Linear Equations in One Variable: ICSE Class 7 Guide

A boy in my Saturday class once told me his school paper had a 'really hard' question. It said: 'The sum of three consecutive numbers is 54. Find them.' He had written half a page of trial and error, crossed everything out, and still got the wrong answer. I showed him one line, $x + (x + 1) + (x + 2) = 54$, and watched his face light up when he got $x = 17$ and the three numbers 17, 18 and 19 in under a minute.

That is the magic of linear equations. They take word problems that look intimidating and turn them into a clean, mechanical recipe: name the unknown, write the equation, solve, answer in words. Four steps. Same four steps for every problem.

ICSE Class 7 leans heavily on this chapter because it is the foundation for everything algebraic that follows. The Selina textbook has exercises full of age problems, number problems and perimeter problems, and once your child gets the rhythm of the four step method they will start looking forward to these questions.

An equation is any statement that says two expressions are equal. $2 + 3 = 5$ is an equation. $x + 4 = 10$ is also an equation, but this one has an unknown called $x$ that we need to find.

A linear equation is an equation where the variable appears only to the first power. No $x^2$, no $x^3$, no square roots or fractions of $x$. The word linear comes from the fact that if you draw such an equation on a graph, you get a straight line. For Class 7 we only work with equations in one variable, so the graph idea is saved for Class 9.

Examples of linear equations in one variable:

* $x + 7 = 12$
* $3y - 5 = 10$
* $2(z - 3) = 8$

Not linear:

* $x^2 + 1 = 5$ (has $x^2$)
* $\frac{1}{x} = 3$ (has $x$ in denominator)

The solution of an equation is the value of the variable that makes both sides equal. For $x + 7 = 12$, the solution is $x = 5$ because $5 + 7$ does equal $12$.

Imagine a see-saw in perfect balance. On one side sits $x + 7$, on the other sits $12$. Because they are equal, the see-saw is level. Now here is the golden rule of equation solving: whatever you do to one side, you must do to the other, otherwise the see-saw tilts.

You can add the same thing to both sides, subtract the same thing from both sides, multiply both sides by the same number or divide both sides by the same non zero number. The see-saw stays balanced and the solution does not change.

To solve $x + 7 = 12$ using the balance method, we want $x$ alone on one side. So we subtract $7$ from both sides.

Done. This method is extremely safe for beginners because it makes every step visible. Selina uses it in the first few pages of the chapter before moving to the faster transposition method.

Some equations have the variable appearing on both sides of the equal sign. The trick is to first bring all variable terms to one side and all constants to the other side, then solve as usual.

Worked example 1:
Solve $5x - 3 = 2x + 9$.

Step 1: Move $2x$ from the right side to the left. It becomes $-2x$.

Step 2: Move $-3$ from the left side to the right. It becomes $+3$.

Step 3: Simplify both sides.

Step 4: Divide both sides by $3$.

Check: Put $x = 4$ back into the original equation. Left side: $5(4) - 3 = 17$. Right side: $2(4) + 9 = 17$. Both sides match, so the answer is correct.

Always check your answer. It takes ten seconds and saves you the heartbreak of a wrong mark on a silly arithmetic error.

Selina Exercise 12C and beyond loves to throw brackets and fractions at you. The strategy is always the same: open the brackets first, clear the fractions second, then transpose.

Worked example 2:
Solve $3(2x - 5) = 4x + 1$.

Step 1: Expand the bracket on the left.

Step 2: Transpose.

Step 3: Simplify.

Step 4: Divide.

Worked example 3 (with fractions):
Solve $\frac{x}{2} + \frac{x}{3} = 10$.

Step 1: Find the LCM of the denominators. LCM of $2$ and $3$ is $6$. Multiply every term on both sides by $6$ to clear fractions.

Step 2: Simplify.

Step 3: Divide.

The LCM trick is a lifesaver. Never try to add fractions term by term in an equation, it just creates more work.

Every word problem in this chapter can be cracked with four steps:

1. Read twice. Once for the story, once for the numbers.
2. Name the unknown. Assign a letter to the thing you are asked to find.
3. Write the equation. Convert the story into a mathematical sentence.
4. Solve and answer in words. Find the value of the letter, then write your final answer as a full sentence.

Worked example 4 (Age problem):
Ravi is three times as old as his son Arjun. Six years ago, Ravi was five times as old as Arjun. Find their present ages.

Step 1: Read twice.

Step 2: Let Arjun's present age be $x$ years. Then Ravi's present age is $3x$ years.

Step 3: Six years ago, Arjun was $x - 6$ and Ravi was $3x - 6$. The problem says Ravi was five times Arjun at that time.

Step 4: Solve.

So Arjun is $12$ years old and Ravi is $3 \times 12 = 36$ years old.

Check: Six years ago Arjun was $6$ and Ravi was $30$. Is $30 = 5 \times 6$? Yes. Correct.

Worked example 5 (Number problem):
Five more than twice a number equals three less than four times the same number. Find the number.

Step 1: Let the number be $x$.

Step 2: 'Five more than twice a number' is $2x + 5$. 'Three less than four times the number' is $4x - 3$. These are equal.

Step 3:

The number is $4$.

Worked example 6 (Perimeter problem):
The length of a rectangle is $5$ cm more than its breadth. If the perimeter is $50$ cm, find the length and breadth.

Step 1: Let breadth $= x$ cm. Then length $= x + 5$ cm.

Step 2: Perimeter $= 2(\text{length} + \text{breadth}) = 50$.

Step 3: Solve.

Breadth is $10$ cm, length is $15$ cm.

Check: Perimeter $= 2(15 + 10) = 50$ cm. Correct.

Watch for these errors, they are the top six reasons students lose marks in school exams on this chapter.

1. Forgetting to change sign during transposition. Writing $x + 7 = 12$ becomes $x = 12 + 7 = 19$. The $7$ should become $-7$ when it crosses over.

2. Dividing only one term. Writing $2x + 4 = 10$ becomes $x + 4 = 5$. When dividing, you must divide every term on both sides, not just one.

3. Opening brackets incorrectly. Writing $3(2x - 5) = 6x - 5$. The $3$ must multiply both terms inside, so it is $6x - 15$.

4. Losing the negative sign. Solving $-2x = 10$ and writing $x = 5$. The correct answer is $x = -5$.

5. Skipping the verification. Not checking the answer by substituting back. This is the cheapest insurance against silly mistakes.

6. Wrong translation in word problems. Writing 'five more than twice a number' as $5 + 2 + x$ instead of $2x + 5$. Read the phrase slowly, identify 'twice a number' first, then the 'five more' part.

* A linear equation in one variable has the form $ax + b = c$ where the variable power is $1$.
* The balance method keeps both sides equal by performing the same operation on both.
* Transposition is a shortcut where terms crossing the equal sign flip their operation.
* For equations with brackets, open brackets first. For fractions, multiply by the LCM to clear them.
* For word problems follow the four step method: read, name the unknown, write the equation, solve and answer in words.
* Always verify your answer by substituting it back into the original equation.
* Practice word problems daily. Translation from English to maths is a skill built through repetition.