Linear Equations in One Variable

A shopkeeper in Jaipur sells notebooks for ₹40 each. After a full day of sales, the cash register shows ₹2,200 from notebooks alone. How many notebooks were sold?

You already know the answer is 2200 \div 40 = 55. But notice what you did to get it. You set up a relationship — price times quantity equals total — plugged in the known values, and solved for the unknown. That relationship, written out, is

where x is the number of notebooks. This is a linear equation in one variable. The word linear means the unknown x appears to the first power only — no x^2, no \sqrt{x}, no x in a denominator. The word one variable means there is exactly one unknown letter.

Every linear equation in one variable, no matter how tangled it looks on the page, boils down to the same skeleton: isolate x on one side, everything else on the other. The tools for doing this are exactly the properties from Operations and Properties: the additive inverse to cancel constants, the multiplicative inverse to cancel coefficients, and the distributive law to clear brackets. This article is the manual for using those tools fluently.

What makes an equation linear

An equation is linear in x when, after all simplification, the highest power of x is 1. Here are some examples and non-examples.

Equation	Linear?	Why
3x + 7 = 22	Yes	x appears to the first power
5(x - 3) = 2x + 9	Yes	after expanding, every x term is degree 1
\frac{x}{4} + 1 = 3	Yes	\frac{x}{4} = \frac{1}{4} \cdot x is still degree 1
x^2 - 5x + 6 = 0	No	x^2 makes it quadratic
\frac{1}{x} = 5	No	x is in the denominator, which is x^{-1}

The condition a \neq 0 is load-bearing. If a = 0, the equation collapses to b = 0, which is either always true (if b = 0) or never true (if b \neq 0). Neither case gives a single solution for x, so neither is a proper linear equation.

Solving by isolating the variable

The strategy is always the same: use the do-the-same-to-both-sides rule to peel away everything surrounding x, one layer at a time.

Take 7x - 11 = 24.

Peel away the constant. Add 11 to both sides:

Peel away the coefficient. Divide both sides by 7:

Check: 7(5) - 11 = 35 - 11 = 24. Correct.

The order of peeling — constants first, then the coefficient — is not a rigid rule but a habit that keeps the arithmetic clean. You could divide first, but that usually introduces fractions early and makes the next step messier.

When both sides have the variable

If x appears on both sides, collect all x-terms on one side and all constants on the other.

Take 5x + 3 = 2x + 18.

Subtract 2x from both sides: 3x + 3 = 18.

Subtract 3 from both sides: 3x = 15.

Divide both sides by 3: x = 5.

Check: left side = 5(5) + 3 = 28, right side = 2(5) + 18 = 28. Both sides match.

When brackets appear

Brackets mean the distributive law comes first.

Take 4(2x - 1) - 3(x + 5) = 2.

Distribute: 8x - 4 - 3x - 15 = 2.

Combine like terms on the left: 5x - 19 = 2.

Add 19 to both sides: 5x = 21.

Divide by 5: x = \frac{21}{5}.

Check: 4(2 \cdot \frac{21}{5} - 1) - 3(\frac{21}{5} + 5) = 4(\frac{42}{5} - \frac{5}{5}) - 3(\frac{21}{5} + \frac{25}{5}) = 4 \cdot \frac{37}{5} - 3 \cdot \frac{46}{5} = \frac{148}{5} - \frac{138}{5} = \frac{10}{5} = 2. Correct.

Equations with fractions

When fractions appear, the cleanest first move is to clear all denominators by multiplying both sides by the LCM of the denominators. This converts the equation into one with only integers, which is easier to handle.

Take \frac{x}{3} + \frac{x}{4} = 7.

The denominators are 3 and 4, whose LCM is 12. Multiply every term by 12:

Check: \frac{12}{3} + \frac{12}{4} = 4 + 3 = 7. Correct.

Here is a more tangled one: \frac{3x + 1}{5} - \frac{2x - 3}{7} = 1.

The LCM of 5 and 7 is 35. Multiply every term by 35:

Check: \frac{3(13/11) + 1}{5} - \frac{2(13/11) - 3}{7} = \frac{39/11 + 11/11}{5} - \frac{26/11 - 33/11}{7} = \frac{50/11}{5} - \frac{-7/11}{7} = \frac{10}{11} + \frac{1}{11} = \frac{11}{11} = 1. Correct.

The key insight is that multiplying by the LCM is not a special trick — it is the multiplicative version of the same rule you already know. You are multiplying both sides by the same number, which preserves equality. The only thing that changes is that the denominators vanish, because each denominator divides evenly into the LCM.

Word problems

The hard part of a word problem is never the algebra. It is the translation — converting a sentence in English (or Hindi) into a sentence in algebra. Once the equation is written, solving it is the same mechanical process you have already practised.

Here is a reliable approach:

1. Name the unknown. Write down what x stands for, with units if applicable.
2. Write the relationship. Translate the problem's conditions into an equation involving x.
3. Solve the equation.
4. Interpret and check. Make sure the answer makes sense in the original context.

Example. Priya is three years older than twice Ravi's age. If the sum of their ages is 27, find Ravi's age.

Let Ravi's age be x years. Then Priya's age is 2x + 3 years. Their sum is 27:

Ravi is 8 years old; Priya is 2(8) + 3 = 19 years old. Check: 8 + 19 = 27. Correct.

Example. A train covers a certain distance at 60 km/h. If it had gone 20 km/h faster, the journey would have taken 1 hour less. Find the distance.

Let the distance be d km. At 60 km/h, the time is \frac{d}{60} hours. At 80 km/h, the time is \frac{d}{80} hours. The second time is 1 hour less:

LCM of 60 and 80 is 240. Multiply both sides by 240:

The distance is 240 km. Check: at 60 km/h the time is 4 hours, at 80 km/h the time is 3 hours. The difference is indeed 1 hour.

Extraneous solutions

Sometimes the algebra hands you a value that does not actually work in the original equation. This happens most often when you clear denominators involving the variable, because multiplying both sides by an expression that might be zero can introduce a fake solution.

Take the equation \frac{x^2 - 4}{x - 2} = x + 3.

Factor the numerator: x^2 - 4 = (x - 2)(x + 2). So the left side becomes \frac{(x-2)(x+2)}{x-2}, which simplifies to x + 2 — but only when x \neq 2, because at x = 2 the denominator is zero and the original expression is undefined.

After simplification, the equation reads x + 2 = x + 3, which gives -1 = 0 — a contradiction. No value of x satisfies the equation.

But suppose you had instead cross-multiplied without noticing the restriction: (x^2 - 4) = (x - 2)(x + 3), expanding the right side to x^2 + x - 6, and then solving x^2 - 4 = x^2 + x - 6 gives x = 2. This is the extraneous solution — the algebra says x = 2, but plugging it back in produces \frac{0}{0} on the left side, which is undefined. The answer does not survive the check.

The rule is simple: always check your answer in the original equation, especially when you have multiplied both sides by an expression containing the variable. If the value makes any denominator zero, it is extraneous and must be discarded.

Here is a subtler case. Take \frac{5}{x - 1} + 2 = \frac{x + 9}{x - 1}.

Multiply both sides by (x - 1):

Check: left side = \frac{5}{5} + 2 = 1 + 2 = 3. Right side = \frac{15}{5} = 3. Both sides equal 3, so x = 6 is a valid solution — not extraneous.

The difference between the two cases: in the first, the candidate x = 2 made the denominator zero; in the second, the candidate x = 6 did not. The check catches the bad one and confirms the good one.

The solution on the number line

Every linear equation ax + b = 0 (with a \neq 0) has exactly one solution: x = -b/a. On the number line, this solution is a single point. You can see this by plotting the left-hand side as a function and asking where it crosses zero.

The interactive figure below shows the line y = ax + b. Drag the red point along the line and watch the readout. The solution is the x-value where y hits zero — the point where the line crosses the horizontal axis.

Two worked examples

Common confusions

• "I can cancel x from both sides of 3x = 5x." You can, but only by dividing both sides by x — and that is legal only if x \neq 0. If you do it blindly, you lose the solution x = 0. The safer move is to subtract 3x from both sides: 0 = 2x, so x = 0.

• "Clearing fractions means cross-multiplying." Cross-multiplication works only for equations of the form \frac{A}{B} = \frac{C}{D}. For equations with more than two fractions, or with additions of fractions, you need to multiply by the LCM of all the denominators, not cross-multiply.

• "The answer is x = 2, so I'm done." Not always. If the original equation had x in a denominator, x = 2 might make that denominator zero. You must check by substituting back into the original equation, not the simplified one.

• "(2x + 5)/3 - (x - 1)/4 = 2 has two variables because there are two fractions." No. Both fractions involve the same variable x. The number of fractions is not the number of variables — count the distinct letters, not the distinct terms.

• "In the word problem, 'twice Anil's age' is 2x." Only if the problem means twice his current age. If it says "twice his age in 5 years," the quantity being doubled is x + 5, and the expression is 2(x + 5) = 2x + 10, not 2x + 5.

• "A linear equation can have no solution." It depends. If after simplification you get 0 = 0, the equation is true for all x — it is an identity, not a proper linear equation. If you get 0 = 5, the equation has no solution — it is a contradiction. A genuine linear equation ax + b = 0 with a \neq 0 always has exactly one solution.

Going deeper

Why the solution is unique

A linear equation ax + b = 0 has exactly one solution because the function f(x) = ax + b is a straight line with slope a \neq 0. A non-horizontal line crosses the x-axis exactly once. This geometric fact — one crossing, one solution — is the reason linear equations are the simplest type of equation to solve, and why the whole process is so mechanical.

In contrast, a quadratic equation ax^2 + bx + c = 0 corresponds to a parabola, which can cross the x-axis zero, one, or two times. That is why quadratics need the quadratic formula and have a discriminant — the parabola might miss the axis entirely.

Linear equations as functions

When you plot y = ax + b, you get a straight line. The solution of ax + b = 0 is the x-intercept of that line. But the line also tells you something else: the sign of ax + b at every other point. To the left of the solution (if a > 0), ax + b is negative; to the right, it is positive. This is the idea behind solving linear inequalities — you find the root of the corresponding equation, then pick the correct side of the root based on the sign of a.

The general strategy: inverse operations

Every solving step is an inverse operation. Adding is undone by subtracting. Multiplying is undone by dividing. Squaring is undone by taking a square root. This is why the properties from Operations and Properties — additive inverse, multiplicative inverse — are the foundation of all equation-solving. Linear equations use only the first two inverses. Quadratics bring in the third. As equations get more complex, you need more kinds of inverse — but the fundamental idea never changes: peel off layers by applying the inverse of each operation, one at a time.

Equations with parameters

Sometimes a linear equation contains a parameter — a letter that is not the variable you are solving for. Take ax + b = cx + d, and solve for x. Subtract cx from both sides: (a - c)x + b = d. Subtract b: (a - c)x = d - b. Divide by (a - c): x = \frac{d - b}{a - c}, provided a \neq c. If a = c, the equation has no x-term, and it either has no solution (if b \neq d) or infinitely many (if b = d). The algebra is the same as before — the only new wrinkle is the condition a \neq c for a unique solution.

Where this leads next

Linear equations in one variable are the first rung of a ladder that extends through all of algebra.

• Algebraic Expressions — the building blocks that make up both sides of every equation you solve.
• Operations and Properties — the foundational rules (additive inverse, multiplicative inverse, distributive law) that justify every solving step.
• Linear Equations in Two Variables — what happens when the equation has two unknowns and the solution is a line, not a point.
• Systems of Linear Equations — solving two or more linear equations simultaneously to pin down two or more unknowns.
• Polynomials — Introduction — the broader family of expressions that includes linear equations as the simplest case.