no square roots of the unknown, and no unknown in a denominator. The unknown is a variable, usually written $x$, and $2x + 3 = 11$ is a typical example.

The name comes from graphs: an expression such as $2x + 3$, graphed as $y = 2x + 3$, is a straight line.

Standard Form: $ax + b = 0$

The standard form of a linear equation in one variable is $ax + b = 0$, where $a$ and $b$ are real numbers and $a \neq 0$. The number $a$ is the coefficient of $x$, the number multiplying it.

Solving the standard form takes two steps:

$$ax + b = 0$$

$$ax = -b \quad (\text{subtract } b \text{ from both sides})$$

$$x = -\frac{b}{a} \quad (\text{divide both sides by } a, \text{ allowed because } a \neq 0)$$

Any linear equation can be moved into this shape. For $3x + 4 = x + 12$, subtract $x$ and $12$ from both sides to get $2x - 8 = 0$, so $a = 2$, $b = -8$ and $x = -\frac{-8}{2} = 4$. The page on the standard form of linear equations covers the form in more detail.

How Many Solutions Can a Linear Equation in One Variable Have?

After you collect all the $x$ terms on one side and all the numbers on the other, every equation of this kind becomes $ax + b = 0$ for some numbers $a$ and $b$. What happens next depends on $a$:

• If $a \neq 0$: exactly one solution, $x = -\frac{b}{a}$.

• If $a = 0$ and $b \neq 0$: the $x$ terms cancel and you are left with a false statement such as $-1 = 0$. No number works, so there is no solution.

• If $a = 0$ and $b = 0$: the $x$ terms cancel and you are left with $0 = 0$. Every number works.

Strictly, the standard form needs $a \neq 0$, so the last two cases are equations that look linear until the $x$ terms cancel. Many textbooks still treat all three as types of linear equation in one variable: OpenStax College Algebra 2e, section 2.2 calls them conditional, inconsistent and identity equations.

The graph shows the same three cases. Move everything to one side and graph $y = ax + b$: the solutions are the points where this line meets the x-axis. A slanted line ($a \neq 0$) crosses the x-axis exactly once, at $x = -\frac{b}{a}$; a horizontal line ($a = 0$) either runs parallel to the x-axis and never meets it, or lies on it.

You can also graph each side of the original equation as its own line; the solution is where the two lines meet. Lines with different slopes meet once, parallel lines (same slope, different intercept) never meet, and two copies of the same line meet everywhere. To try this, type $y = 2x - 8$ into the Desmos graphing calculator and change the numbers.

How to Solve a Linear Equation in One Variable

An equation is a balance: the two sides are equal, and they stay equal as long as you do the same thing to both sides. That rule comes from the properties of equality.

1. Clear brackets by multiplying them out.

2. Clear fractions by multiplying every term by the lowest common denominator.

3. Collect the $x$ terms on one side and the numbers on the other.

4. Divide both sides by the coefficient of $x$.

5. Check the answer by substituting it into the original equation.

For one-step and two-step cases, see two-step equations and how to solve for x.

Examples of Linear Equations in One Variable

Example 1

Solve $5x - 7 = 2x + 8$.

$$5x - 7 = 2x + 8$$

$$3x - 7 = 8 \quad (\text{subtract } 2x \text{ from both sides})$$

$$3x = 15 \quad (\text{add } 7 \text{ to both sides})$$

$$x = 5 \quad (\text{divide both sides by } 3)$$

Check: $5(5) - 7 = 18$ and $2(5) + 8 = 18$.

Final answer: $x = 5$

Example 2

Solve $4 - 2(x - 3) = 12$.

A first attempt often multiplies out the bracket as $4 - 2x - 6 = 12$. That gives $-2 - 2x = 12$, so $-2x = 14$ and $x = -7$. Check it: $4 - 2(-7 - 3) = 4 + 20 = 24$, not $12$, so something went wrong.

The $-2$ multiplies both terms in the bracket, and $(-2) \times (-3) = +6$:

$$4 - 2(x - 3) = 12$$

$$4 - 2x + 6 = 12 \quad (\text{multiply out the bracket})$$

$$10 - 2x = 12 \quad (\text{combine } 4 + 6)$$

$$-2x = 2 \quad (\text{subtract } 10 \text{ from both sides})$$

$$x = -1 \quad (\text{divide both sides by } -2)$$

Check: $4 - 2(-1 - 3) = 4 + 8 = 12$.

Final answer: $x = -1$

Example 3

Solve $\frac{x}{3} - \frac{1}{4} = \frac{x}{6}$.

The lowest common denominator of 3, 4 and 6 is 12, so multiply every term by 12.

$$12 \times \frac{x}{3} - 12 \times \frac{1}{4} = 12 \times \frac{x}{6}$$

$$4x - 3 = 2x$$

$$2x - 3 = 0 \quad (\text{subtract } 2x \text{ from both sides})$$

$$2x = 3 \quad (\text{add } 3 \text{ to both sides})$$

$$x = \frac{3}{2} \quad (\text{divide both sides by } 2)$$

Check: $\frac{3}{2} \div 3 = \frac{1}{2}$, and $\frac{1}{2} - \frac{1}{4} = \frac{1}{4}$; also $\frac{3}{2} \div 6 = \frac{1}{4}$.

Final answer: $x = \frac{3}{2}$

Example 4

Solve $3(x + 2) = 3x + 5$.

$$3(x + 2) = 3x + 5$$

$$3x + 6 = 3x + 5 \quad (\text{multiply out the bracket})$$

$$6 = 5 \quad (\text{subtract } 3x \text{ from both sides})$$

The $x$ terms cancelled and left a false statement. Here $a = 0$ and $b = 1$, so no value of $x$ can make the two sides equal.

Final answer: no solution

Solving Word Problems With Linear Equations

A word problem becomes a linear equation once each phrase is translated into algebra. Say what the variable stands for before you write anything else.

Example 5

A taxi charges a fixed 50 rupees plus 12 rupees for each kilometre. A ride costs 146 rupees. How long was the ride?

Let $x$ be the distance in kilometres.

$$50 + 12x = 146$$

$$12x = 96 \quad (\text{subtract } 50 \text{ from both sides})$$

$$x = 8 \quad (\text{divide both sides by } 12)$$

Check: $50 + 12 \times 8 = 50 + 96 = 146$.

Final answer: 8 kilometres

Example 6

Five less than twice a number is 17. Find the number.

Let $x$ be the number. "Five less than twice a number" means start with $2x$ and take 5 away, so it is $2x - 5$, not $5 - 2x$.

$$2x - 5 = 17$$

$$2x = 22 \quad (\text{add } 5 \text{ to both sides})$$

$$x = 11 \quad (\text{divide both sides by } 2)$$

Check: $2 \times 11 - 5 = 17$.

Final answer: 11

More real situations, such as costs, speeds and mixtures, are worked through on the applications of linear equations page.

Common Mistakes With Linear Equations in One Variable

Mistake 1: Moving a term without changing its sign

Where it slips in: Moving a number across the equals sign, as in $4x + 5 = 21$.

Don't do this: Writing $4x = 21 + 5$, which gives $x = 6.5$.

The correct way: Subtract 5 from both sides: $4x = 16$, so $x = 4$. "Moving" a term is shorthand for doing the opposite operation to both sides, which is why its sign changes.

Mistake 2: Losing the sign when dividing by a negative coefficient

Where it slips in: The last step, when the coefficient is negative, as in $-2x = 2$.

Don't do this: Dividing by 2 instead of $-2$ and writing $x = 1$.

The correct way: Divide by the whole coefficient, sign included: $x = \frac{2}{-2} = -1$. Substituting back catches this at once, because $-2 \times 1 = -2$, not $2$.

Mistake 3: Combining unlike terms

Where it slips in: Simplifying one side before solving, as in $2x + 3 + 2x$.

Don't do this: Writing $7x$, as if the $3$ were $3x$.

The correct way: Add only the $x$ terms: $2x + 3 + 2x = 4x + 3$. The 3 stays a plain number.

Mistake 4: Writing $x = 0$ when the $x$ terms cancel

Where it slips in: Equations such as $3(x + 2) = 3x + 5$, where both sides have the same $x$ term.

Don't do this: Deciding that $x = 0$, or that an arithmetic error must have happened. When the $x$ terms vanish, the usual reaction is to go back and hunt for a mistake that is not there.

The correct way: Read the statement that is left. A false statement such as $6 = 5$ means no solution; a true statement such as $0 = 0$ means every number is a solution.

Practice Problems

1. Solve $7x - 4 = 3x + 8$.

2. Solve $5(x + 4) = 2x + 11$.

3. Solve $6x + 1 = 2x + 4$.

4. Solve $\frac{x}{2} + 1 = \frac{x}{3} + 3$.

5. Solve $2(3x - 1) = 6x + 4$.

6. Solve $4(x - 2) + 3 = 4x - 5$.

7. A number is tripled and then decreased by 7, giving 20. Find the number.

Answers

1. $4x = 12$, so $x = 3$. Check: $7(3) - 4 = 17$ and $3(3) + 8 = 17$.

2. $5x + 20 = 2x + 11$, so $3x = -9$ and $x = -3$. Check: $5(1) = 5$ and $2(-3) + 11 = 5$.

3. $4x = 3$, so $x = \frac{3}{4}$. Check: $6 \times \frac{3}{4} + 1 = \frac{11}{2}$ and $2 \times \frac{3}{4} + 4 = \frac{11}{2}$.

4. Multiply by 6: $3x + 6 = 2x + 18$, so $x = 12$. Check: $6 + 1 = 7$ and $4 + 3 = 7$.

5. $6x - 2 = 6x + 4$ leaves $-2 = 4$, which is false, so there is no solution.

6. $4x - 5 = 4x - 5$ is true for every $x$, so every number is a solution (an identity).

7. $3x - 7 = 20$, so $3x = 27$ and $x = 9$. Check: $3 \times 9 - 7 = 20$.

Key Takeaways

• Linear equations in one variable can be written as $ax + b = 0$; when $a \neq 0$ the only solution is $x = -\frac{b}{a}$.

• If the $x$ terms cancel, a false statement means no solution and a true statement means every number is a solution.

• Solving keeps the balance: do the same thing to both sides until $x$ is alone, then check in the original equation.

• The most common errors are sign errors: when moving a term, multiplying out a bracket after a minus sign, and dividing by a negative coefficient.

What to Learn Next

1. Graphing linear equations: see where $y = ax + b$ crosses the x-axis.

2. Linear inequalities: replace the equals sign with $<$ or $>$ and the answer becomes a range of numbers.

3. Practice the seven problems above without looking at the answers. If a sign goes wrong, return to Example 2.

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