| Primary source | OpenStax College Algebra 2e, Section 2.2: "Linear Equations in One Variable" |
| Direct link | https://openstax.org/books/college-algebra-2e/pages/2-2-linear-equations-in-one-variable |
| Supplementary | OpenStax Precalculus 2e, Section 2.1: "Linear Functions" (graphical perspective) |
| Supplementary link | https://openstax.org/books/precalculus-2e/pages/2-1-linear-functions |
Both sources are free and openly licensed. If you are using a printed copy, College Algebra 2e Section 2.2 begins on page 139 of the PDF.
An equation is a claim, not an instruction.
When you write $3x + 5 = 14$, you are claiming: "there exists a value of $x$ that makes both sides equal." Solving the equation means finding that value. When you check your answer by substituting it back in, you are verifying the claim. Students who keep this framing in mind catch their own errors. Students who treat solving as a ritual of symbol-manipulation tend to drift and lose track of what they are doing.
Everything in this lesson follows from one principle: whatever you do to one side of an equation, do the same thing to the other side. This keeps the claim true while changing its form, step by step, until the form is simple enough to read off the answer.
Before this lesson, confirm you can:
Definition. A linear equation in one variable is any equation that can be written in the form $ax + b = 0$ where $a$ and $b$ are constants and $a \neq 0$.
General solution algorithm.
Special cases.
The word linear means "degree one in the variable." The variable $x$ appears to the first power only. It is never squared, cubed, inside a radical, or in a denominator.
| Equation | Linear? | Reason |
|---|---|---|
| $3x + 7 = 19$ | Yes | $x$ appears only to the first power |
| $x^2 + 3 = 7$ | No | $x$ is squared |
| $\sqrt{x} = 4$ | No | $x$ is inside a radical |
| $\frac{1}{x} = 5$ | No | $x$ is in a denominator |
| $3(x - 2) + x = 1$ | Yes | After distribution, only $x^1$ terms appear |
The solution to a linear equation is a single value (or a special case: all reals or no solution).
Imagine a balance scale that is perfectly level. Each side holds a certain weight. The equation $3x + 5 = 14$ says: the left side weighs the same as the right side.
If you add 3 pounds to one side, you must add 3 pounds to the other side to keep the scale level. If you divide one side by 2, you must divide the other side by 2.
This is the entire foundation of algebra:
Properties of Equality. If $A = B$, then $A + C = B + C$ (addition property). If $A = B$, then $A \cdot C = B \cdot C$ for $C \neq 0$ (multiplication property).
Subtraction is adding a negative. Division is multiplying by a reciprocal. So these two properties cover every legal move.
Example 1. Solve $x - 7 = 12$.
The variable is already isolated except for the $-7$. Add $7$ to both sides.
\[ x - 7 + 7 = 12 + 7 \] \[ x = 19 \]
Check: $19 - 7 = 12$. Correct.
Example 2. Solve $4x + 3 = 23$.
Step 1. Subtract $3$ from both sides to move the constant away from $x$. \[ 4x + 3 - 3 = 23 - 3 \] \[ 4x = 20 \]
Step 2. Divide both sides by $4$. \[ x = 5 \]
Check: $4(5) + 3 = 20 + 3 = 23$. Correct.
Example 3. Solve $5x - 3 = 2x + 9$.
Step 1. Subtract $2x$ from both sides to bring all $x$-terms to the left. \[ 5x - 2x - 3 = 9 \] \[ 3x - 3 = 9 \]
Step 2. Add $3$ to both sides. \[ 3x = 12 \]
Step 3. Divide by $3$. \[ x = 4 \]
Check: $5(4) - 3 = 17$ and $2(4) + 9 = 17$. Correct.
Student note. You can move the $x$-terms to either side -- left or right. The algebra works either way. Convention prefers a positive leading coefficient, so subtract the smaller $x$-coefficient from the larger side.
Example 4. Solve $3(2x - 4) = 2(x + 1) + 6$.
Step 1. Distribute. \[ 6x - 12 = 2x + 2 + 6 \] \[ 6x - 12 = 2x + 8 \]
Step 2. Subtract $2x$ from both sides. \[ 4x - 12 = 8 \]
Step 3. Add $12$ to both sides. \[ 4x = 20 \]
Step 4. Divide by $4$. \[ x = 5 \]
Check: $3(2 \cdot 5 - 4) = 3(6) = 18$. Right side: $2(5+1) + 6 = 12 + 6 = 18$. Correct.
Missing Distribution. Students often write $3(2x - 4) = 6x - 4$ instead of $6x - 12$. The factor of $3$ must multiply every term inside the parentheses, including the constant. Read it as: $3 \times 2x$ AND $3 \times (-4)$.
Fractions do not have to make the solving harder. The trick is to clear them at the start by multiplying every term on both sides by the least common denominator (LCD) of all fractions in the equation.
Example 5. Solve $\dfrac{x}{3} + \dfrac{1}{4} = \dfrac{5}{6}$.
Step 1. Find the LCD. The denominators are $3$, $4$, and $6$. The LCD is $12$.
Step 2. Multiply every term by $12$. \[ 12 \cdot \frac{x}{3} + 12 \cdot \frac{1}{4} = 12 \cdot \frac{5}{6} \] \[ 4x + 3 = 10 \]
Step 3. Solve the resulting integer equation. \[ 4x = 7 \] \[ x = \frac{7}{4} \]
Check: $\frac{7/4}{3} + \frac{1}{4} = \frac{7}{12} + \frac{3}{12} = \frac{10}{12} = \frac{5}{6}$. Correct.
Partial Multiplication. Students multiply the terms with fractions by the LCD but forget to multiply the whole-number terms. Multiply every single term, on both sides.
Not every linear equation has exactly one solution. There are two other possibilities.
Example 6 (Identity). Solve $2(x + 3) = 2x + 6$.
Distribute: $2x + 6 = 2x + 6$.
Subtract $2x$: $6 = 6$.
The variable has vanished and the result is a true statement. This equation is true for every real number $x$. The solution set is all real numbers, written $(-\infty, \infty)$ or $\mathbb{R}$.
Why this happens. Both sides were the same expression written in different forms. No value of $x$ makes it false, and no value makes it special.
Example 7 (Contradiction). Solve $3(x + 1) = 3x + 7$.
Distribute: $3x + 3 = 3x + 7$.
Subtract $3x$: $3 = 7$.
The variable has vanished and the result is a false statement. There is no value of $x$ that makes $3 = 7$ true. The equation has no solution, written $\emptyset$ or "no solution."
Why this happens. The equation made an impossible promise: it claimed two expressions that always differ by $4$ were equal.
The skill of solving a linear equation is useful only when paired with the skill of writing one. Most real-world problems come as sentences, not as equations.
Translation vocabulary.
| English phrase | Algebraic meaning |
|---|---|
| "a number" | $x$ (or another variable) |
| "is", "equals", "is the same as" | $=$ |
| "more than", "added to" | $+$ |
| "less than", "decreased by" | $-$ |
| "times", "of", "product" | $\times$ (multiplication) |
| "divided by", "quotient of" | $\div$ |
| "twice", "double" | $2x$ |
| "three times as much" | $3x$ |
A four-step word-problem process.
Example 8. A phone plan charges a flat monthly fee plus a per-minute rate. The total charge for $40$ minutes is $\$22.00$. The total charge for $100$ minutes is $\$40.00$. Find the per-minute rate.
Let $r$ = per-minute rate in dollars, and $F$ = monthly flat fee.
The two conditions give: \[ 40r + F = 22 \] \[ 100r + F = 40 \]
Subtract the first equation from the second: \[ 60r = 18 \implies r = 0.30 \]
The per-minute rate is $\$0.30$ per minute.
Check: $40(0.30) + F = 22 \implies F = 10$. Then $100(0.30) + 10 = 30 + 10 = 40$. Correct.
This problem used two equations, but the core skill was recognizing that subtracting one equation from another creates a single equation in one unknown. That is the method of elimination, covered in the Systems of Linear Equations lesson.
| Error | Example | Correction |
|---|---|---|
| Forgetting to distribute | $3(x-4) = 3x - 4$ | Must be $3x - 12$ |
| Adding to one side only | $x - 7 = 12 \to x = 12$ | Add $7$ to both sides: $x = 19$ |
| Sign error when moving terms | $5 = x + 3 \to x = 8$ | Subtract $3$: $x = 2$ |
| Wrong LCD | Multiplying by $6$ when LCD is $12$ | Find LCM of all denominators |
| Stopping at $4x = 20$ without dividing | $x = 4x = 20$ | Divide both sides by $4$: $x = 5$ |
| Skipping the check step | No error detected | Always substitute back in |
Problem 1. Solve: $6x - 5 = 25$.
Add $5$ to both sides: $6x = 30$. Divide by $6$: $x = 5$.
Check: $6(5) - 5 = 30 - 5 = 25$. Correct.
Problem 2. Solve: $\dfrac{2x}{5} = 8$.
Multiply both sides by $5$: $2x = 40$. Divide by $2$: $x = 20$.
Check: $\frac{2 \cdot 20}{5} = \frac{40}{5} = 8$. Correct.
Problem 3. Solve: $-3(x + 2) = 15$.
Distribute: $-3x - 6 = 15$. Add $6$: $-3x = 21$. Divide by $-3$: $x = -7$.
Check: $-3(-7+2) = -3(-5) = 15$. Correct.
Problem 4. Solve: $7x - 4 = 3x + 16$.
Subtract $3x$: $4x - 4 = 16$. Add $4$: $4x = 20$. Divide by $4$: $x = 5$.
Check: $7(5) - 4 = 31$ and $3(5) + 16 = 31$. Correct.
Problem 5. Solve: $\dfrac{x+1}{3} - \dfrac{x-2}{6} = 1$.
LCD $= 6$. Multiply every term by $6$:
$2(x+1) - (x-2) = 6$
$2x + 2 - x + 2 = 6$
$x + 4 = 6$
$x = 2$
Check: $\frac{3}{3} - \frac{0}{6} = 1 - 0 = 1$. Correct.
Problem 6. Solve: $4(2x - 1) - 3(x + 2) = 5(x - 1)$.
Distribute each group:
$8x - 4 - 3x - 6 = 5x - 5$
$5x - 10 = 5x - 5$
Subtract $5x$: $-10 = -5$.
This is a false statement. No solution (contradiction).
Problem 7. A landscaping company charges a flat visit fee plus an hourly rate. A 2-hour job costs $\$85$. A 5-hour job costs $\$160$. Find the flat visit fee.
Let $F$ = flat fee, $r$ = hourly rate.
$2r + F = 85$
$5r + F = 160$
Subtract the first from the second: $3r = 75 \implies r = 25$.
Substitute back: $2(25) + F = 85 \implies F = 35$.
The flat visit fee is $\$35$.
Check: $5(25) + 35 = 125 + 35 = 160$. Correct.
Problem 8. For what values of $k$ does $kx + 3 = k^2 x - 5$ have no solution?
Rearrange: $(k - k^2)x = -8$, which gives $k(1-k)x = -8$.
The equation has no solution when the coefficient of $x$ is zero AND the constant is nonzero. Here the constant is $-8 \neq 0$.
Set $k(1-k) = 0$: either $k = 0$ or $k = 1$.
When $k = 0$: the equation becomes $3 = -5$, which is false. No solution.
When $k = 1$: the equation becomes $x + 3 = x - 5$, which gives $3 = -5$. No solution.
Both $k = 0$ and $k = 1$ yield no solution.
You have mastered this skill when you can do all of the following without referring to notes:
Think of a linear equation as a locked box on a balance scale. The box holds the number you want. On the other side of the scale is everything else.
To open the box (isolate $x$), you perform "anti-operations" -- operations that undo what was done to $x$ -- one layer at a time, from the outside in.
Each step simplifies the box a little more. When the box is alone on one side of the scale with a number on the other, you are done.
This mental picture also explains special cases:
For students who find math intimidating: Every step in this lesson is reversible. If you get stuck, go back one step and try a different move. There is no situation where you permanently break the problem. The check at the end always tells you whether you succeeded.
For students interested in proof: The two properties of equality stated above are consequences of the field axioms. A field is any number system where addition and multiplication satisfy the usual rules. The real numbers form a field. The technique in this lesson works in any field -- including modular arithmetic (used in cryptography) and rational functions (used in calculus).
For students interested in careers: Linear equations appear in budgeting, unit conversion, break-even analysis, dosage calculations, and every engineering formula where you are solving for one unknown quantity. The algebra here is not an abstract exercise -- it is the calculation layer beneath every spreadsheet formula you will ever write.