Linear equations in one variable

A pan balance stays level only when both sides weigh the same, and a linear equation is exactly that idea written in symbols: a statement that two expressions, one of them holding an unknown, are equal. Solving the equation means finding the single value of that unknown — call it $x$ — that keeps the two sides balanced. Add a weight to one pan and the balance tips unless you add the same weight to the other; that single habit, done algebraically instead of physically, is the whole technique of transposition. This article builds the ICSE Class 8 picture from the general form $ax+b=c$ up: how transposition moves terms across the equals sign without breaking the balance, how brackets and fractional coefficients are cleared before you transpose, and how a word problem — ages, consecutive numbers, perimeters — turns into an equation you can actually solve.

The core insight

A linear equation in one variable, $ax+b=c$ , is solved by performing the same operation on both sides until the unknown stands alone, a process carried out in practice through transposition — sign-flipping a term as it crosses the equals sign.

The core idea

The general form. Every linear equation in one variable can be written as

ax+b=c

where: $a,b,c$ are known numbers, $x$ is the unknown, and $a\neq0$ (if $a=0$ there is no $x$ term left to solve for). The equation is "linear" because $x$ appears only to the first power — no $x^2$ , no $1/x$ .

The golden rule. Whatever you do to one side of the equation, you must do to the other side too, or the equality breaks. Add 5 to the left pan of the balance and it dips; add 5 to the right pan as well and it levels again. In practice, instead of adding the same term to both sides and then simplifying, it is quicker to transpose: move a term across the equals sign, flipping its operation as it crosses. A term that was added becomes subtracted, and a term that was multiplying becomes dividing:

x+5=12 \;\Rightarrow\; x=12-5=7

3x=15 \;\Rightarrow\; x=\dfrac{15}{3}=5

Figure

A pan balance in equilibrium labelled a x plus b on the left pan and c on the right pan, with an arrow showing a term moving across and its sign flipping

Brackets and fractions. When an equation has brackets, expand them first using the distributive law before transposing anything. When it has fractional coefficients, multiply every single term on both sides by the LCM of the denominators to clear the fractions — leaving out even one term is the single most common slip.

Setting up word problems. Read the problem, decide exactly what is unknown, and represent it with one letter, stating clearly what it stands for — "let the son's present age be $x$ years," not just "let $x$ be the number." Then express every other quantity mentioned in terms of that same letter, translate the relationship in words into an equation, solve it, and finally translate the numeric answer back into the story, checking it makes sense (an age or a length should never come out negative). Common word-problem families follow recognisable patterns worth knowing on sight. Age problems usually give a relationship between two people's ages now, and sometimes a relationship a few years in the past or future — remember that if the present age is $x$ , an age " $n$ years ago" is $x-n$ and an age " $n$ years hence" is $x+n$ , for both people at once. Consecutive number problems rely on the fact that consecutive integers differ by 1, consecutive even or odd integers differ by 2, so if the first is $x$ the next is $x+1$ (or $x+2$ for evens/odds). Perimeter and geometry problems turn a verbal description of sides into an algebraic expression for each side, then apply the relevant perimeter or angle-sum formula. In every family, the equation itself is usually short once the unknown and the other quantities are correctly expressed — the setup, not the solving, is where marks are lost.

Derivation ASolving ax + b = c for x

The goal is to show that the familiar shortcut of "moving a term across and flipping its sign" is not a trick to memorise but a direct, provable consequence of the golden rule that both sides of an equation must always receive the same operation, so that transposition can be trusted as a genuine algebraic law rather than a guess.

Step 1. Start from the general equation and apply the golden rule to remove $b$ from the left side: subtract $b$ from both sides.

ax+b-b=c-b

Step 2. Simplify the left side, since $b-b=0$ .

ax=c-b

Notice the outcome: $b$ , which was added on the left, now appears subtracted on the right — exactly the sign-flip rule for transposing an added term, now shown to follow from subtracting $b$ from both sides rather than assumed outright.

Step 3. Apply the golden rule again to remove the coefficient $a$ from the left side: divide both sides by $a$ (valid since $a\neq0$ ).

\frac{ax}{a}=\frac{c-b}{a}

Step 4. Simplify the left side, since $\dfrac{a}{a}=1$ .

x=\frac{c-b}{a}

This is the general solution of every equation of the form $ax+b=c$ . It also shows the second transposition rule: a term multiplying one side ( $a$ ) becomes a term dividing the other side once it crosses over, again because dividing both sides by $a$ is simply the golden rule applied a second time. Check: substituting $x=\dfrac{c-b}{a}$ back into $ax+b$ gives $a\left(\dfrac{c-b}{a}\right)+b=(c-b)+b=c$ , which matches the right side, confirming the formula.

Worked example

Question: Solve for $x$ : $3(x-2) = 2(x+5) - 4$ .

Step 1. Expand both brackets using the distributive law.

3x - 6 = 2x + 10 - 4

Step 2. Simplify the right side by combining the constant terms.

3x - 6 = 2x + 6

Step 3. Transpose $2x$ to the left side (it flips from $+2x$ to $-2x$ ) and transpose $-6$ to the right side (it flips from $-6$ to $+6$ ).

3x - 2x = 6 + 6

Step 4. Simplify both sides.

x = 12

Step 5. Check by substituting $x=12$ into the original equation: LHS $=3(12-2)=3(10)=30$ ; RHS $=2(12+5)-4=2(17)-4=34-4=30$ . LHS equals RHS, so $x=12$ is confirmed correct.

Competitive stretch

The same routine handles a word problem. A rectangle's length is 5 cm more than its breadth, and its perimeter is 50 cm. Let the breadth be $x$ cm, so the length is $(x+5)$ cm. The perimeter condition gives $2\big(x+(x+5)\big)=50$ , which simplifies to $4x+10=50$ ; transposing 10 gives $4x=40$ , so $x=10$ . The breadth is 10 cm and the length is 15 cm — check: $2(10+15)=2(25)=50$ ✓.

Common traps

Forgetting to flip the sign of a term when transposing it — a term that was $+7$ on the left must become $-7$ when moved to the right.
Not distributing a bracket's outer number over every term inside, e.g. expanding $3(x-2)$ as $3x-2$ instead of the correct $3x-6$ .
Multiplying only some terms by the LCM when clearing fractions, instead of every single term on both sides.
Defining a word-problem variable vaguely, such as "let $x$ be the number," without pinning down exactly what quantity it represents, which leads to setting up the wrong equation.

Ask the tutor: Want more practice equations with brackets and fractions, or help setting up a tricky age or perimeter word problem? Ask the AI tutor for extra worked examples.

Why do we flip the sign when a term crosses the equals sign, instead of just erasing it?

Transposition is really the golden rule in disguise. Moving $+b$ from the left side to become $-b$ on the right is exactly the same as subtracting $b$ from both sides of the equation and simplifying — the derivation above proves this line by line. The "flip the sign" habit is just a fast way of writing that subtraction step without redoing the full working every time.

Why must I multiply every term by the LCM when clearing a fraction, not just the fractional one?

The golden rule again: whatever you do to the equation, you must do to every term on both sides, or the balance breaks. If the equation is $\dfrac{x}{2}+3=7$ and you multiply only the $\dfrac{x}{2}$ term by 2, you get $x+3=7$ , which is a different, wrong equation — the correct step multiplies all three terms, giving $x+6=14$ .

How do I know what letter to use and what it should stand for in a word problem?

Pick a single letter for the one quantity you genuinely do not know, and write a short sentence pinning it down, such as "let the smaller number be $x$ ." Every other quantity in the problem should then be written in terms of that same $x$ — for consecutive numbers, the next one is $x+1$ or $x+2$ for odd or even numbers; for a rectangle, if the breadth is $x$ , a length "5 more" is $x+5$ . Being precise here is usually the harder half of the problem; the algebra afterwards is routine.

Linear equations in one variable

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads