Solving linear equations

A weighing balance stays level only if both pans hold equal weight — add a stone to one side and the balance tips unless you add the same stone to the other. A linear equation in one variable, such as $3(x-4)+2 = 2(x+1)-5$ , is exactly this kind of balance written in symbols: two expressions in a single unknown $x$ (never $x^2$ or $x^3$ ) that are declared equal, and "solving" it means finding the one number $x$ that keeps the pans level. This chapter builds on Algebraic Expressions and Rational Numbers, and teaches one reliable method — transposition — that handles brackets, fractional coefficients, and word problems about ages, consecutive numbers, and perimeters alike. Equations with two unknowns, a variable in the denominator, or $x^2$ terms belong to later classes and are outside this scope.

The core insight

An equation is a balance between two expressions in $x$ ; adding, subtracting, multiplying, or dividing both sides identically — most efficiently done by transposing terms across the "=" sign with a flipped sign — isolates $x$ and reveals the unique value that keeps the two sides equal.

The core idea

Think of the "=" sign as the fulcrum of a balance. The two operations that never disturb the balance are: adding or subtracting the same quantity from both sides, and multiplying or dividing both sides by the same non-zero quantity. Doing this term-by-term, over and over, is slow, so we shortcut it with transposition: a term crossing the "=" sign has its sign reversed, and a factor multiplying an entire side becomes a divisor on the other side (or vice versa).

ax + b = cx + d \quad\Longrightarrow\quad (a-c)x = d - b

where: $a$ and $c$ are the coefficients of $x$ on the left and right, $b$ and $d$ are the constant terms, and the arrow shows the result after transposing the $x$ -terms to the left and the constants to the right — the equation is now one step from solved. Two complications need handling before transposition can start. If the equation has brackets, expand them first with the distributive law $a(b+c) = ab + ac$ — a minus sign outside a bracket flips the sign of every term inside. If the equation has fractions, multiply every single term, on both sides, by the LCM of all the denominators; this clears the fractions in one stroke and turns the equation into one with only whole-number (or simple) coefficients. Skipping either step, or applying it to only some terms, is the single biggest source of wrong answers in this chapter. For word problems, an extra first step — translation — comes before any algebra: read the sentence, assign a letter (usually $x$ ) to the unknown, and write each phrase as an algebraic expression. "5 more than twice a number" is $2x+5$ , not $2(x+5)$ ; getting the translation wrong ruins an otherwise perfect solution.

Figure

A balance scale with algebraic terms on each pan staying level after equal terms are removed from both sides

Derivation: general method for solving a linear equation

Derivation AGeneral method for solving a linear equation in one variable

The goal is to derive a fixed, reliable sequence of steps that turns any linear equation in one variable — however many brackets, fractions, or terms it starts with — into the simple form $x = \text{(a number)}$ , and to show, using the equation $3(x-4)+2 = 2(x+1)-5$ as the running example, exactly why each step is legal and where students most often slip.

Step 1. Expand every bracket using the distributive law $a(b+c) = ab+ac$ , watching the sign carefully when a minus sign sits outside a bracket. In the example, $3(x-4)+2 = 2(x+1)-5$ becomes $3x - 12 + 2 = 2x + 2 - 5$ . If any denominators are present instead of (or alongside) brackets, this is also the moment to multiply every term of the equation by the LCM of all denominators, since an equation with fractions cannot be transposed cleanly until they are cleared.

Step 2. Combine like terms separately on the left side and on the right side; this is legal because each side is just being simplified internally, with no operation applied across the "=" sign yet. Here, $3x - 12 + 2$ simplifies to $3x - 10$ , and $2x + 2 - 5$ simplifies to $2x - 3$ , giving $3x - 10 = 2x - 3$ .

Step 3. Transpose to gather all $x$ -terms on one side (conventionally the left) and all constant terms on the other, flipping the sign of every term that crosses the "=" sign — this is equivalent to subtracting $2x$ from both sides and adding $10$ to both sides, but is faster to write. From $3x - 10 = 2x - 3$ , moving $2x$ left (it becomes $-2x$ ) and $-10$ right (it becomes $+10$ ) gives $3x - 2x = -3 + 10$ .

Step 4. Combine the like terms produced by transposition, then divide both sides by the coefficient of $x$ to isolate it. Here $3x - 2x = x$ and $-3+10 = 7$ , so directly $x = 7$ ; in general, if the coefficient were not $1$ , this last step would be a genuine division of both sides by that coefficient.

Step 5. Verify by substituting the solution into the original, unsimplified equation — never into an intermediate line, since an error made earlier would otherwise go undetected. Substituting $x=7$ : left side $3(7-4)+2 = 3(3)+2 = 11$ ; right side $2(7+1)-5 = 2(8)-5 = 11$ . Both sides equal $11$ , confirming $x=7$ is correct.

Worked example

Solve for $x$ : $\dfrac{2(x+1)}{3} + 1 = \dfrac{x-2}{2}$ .

Step 1. Clear the fractions by multiplying every term on both sides by the LCM of the denominators $3$ and $2$ , which is $6$ : $6 \cdot \dfrac{2(x+1)}{3} + 6 \cdot 1 = 6 \cdot \dfrac{x-2}{2}$ , giving $4(x+1) + 6 = 3(x-2)$ .

Step 2. Expand the brackets: $4x + 4 + 6 = 3x - 6$ , which simplifies to $4x + 10 = 3x - 6$ .

Step 3. Transpose the $x$ -terms to the left and the constants to the right: $4x - 3x = -6 - 10$ , so $x = -16$ .

Step 4. Check in the original equation. Left side: $\dfrac{2(-16+1)}{3} + 1 = \dfrac{2(-15)}{3} + 1 = -10 + 1 = -9$ . Right side: $\dfrac{-16-2}{2} = \dfrac{-18}{2} = -9$ . Both sides equal $-9$ , so $\boxed{x = -16}$ is correct.

Competitive stretch

Word problem: The sum of three consecutive odd numbers is 63. Find the numbers. Let the smallest be $x$ ; the numbers are $x$ , $x+2$ , $x+4$ , so $x + (x+2) + (x+4) = 63$ gives $3x + 6 = 63$ , then $3x = 57$ , so $x = 19$ . The numbers are $19$ , $21$ , $23$ , and indeed $19+21+23=63$ . ✓

Common traps

Forgetting the sign flip on transposition — a "+3" crossing the "=" must become "−3", and a "−5x" must become "+5x"; dropping this flip is the most common single error.
Distributing incorrectly through a bracket with a minus sign outside it — $5 - 2(x-3)$ expands to $5 - 2x + 6$ , not $5 - 2x - 6$ .
Multiplying only some terms by the LCM when clearing fractions — every term on both sides, including whole numbers with no visible denominator, must be multiplied, or the equation is no longer equivalent to the original.
Skipping the check, or checking against a simplified intermediate line instead of the original equation, which lets arithmetic slips through undetected.

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Why does a term change sign when it moves across the equals sign?

Transposition is just a shortcut for a legal balance operation. Moving "+3" from the right side to the left as "−3" is really the same as subtracting 3 from both sides of the original equation and then simplifying — the 3 cancels on the right and appears as −3 on the left. The sign flip is not a special rule for equations; it is ordinary subtraction written more quickly.

Do I always have to clear fractions before transposing?

It is strongly recommended, because transposing with fractions still present makes it easy to move only part of a fractional term or to forget to apply an operation to a denominator. Multiplying every term by the LCM of all denominators first turns the equation into one with simple coefficients, after which transposition is far less error-prone. You technically can transpose fractional terms directly, but clearing fractions first is safer and is the method examiners expect.

My answer to a word problem came out negative — did I make a mistake?

Not necessarily; a negative solution to the equation is not automatically wrong. Check it algebraically first by substituting back into the equation you formed. But then look at the context: if $x$ represents an age, a length, or a count of people, a negative value is not physically sensible, and you should re-read the problem to see whether you translated a phrase incorrectly. If the equation genuinely models a signed quantity (like a temperature change or a position on a number line), a negative answer can be perfectly valid.