How to solve linear equations

The balance principle#

An equation is a pair of scales in balance. Two moves are allowed: add (or subtract) the same number on both sides, and multiply (or divide) both sides by the same non-zero number. Every technique you learn at school is a special case of these two.

Moving a term across the equals sign and changing its sign is not a separate rule — it is shorthand for the first move. In $x + 5 = 12$ you subtract 5 from both sides: $x$ is left on the left, and 7 appears on the right. On paper it looks as if "the five moved across and became minus five".

The order of work is always the same: expand brackets, clear fractions, collect the variable terms on the left and the numbers on the right, combine like terms, divide by the coefficient, check.

Example 1: the variable on both sides#

Solve $5x - 7 = 3x + 9$.

Step 1. Move $3x$ to the left — its sign changes as it crosses: $5x - 3x - 7 = 9$.

Step 2. Move $-7$ to the right, where it becomes $+7$: $5x - 3x = 9 + 7$.

Step 3. Combine like terms on both sides: $2x = 16$.

Step 4. Divide both sides by the coefficient of the variable: $x = 8$.

Step 5. Check by substituting into the original equation. The left side is $5 \cdot 8 - 7 = 33$, the right side is $3 \cdot 8 + 9 = 33$. They are equal, so the solution is right.

Example 2: with brackets#

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

Step 1. Expand the brackets: the 3 multiplies every term inside, not just the first: $3x - 6 = 2x + 5$.

Step 2. Variables to the left, numbers to the right: $3x - 2x = 5 + 6$.

Step 3. Combine like terms: $x = 11$.

Step 4. Check: the left side is $3 \cdot (11 - 2) = 27$, the right side is $2 \cdot 11 + 5 = 27$. It matches.

If there is a minus sign in front of the bracket, every term inside changes sign: $-(x - 4) = -x + 4$. This is where the most marks are lost in tests.

Example 3: with fractions#

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

Step 1. Find the common denominator of the fractions — 4 — and multiply both whole sides of the equation by it. How to find a common denominator is covered in adding fractions with unlike denominators.

Step 2. On the left the fraction cancels and $x + 1$ is left; on the right $\frac{4(x-3)}{2} = 2(x - 3)$. The equation no longer has fractions: $x + 1 = 2(x - 3)$.

Step 3. Expand: $x + 1 = 2x - 6$.

Step 4. Move terms: $x - 2x = -6 - 1$, that is $-x = -7$.

Step 5. Divide both sides by $-1$: $x = 7$.

Step 6. Check: the left side is $\frac{8}{4} = 2$, the right side is $\frac{4}{2} = 2$. Correct.

You have to multiply every term of the equation by the common denominator, including terms that have no fraction. A forgotten term is the most common reason for a wrong answer in equations of this kind.

When there is no solution or infinitely many#

Sometimes, after combining like terms, the variable disappears from both sides.

If you are left with a false statement — for example $0 \cdot x = 5$ or $3 = 8$ — there is no solution: no number works.

If you are left with a true statement such as $0 = 0$, any number works: the equation turned out to be an identity. That happens with $2(x + 3) = 2x + 6$ — both sides say the same thing.

Both cases are legitimate answers, not a sign that the solution broke. Writing $x = 0$ here is wrong: zero either does not work, or works just like every other number.

Typical mistakes#

Moving a term without changing its sign. Substitution catches it: if the answer does not fit, look for the sign.

Dividing only one side by the coefficient. You have to divide both.

Multiplying only the first term in the brackets: turning $3(x - 2)$ into $3x - 2$ instead of $3x - 6$.

Cancelling the variable by dividing by an expression that could be zero. In linear equations you never need to divide by an expression with the variable — moving terms is enough.

Checking and what comes next#

The substitution check is not optional: it takes one line and catches every mistake listed above. Always substitute into the original equation, not into an intermediate line — otherwise a mistake made along the way is repeated in the check.

Once single equations stop causing trouble, the next topic is systems of linear equations: there you reduce two unknowns to one and solve with exactly the same moves.

Step-by-step plan

1. Step 1 — the simplest onesTen equations of the form x + a = b and ax = b, each checked by substitution.
2. Step 2 — variable on both sidesCollect variables on the left and numbers on the right; watch the sign change when you move a term.
3. Step 3 — bracketsEquations with brackets, including a minus sign in front of a bracket.
4. Step 4 — fractionsMultiply both sides by the common denominator without skipping terms that have no fraction.
5. Step 5 — special casesRecognise equations with no solution and identities, and write the answer in words.

Check yourself

Sources

• Khan Academy — Algebra 1Linear equations: video explanations and practice with answer checking
  free
• OpenStax Elementary Algebra 2eFree textbook; chapter 2 solves linear equations step by step
  free

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