Linear Equations in One Variable

You have solved equations since Class VI — little puzzles like $x + 5 = 12$ where a number on one side sat alone. This chapter grows that idea in two directions at once. First, the unknown $x$ is now allowed to appear on both sides of the equality sign, as in $2x - 3 = x + 2$ , so you can no longer just "read off" the answer — you need a method. Second, the expressions themselves may need tidying first: brackets to open, fractions to clear, like terms to combine, before the equation is even in a shape you recognise. Both problems have the same solution, one of the most useful ideas in all of school algebra: the balancing principle. An equation is a perfectly balanced weighing scale — whatever you do to one pan, you must do to the other, and the scale stays level. From that single principle falls a shortcut called transposition — moving a term from one side of the equation to the other by flipping its sign — which turns a slow, two-step balancing argument into one quick line of algebra. This chapter shows you exactly why that shortcut is allowed, not just how to use it, and then applies it to increasingly untidy equations: ones with the variable on both sides, ones buried inside brackets, and ones dressed up in fractions with different denominators. By the end you will be able to take almost any linear equation in one variable — however cluttered it looks — and reduce it, step by step, to the simple form $x = \text{a number}$ , then check your own answer without anyone else marking it. These are the exact algebraic muscles you will lean on in every later chapter, from factorisation to graphs to mensuration problems that translate word situations into equations.

Key takeaways

An algebraic equation is an equality between two expressions; left of "=" is the LHS, right of "=" is the RHS.
A linear equation in one variable has exactly one variable of highest power $1$ ; general form $ax + b = 0$ , $a \neq 0$ .
A solution (or root) is the value of the variable for which LHS equals RHS; always check by substituting it back in.
The balancing principle: adding, subtracting, multiplying or dividing both sides by the same quantity leaves the solution unchanged.
Transposition is the balancing principle in shorthand — a term moved across "=" has its sign (or operation) reversed.
Equations may have the variable on both sides; collect variable terms on one side, constants on the other, then divide out the coefficient.
Equations with brackets are solved by opening them and combining like terms first, reducing to the variable-on-both-sides form.
Equations with fractions are solved by multiplying both sides by the LCM of all the denominators, clearing every fraction at once.
Linear equations model real situations — ages, numbers, perimeters, money — by translating a word statement into an equation.
Key idea: any linear equation in one variable, however cluttered with brackets, fractions or variables on both sides, can always be reduced by the balancing principle (transposition, opening brackets, clearing denominators with the LCM) to the single form x = a number.

The core insight

Every linear equation in one variable can be reduced, by repeatedly performing the same operation on both sides (transposition, opening brackets, clearing fractions with the LCM), to the simple form $x = a$ number, which you then verify by substituting back into the original LHS and RHS.

What this chapter covers

This chapter has three connected ideas. It opens by revising what an equation actually is — an equality of two expressions, with an LHS and an RHS — and what makes it linear in one variable. It then tackles the main new skill of Class VIII: equations with the variable on both sides, solved using the balancing principle and its shortcut, transposition. Finally it handles equations that look messier than they are — brackets to open, fractions to clear using the LCM — showing these always reduce to the same variable-on-both-sides method. As the chapter's closing remarks note, linear equations power word problems on ages, numbers, perimeters and money.

The core ideas

Algebraic equations: LHS and RHS

An algebraic expression — such as $5x$ , $2x - 3$ , or $3y + 7$ — has no equality sign; it is just a combination of numbers, variables and operations. An algebraic equation is different: it asserts that two expressions have the same value, joined by an equality sign. In the equation

2x - 3 = 7,

the expression on the left, $2x - 3$ , is the Left Hand Side (LHS), and the expression on the right, $7$ , is the Right Hand Side (RHS). An equation is true only for particular values of the variable; an expression, having no equality sign, is never "true" or "false" at all.

What makes an equation linear in one variable

An equation is linear in one variable when it contains only a single variable (say $x$ ) and the highest power of that variable appearing anywhere in the equation is $1$ . Its most general form, after every term has been collected, is

ax + b = 0, \qquad a \neq 0,

where: $a$ and $b$ are fixed numbers (rational numbers, possibly written as decimals or fractions) and $x$ is the unknown. The condition $a \neq 0$ matters — if $a$ were $0$ there would be no variable term left, and the sentence would just be a statement about numbers, true or false regardless of $x$ . Expressions such as $x^2 + 1$ or $y + y^2$ are not linear, because the variable's highest power there is $2$ ; equations built from them are quadratic, not linear, and are studied later.

Solutions and the balancing principle

A value of the variable that makes LHS equal RHS is called a solution (or root) of the equation. For $2x - 3 = 7$ , substituting $x = 5$ gives LHS $= 2 \times 5 - 3 = 7 =$ RHS, so $x = 5$ is a solution; substituting $x = 10$ gives LHS $= 17 \neq$ RHS, so $x = 10$ is not. How do you find a solution rather than just check one? Picture the equation as a two-pan weighing balance that is perfectly level: LHS sits in one pan, RHS in the other, and equality means the pans are exactly level. The balancing principle says that if you perform the same operation on both pans, the scale stays level:

\text{if } A = B, \text{ then } A \pm k = B \pm k, \quad A \times k = B \times k, \quad \frac{A}{k} = \frac{B}{k} \ (k \neq 0).

where: $A$ and $B$ are the two sides of the equation and $k$ is any number, or even any algebraic expression, applied identically to both sides. Repeatedly choosing a convenient $k$ — to cancel a constant, or to remove a coefficient — walks the equation, step by balanced step, down to $x = \text{a number}$ .

Figure

A two pan balance scale showing the equation two x minus three equals seven staying level after three is added to both pans and again after both pans are divided by two, ending with x alone equal to five

Transposition: the balancing principle in shorthand

Applying the balancing principle one careful step at a time is safe but slow. In practice, you use a faster shortcut called transposition: a term can be moved from one side of the equation to the other, provided its sign (or operation) is reversed.

A + t = B \quad \Longleftrightarrow \quad A = B - t.

where: $t$ is the term being moved; a term added on one side becomes subtracted on the other, a subtracted term becomes added, a multiplying factor becomes a dividing factor, and a dividing factor becomes a multiplying one. Transposition is not a separate rule to memorise blindly — it is exactly the balancing principle with the "subtract $t$ from both sides" step already carried out and written in one line. The first derivation below proves this equivalence precisely.

Figure

A diagram of the equation two x minus three equals x plus two with a curved arrow showing the term x crossing from the right side to the left side and its sign flipping from plus to minus

Equations with the variable on both sides

When the RHS is not a bare number but another expression containing the variable, first move all variable terms to one side and all constant terms to the other, then divide by the coefficient that remains. For the general linear-in-one-variable equation

ax + b = cx + d \quad \Longrightarrow \quad x = \frac{d - b}{a - c}, \qquad a \neq c,

where: $a$ and $c$ are the coefficients of $x$ on the LHS and RHS, $b$ and $d$ are the constant terms, and $a \neq c$ is essential — otherwise the variable terms cancel and no unique solution exists. This is the pattern of NCERT's Example 1 ( $2x - 3 = x + 2$ ) and Example 2: collect the $x$ 's, collect the numbers, divide.

Reducing equations to simpler form

Some equations are not yet in the tidy $ax + b = cx + d$ shape — they hide behind brackets or fractions. Two habits reduce them. Opening brackets first: multiply out every bracketed expression using the distributive law before collecting like terms, so $5(2x + 1) - 2 = 3(x - 4)$ becomes $10x + 5 - 2 = 3x - 12$ , an ordinary variable-on-both-sides equation. Clearing fractions by the LCM: if the equation contains fractional terms such as $\dfrac{6x+1}{3} + 1 = \dfrac{x-3}{6}$ , multiply every term on both sides by the LCM of all the denominators, since each denominator divides the LCM exactly.

\text{LCM}(m, n, k) \times \left(\frac{px}{m} + \frac{q}{n}\right) = \text{LCM}(m,n,k) \times \left(\frac{rx}{k} + s\right)

where: $m$ , $n$ , $k$ are the denominators appearing anywhere in the equation and $p, q, r, s$ are the numerators or free constants — this is precisely how NCERT's Example 16 (multiplying by $6$ ) and Example 17 (multiplying by $2$ ) turn fractional equations into whole-number ones.

Applications: turning words into equations

The real payoff of this chapter is translating a described situation — ages, consecutive numbers, a perimeter, or a combination of currency notes — into a linear equation, then solving it with the same balancing and transposition steps. Name the unknown quantity $x$ , write every other quantity in terms of $x$ , translate the connecting sentence into an equation, then check the answer against the original situation (an age cannot come out negative). The Higher-order-thinking worked example below carries out this translation in full for an age problem.

Derivation AThe transposition rule, derived from the balancing principle

The goal of this derivation is to show, in full generality, that "transposing" a term across the equals sign — moving it to the other side and reversing its sign or operation — is not an independent rule to be memorised, but is simply the balancing principle with one obvious step already carried out. Once this is proved for addition and subtraction, the same argument run with multiplication and division justifies moving a multiplying or dividing factor across the equals sign too. A rule you understand cannot be misapplied the way a rule you merely memorise can.

Step 1. Start from a general equation of the form $A + t = B$ , where $A$ is whatever remains on the LHS once the term $t$ is set aside, and $B$ is the RHS. This covers every case you will meet: $t$ might be a constant like $3$ , or a variable term like $x$ .

Step 2. Apply the balancing principle by subtracting $t$ from both sides — a legal operation, since subtracting the same quantity from both pans of a balanced scale keeps it level. This gives $A + t - t = B - t$ .

Step 3. Simplify the LHS. Since $t - t = 0$ , the left side collapses to $A + 0 = A$ , so the equation becomes $A = B - t$ .

Step 4. Compare the starting equation $A + t = B$ with the result $A = B - t$ . The term $t$ , added on the left, now appears subtracted on the right — it has "crossed" the equals sign and flipped sign, purely from the balancing step in Step 2, not by any new rule. The identical argument with $A - t = B$ (add $t$ to both sides) shows a subtracted term becomes added, and with $At = B$ (divide both sides by $t \neq 0$ ) shows a multiplying factor becomes a dividing one. This is also why transposition can never apply to a bare expression that has no equals sign to balance across.

Go deeper · Derivation B · 1 minClearing fractions by multiplying both sides by the LCM

Derivation BClearing fractions by multiplying both sides by the LCM

The goal of this derivation is to justify the standard method for equations with fractional terms of different denominators: multiply every term on both sides by the LCM of the denominators, and the fractions disappear while the solution stays exactly the same. This is the method behind NCERT's Example 16 and Example 17, and it shows why the smallest common multiple, rather than any other common multiple, is the efficient choice.

Step 1. Take a representative fractional equation, $\dfrac{6x+1}{3} + 1 = \dfrac{x-3}{6}$ . The two denominators appearing are $3$ and $6$ ; their LCM is $6$ , since $6$ is the smallest number that both $3$ and $6$ divide into exactly ( $6 = 3 \times 2 = 6 \times 1$ ).

Step 2. Apply the balancing principle by multiplying both sides of the equation by this LCM, $6$ . This is legal for exactly the same reason multiplying both pans of a balanced scale by the same number keeps it level: $6 \times \left(\dfrac{6x+1}{3} + 1\right) = 6 \times \dfrac{x-3}{6}$ .

Step 3. Distribute the $6$ across every term on the LHS, and simplify each fraction using the fact that the LCM is an exact multiple of each denominator: $6$ divided by $3$ is $2$ , so $6 \times \dfrac{6x+1}{3} = 2(6x+1) = 12x + 2$ ; and $6 \times 1 = 6$ . On the RHS, $6$ divided by $6$ is $1$ , so $6 \times \dfrac{x-3}{6} = 1 \times (x - 3) = x - 3$ . The equation is now $12x + 2 + 6 = x - 3$ , with no fractions anywhere.

Step 4. Solve the resulting whole-number equation by the ordinary variable-on-both-sides method: combine $12x + 8 = x - 3$ , transpose $x$ to the left and $8$ to the right, giving $11x = -11$ , so $x = -1$ . Because every step used only balancing-principle moves, this $x = -1$ is guaranteed to satisfy the original fractional equation too, which is exactly why substituting it back in gives LHS $=$ RHS. Choosing the LCM rather than a larger common multiple keeps every resulting coefficient as small as possible.

Worked examples

Board level - variable on both sides (Exercise 2.1 pattern)

Solve $5t - 3 = 3t - 5$ and check your result.

Step 1. Transpose the variable term $3t$ from the RHS to the LHS, flipping its sign: $5t - 3t - 3 = -5$ , which simplifies to $2t - 3 = -5$ .

Step 2. Transpose the constant $-3$ from the LHS to the RHS, flipping its sign: $2t = -5 + 3 = -2$ .

Step 3. Divide both sides by the coefficient of $t$ , which is $2$ : $t = \dfrac{-2}{2} = -1$ .

Step 4. Check by substituting $t = -1$ . LHS $= 5(-1) - 3 = -8$ . RHS $= 3(-1) - 5 = -8$ . LHS $=$ RHS. Answer: $t = -1$ .

Board level - clearing fractions with the LCM (Exercise 2.2 pattern)

Solve $\dfrac{x}{2} - \dfrac{1}{5} = \dfrac{x}{3} + \dfrac{1}{4}$ .

Step 1. Find the LCM of the four denominators $2$ , $5$ , $3$ , $4$ : $\text{LCM}(2,5,3,4) = 60$ . Multiply every term on both sides by $60$ .

Step 2. Simplify each term using $60 \div 2 = 30$ , $60 \div 5 = 12$ , $60 \div 3 = 20$ , $60 \div 4 = 15$ : the equation becomes $30x - 12 = 20x + 15$ , with no fractions left.

Step 3. Transpose $20x$ to the left and $-12$ to the right: $30x - 20x = 15 + 12$ , giving $10x = 27$ , so $x = \dfrac{27}{10} = 2.7$ .

Step 4. Check by substituting $x = 2.7$ : LHS $= \dfrac{2.7}{2} - \dfrac{1}{5} = 1.35 - 0.2 = 1.15$ ; RHS $= \dfrac{2.7}{3} + \dfrac{1}{4} = 0.9 + 0.25 = 1.15$ . LHS $=$ RHS. Answer: $x = 2.7$ .

Higher-order thinking (HOTS / olympiad-style) - an age problem translated into an equation

A father's present age is three times his daughter's present age. Five years ago, the father's age was four times the daughter's age at that time. Find their present ages.

Step 1. Translate the words into algebra. Let the daughter's present age be $x$ years; the father's present age is then $3x$ years.

Step 2. Write both ages "five years ago" by subtracting $5$ : daughter $x - 5$ , father $3x - 5$ .

Step 3. Translate the second sentence into an equation: five years ago the father's age was four times the daughter's age then, so $3x - 5 = 4(x - 5)$ .

Step 4. Open the bracket and transpose. The RHS becomes $4x - 20$ , so $3x - 5 = 4x - 20$ . Transposing $4x$ left and $-5$ right gives $3x - 4x = -20 + 5$ , so $-x = -15$ , and $x = 15$ .

Step 5. Interpret and check. The daughter is $x = 15$ and the father is $3x = 45$ . Five years ago they were $10$ and $40$ , and indeed $40 = 4 \times 10$ . Answer: daughter is 15 years old, father is 45 years old. Timing note: aim for under three minutes. The only new skill beyond a routine equation is Step 1 to Step 3 — naming one unknown, writing every other quantity in terms of it, and turning the sentence into an equation. Once the equation is written, it is an ordinary bracket-and-transpose problem. Always finish, as in Step 5, by checking the numbers make sense in the real situation, not only in the equation — a negative or fractional age would signal a translation error, not an arithmetic one.

Common traps

Sign errors during transposition are the single most common slip: a term that was subtracted must arrive on the other side added, and vice versa. When clearing fractions, students often multiply the fractional terms by the LCM but forget to multiply a lone whole-number term (like the $+1$ in Example 16), which silently changes the equation. When opening brackets preceded by a minus sign, such as $-2(3x - 4)$ , every term inside must have its sign reversed ( $-6x + 8$ ), not just the first one. Dividing both sides by an expression that could be zero is not a valid step. Finally, many students solve correctly but skip the final check — substituting the answer back into the original LHS and RHS.

Common trapTransposition only works because it is the balancing principle in disguise

The single biggest misconception in this chapter is treating transposition as a free-standing trick — "just move it and flip the sign" — memorised without understanding why it is allowed. That is dangerous because it invites two very specific errors: applying "transposition" to a bare algebraic expression that has no equals sign to balance across (expressions cannot be transposed, only equated quantities can), and getting the flipped operation wrong when a multiplying or dividing factor, rather than an added or subtracted term, crosses the sign. Remembering that transposition is always shorthand for "do the same thing to both sides" fixes both errors at once: you would never try to balance a scale that has nothing on the other pan, and you would never confuse adding with multiplying when deciding what operation to reverse.

You should be able to…

Distinguish an algebraic expression from an algebraic equation, and identify the LHS and RHS of an equation.
Recognise when an equation is linear in one variable, using the general form $ax + b = 0$ , $a \neq 0$ .
State the balancing principle and explain why performing the same operation on both sides preserves the solution.
Use transposition confidently and correctly, reversing the sign or operation of any term that crosses the equals sign.
Solve linear equations that have the variable on both sides by collecting variable terms on one side and constants on the other.
Open brackets correctly, including brackets preceded by a minus sign, before combining like terms.
Clear all fractions in an equation by multiplying every term on both sides by the LCM of the denominators.
Verify any solution by substituting it back into the original LHS and RHS separately.
Translate a simple word problem, such as one about ages, numbers or perimeters, into a linear equation and solve it.

Ask the tutor: Want to work through a tricky word problem — ages, numbers, or a perimeter puzzle — step by step together?

Why does the sign of a term change when I move it to the other side of the equation?

Because moving a term across the equals sign is never a separate rule — it is a shortcut for performing the identical operation on both sides, the one thing that is always allowed. If a term $t$ is added on the LHS, subtracting $t$ from both sides removes it from the LHS entirely and makes it appear subtracted on the RHS. Writing that subtraction out every time is correct but slow, so transposition skips straight to the result: the term reappears on the other side with its sign reversed. Forget which way to flip? Fall back on the slower balancing-principle version — it always gives the right answer.

Why do I have to multiply every single term by the LCM, even the ones that are not fractions?

Because the balancing principle demands that both entire sides be multiplied by the same number, and "both sides" means every term you have written down, fractional or not. If you multiply only the fractional terms and leave a whole-number term untouched, you have secretly changed the equation into a different one with a different solution. A safe habit is to bracket each full side first, $\text{LCM} \times (\text{whole LHS}) = \text{LCM} \times (\text{whole RHS})$ , and only then distribute term by term, so nothing is left out by accident.

Is an equation like 2x + 3y = 7 also a linear equation in one variable?

No. Although $2x + 3y = 7$ is linear — the highest power of any variable is $1$ — it contains two variables, $x$ and $y$ , not one. This chapter deals only with equations that have a single variable, so $2x + 3y = 7$ falls outside it; it is called a linear equation in two variables and needs a second equation to pin down unique values of $x$ and $y$ , a topic for a later class. Always check first how many different letters appear before applying anything from this chapter.

How can I be sure my answer is actually correct without a teacher checking it?

Substitute your solution back into the original equation, before any simplification, and evaluate the LHS and the RHS completely separately. If both come out to exactly the same number, your solution is verified; if they differ, there is an arithmetic or sign error somewhere in your working, and you should redo the steps rather than trust the answer. This is the fastest way to catch a transposition sign slip or a missed LCM term, and both worked examples above show exactly how to lay it out, LHS and RHS side by side.