Linear Equations

What makes an equation linear

A linear equation in one variable is one where the variable only appears to the first power: no $x^2$, no $\sqrt{x}$, no $\dfrac{1}{x}$. After simplifying, such an equation can be written as

so whenever $a \neq 0$ it has exactly one solution. If the $x$ terms cancel completely ($a = 0$), there is either no solution, as with $x + 1 = x + 2$, or every $x$ works, as with $x + 1 = x + 1$.

The four steps

1. Clear fractions by multiplying every term on both sides by the LCM of the denominators.
2. Expand brackets and simplify each side.
3. Collect the variable terms on one side and the numbers on the other. A term changes sign when it crosses the equals sign.
4. Divide by the coefficient of $x$, then substitute back into the original equation to check.

Worked example: the variable on both sides

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

Solution: Subtract $x$ from both sides and add 5 to both sides:

Check: $3(6) - 5 = 13$ and $6 + 7 = 13$.

Worked example: an equation with mixed fractions

Solve $x - 2\frac{1}{3} = 5\frac{1}{2}$.

Solution: Isolate $x$ by adding $2\frac{1}{3}$ to both sides:

Worked example: fractions with the variable on top

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

Solution: The LCM of 3 and 2 is 6. Multiply every term, including the 4, by 6:

Check: $\dfrac{3}{3} + \dfrac{6}{2} = 1 + 3 = 4$.

Turning words into an equation

Worked example: a word problem

One fourth of a number, added to one sixth of it, is 15. Find the number.

Solution: Let the number be $n$:

Using a common denominator of 12: $\dfrac{3n}{12} + \dfrac{2n}{12} = \dfrac{5n}{12} = 15$, so:

Worked example: an age problem

A father is three times as old as his son. In 12 years he will be twice as old as his son. Find their present ages.

Solution: Let the son be $x$ years old, so the father is $3x$. In 12 years they will be $x + 12$ and $3x + 12$:

The son is 12 and the father is 36. Check: in 12 years they will be 24 and 48, and $48 = 2 \times 24$.

Worked example: consecutive integers

Find three consecutive integers whose sum is 78.

Solution: Call them $x$, $x+1$ and $x+2$, as in the table above. Their sum is 78:

So the integers are 25, 26 and 27. Check: $25 + 26 + 27 = 78$. A shortcut: three consecutive integers add up to 3 times the middle one, so the middle one is $78 \div 3 = 26$. The same method gives 110, 111 and 112 for a sum of 333.

Common mistakes

• Multiplying only some terms by the LCM. In the fractions example, forgetting to multiply the 4 by 6 gives $5x - 1 = 4$ and the wrong answer $x = 1$.
• Losing a minus sign across a bracket. $-(x - 3)$ is $-x + 3$, not $-x - 3$.
• Writing "5 less than x" as $5 - x$. It's $x - 5$. Read the phrase as "start with x, then take 5 away".
• Skipping the check. Substituting the answer back into the original equation catches sign slips before they cost marks.

Where this comes up in exams

The 2025-26 CBSE Class IX syllabus opens its linear equations unit with a recall of linear equations in one variable, before moving on to equations in two variables.

Checked against CBSE Mathematics curriculum, Classes IX-X (2025-26). Syllabi can change from year to year, so confirm with the latest official notification for your exam.

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