Simple Equations — Solving & Word Problems

Imagine a weighing balance. When both sides carry the same weight, the balance stays level. If we add something to one side, the balance tips. To make it level again, we must add the same thing to the other side too.

That’s exactly how equations work — and once you understand that idea, solving any equation becomes straightforward.

What Is an Equation?

An equation is a mathematical statement that says two expressions are equal.

3 + 4 = 7 is an equation. x + 5 = 12 is also an equation — one side has an unknown!

Every equation has:

A Left Hand Side (LHS) — the expression on the left of the equals sign
A Right Hand Side (RHS) — the expression on the right
An equals sign (=) in the middle

In a simple equation, there is exactly one variable (like x, y, or n) and the variable has a power of 1 (we don’t have x² or x³ here).

Examples of simple equations:

x + 7 = 10
2y - 3 = 11
4n = 20
3m + 1 = 7

What Is a Variable?

A variable is a letter that stands for an unknown number we need to find.

Think of it as a mystery box. We know: “mystery box + 5 = 12”. Our job is to figure out what’s inside the box.

In the equation x + 5 = 12, the variable x is the mystery number. When we find x = 7, we’ve “solved” the equation.

Any letter can be a variable — x, y, n, m, a, b. By convention, we often use x, but your textbook may use different letters. Treat them all the same way.

Setting Up an Equation

Before solving, we often need to form an equation from a word problem. This is an important skill.

Step 1: Identify the unknown. Give it a variable name. Step 2: Read the condition given and write it as an equation.

Example: “A number increased by 8 gives 15.”

Unknown = some number → call it x “Increased by 8” means we add 8 → x + 8 “Gives 15” means equals 15 → x + 8 = 15

Example: “Three times a number minus 2 is equal to 10.”

Let the number = n Three times the number = 3n Minus 2 = 3n - 2 Equal to 10 → 3n - 2 = 10

Method 1: Trial and Error (for simple ones)

Sometimes we can guess the answer and check.

Equation: x + 3 = 8

Try x = 4: 4 + 3 = 7 ≠ 8 (nope) Try x = 5: 5 + 3 = 8 ✓ (yes!)

So x = 5.

Trial and error works for simple equations, but it’s slow and unreliable for harder ones. We need a better method.

Method 2: The Balancing Method

Think of the equation as a balance scale. As long as we do the same thing to both sides, the balance stays equal.

Equation: x + 5 = 12

We want x alone on the left. To remove the +5, we subtract 5 from both sides:

x + 5 - 5 = 12 - 5 x = 7

Equation: 3x = 18

To find x, divide both sides by 3:

3x ÷ 3 = 18 ÷ 3 x = 6

Equation: x - 4 = 9

Add 4 to both sides:

x - 4 + 4 = 9 + 4 x = 13

Whatever operation you do to one side, you MUST do the same to the other side. This is the golden rule of equations. Never break it!

Method 3: Transposing

Transposing is a shortcut version of the balancing method. When a term moves from one side to the other, its sign changes.

The rule: A term that changes sides also changes its sign.

Addition becomes subtraction
Subtraction becomes addition
Multiplication becomes division
Division becomes multiplication

Example: x + 5 = 12

Move +5 to the right side. It becomes -5:

x = 12 - 5 = 7

Example: x - 4 = 9

Move -4 to the right side. It becomes +4:

x = 9 + 4 = 13

Example: 3x = 18

Move ×3 to the right side. It becomes ÷3:

x = 18 ÷ 3 = 6

Example: 3x - 2 = 13 (two-step equation)

Step 1: Move -2 to the right (becomes +2):

3x = 13 + 2 = 15

Step 2: Move ×3 to the right (becomes ÷3):

x = 15 ÷ 3 = 5

Moving a term across the = sign:

(+a) becomes (−a)
(−a) becomes (+a)
(×a) becomes (÷a)
(÷a) becomes (×a)

Equations with Brackets

When the equation has brackets, expand them first.

Example: 2(x + 3) = 14

Expand: 2x + 6 = 14

Now transpose:

2x = 14 - 6 = 8 x = 8 ÷ 2 = 4

Alternative: Since both sides are divisible by 2, divide first:

x + 3 = 7 x = 7 - 3 = 4

Both methods give the same answer.

Example: 3(2x - 1) = 15

Expand: 6x - 3 = 15

6x = 15 + 3 = 18 x = 18 ÷ 6 = 3

Always Verify Your Answer

After solving, always substitute your answer back into the original equation to check.

Example: We found x = 7 for x + 5 = 12.

Check: LHS = 7 + 5 = 12 = RHS ✓

Example: We found x = 5 for 3x - 2 = 13.

Check: LHS = 3(5) - 2 = 15 - 2 = 13 = RHS ✓

Many CBSE exam questions ask you to “solve and verify”. Always write the verification step — it can earn you 1 extra mark even if your main working has a small mistake.

Word Problems

Word problems are where equations really shine. Here’s a systematic approach.

Step-by-Step Method for Word Problems

Read the problem carefully.
Identify the unknown — give it a variable.
Write what you know as an equation.
Solve the equation.
Answer the question in a full sentence.
Check if the answer makes sense.

Type 1: Number Problems

Problem: “A number is 4 more than twice another number. The larger number is 14. Find the smaller number.”

Let the smaller number = x. Larger number = 2x + 4

Given: 2x + 4 = 14

2x = 10 x = 5

The smaller number is 5.

Check: 2(5) + 4 = 10 + 4 = 14 ✓

Type 2: Age Problems

Problem: “Ravi is 3 years older than Sita. Together their ages add up to 25. Find their ages.”

Let Sita’s age = x. Ravi’s age = x + 3.

Sum of ages: x + (x + 3) = 25

2x + 3 = 25 2x = 22 x = 11

Sita is 11, Ravi is 14.

Check: 11 + 14 = 25 ✓

Type 3: Perimeter Problems

Problem: “The perimeter of a rectangle is 40 cm. The length is 5 cm more than the breadth. Find the dimensions.”

Let breadth = x cm. Length = (x + 5) cm.

Perimeter = 2(length + breadth)

2(x + 5 + x) = 40 2(2x + 5) = 40 4x + 10 = 40 4x = 30 x = 7.5 cm

Breadth = 7.5 cm, Length = 12.5 cm.

Check: 2(12.5 + 7.5) = 2(20) = 40 ✓

5 Common Mistakes to Avoid

Mistake 1: Wrong sign when transposing

If the equation is x + 8 = 15, many students write x = 15 + 8 = 23. They forgot that +8 becomes -8 when moved to the other side. Correct: x = 15 - 8 = 7.

Mistake 2: Not applying the operation to the entire side

Wrong: 3x + 6 = 15 → 3x + 6 - 6 = 15 - 6 → 3x = 9? (This is actually correct!)

But watch this: 3x + 6 = 15 → divide by 3 → x + 6 = 5? (WRONG!) You must divide EVERY term on both sides: x + 2 = 5. Then x = 3.

Mistake 3: Forgetting to expand brackets

Wrong: 2(x + 3) = 14 → 2x + 3 = 14 Right: 2(x + 3) = 14 → 2x + 6 = 14

Every term inside the bracket gets multiplied by the number outside.

Mistake 4: Skipping verification

Many students solve correctly but skip checking. In exams, verification is often asked explicitly. Also, checking catches silly arithmetic errors.

Mistake 5: Naming two different unknowns with the same variable

In age problems, if Ravi’s age is x, don’t also call the perimeter x. Each unknown needs a different name, or express everything in terms of one variable.

Practice Questions

Question 1: Solve: x + 9 = 16

Transpose +9 to the right side (it becomes -9): x = 16 - 9 = 7

Verification: LHS = 7 + 9 = 16 = RHS ✓

Question 2: Solve: 2y = 26

Transpose ×2 to the right side (it becomes ÷2): y = 26 ÷ 2 = 13

Verification: LHS = 2(13) = 26 = RHS ✓

Question 3: Solve: 5n - 3 = 22

Step 1: Transpose -3 → +3 on the right: 5n = 22 + 3 = 25

Step 2: Transpose ×5 → ÷5 on the right: n = 25 ÷ 5 = 5

Verification: LHS = 5(5) - 3 = 25 - 3 = 22 = RHS ✓

Question 4: Solve: 3(x - 2) = 9

Expand brackets: 3x - 6 = 9

Transpose -6 → +6: 3x = 9 + 6 = 15

Transpose ×3 → ÷3: x = 15 ÷ 3 = 5

Verification: LHS = 3(5 - 2) = 3(3) = 9 = RHS ✓

Question 5: “A number decreased by 7 gives 13. Find the number.”

Let the number = x. “Decreased by 7 gives 13”: x - 7 = 13

Transpose: x = 13 + 7 = 20

The number is 20.

Verification: 20 - 7 = 13 ✓

Question 6: “The sum of two consecutive even numbers is 30. Find the numbers.”

Let the first even number = x. Next consecutive even number = x + 2.

Sum: x + (x + 2) = 30 2x + 2 = 30 2x = 28 x = 14

The two numbers are 14 and 16.

Verification: 14 + 16 = 30 ✓

Question 7: “A boy’s age is 5 less than twice his sister’s age. If the boy is 13, find the sister’s age.”

Let sister’s age = x. Boy’s age = 2x - 5.

Given: 2x - 5 = 13 2x = 18 x = 9

The sister is 9 years old.

Verification: 2(9) - 5 = 18 - 5 = 13 ✓

Question 8: “The perimeter of a square is 52 cm. Find the side of the square.”

Let the side = x cm. Perimeter of a square = 4x.

4x = 52 x = 52 ÷ 4 = 13 cm

Verification: 4(13) = 52 ✓

Frequently Asked Questions

Q1: What is the difference between an expression and an equation?

An expression has no equals sign — for example, 3x + 2 is an expression. An equation has an equals sign and says two expressions are equal — for example, 3x + 2 = 11. You can simplify an expression, but you solve an equation.

Q2: Can the value of the variable be a fraction or a negative number?

Yes! Not all equations give whole number answers. For example, 2x + 1 = 4 gives x = 1.5. And x + 10 = 3 gives x = -7. Both are perfectly valid solutions.

Q3: What if a variable appears on both sides?

Collect all variable terms on one side and all number terms on the other. For example: 5x + 3 = 2x + 12 → 5x - 2x = 12 - 3 → 3x = 9 → x = 3.

Q4: How do we know which operation to do first when transposing?

Think about what’s happening to the variable. First undo addition/subtraction (move constants to the other side), then undo multiplication/division. Work in the reverse order of BODMAS.

Q5: Is transposing and the balancing method the same thing?

They give the same result, but transposing is a shortcut. Balancing method explicitly shows you doing the same operation on both sides. Transposing shows the final result of that process. Either method is accepted in exams.

Q6: What does it mean when the equation has no solution or infinite solutions?

If you end up with something like 5 = 7 (a false statement), the equation has no solution. If you end up with 0 = 0 (always true), it has infinite solutions. But in Class 7, all equations you encounter will have exactly one solution.