Finding the Unknown

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CLASS VII Mathematics ~7 marks/year Ch 15 of 15
Finding the Unknown

Class 7 · Mathematics · NCERT chapter notes · Akanksha Classes

Snapshot
  • This chapter introduces simple linear equations — sentences in maths where a letter (variable) stands for an unknown number we must find.
  • An equation says two expressions are equal, like $2x+3=11$; solving it means finding the value of the variable that makes it true.
  • The master idea is balance: an equation is like a weighing balance — whatever you do to one side you must do to the other.
  • You learn to solve by doing the same operation to both sides and by the shortcut of transposing (moving a term across the $=$ sign and changing its sign).
  • You turn word problems into equations — the real power of algebra.
  • Every solution can be checked by substituting it back into the original equation.
  • Weightage: ~7 marks/year — solving equations (2–3 marks) and forming-and-solving word problems (3–4 marks).
Detailed Notes

1. Variables, expressions and equations

A variable is a letter, like $x$, $y$ or $m$, that stands for an unknown number. An expression combines variables and numbers with operations, such as $2x+3$ or $\dfrac{y}{4}-1$. An equation places an $=$ sign between two expressions, claiming they are equal: $2x+3=11$.

An equation has a left-hand side (LHS), an $=$ sign, and a right-hand side (RHS). It is true only for special value(s) of the variable.

The value of the variable that makes LHS $=$ RHS is called the solution or root of the equation. For $2x+3=11$, the solution is $x=4$, because $2(4)+3=11$.

2. The balance idea

Think of an equation as a weighing balance that is perfectly level. The two pans hold the LHS and RHS, which weigh the same. To keep it level while you simplify, you must treat both pans identically.

Golden rule: you may add, subtract, multiply or divide both sides of an equation by the same number (not $0$ for division), and it stays balanced.
Worked example — keeping the balance

Solve $x+5=12$. Subtract $5$ from both sides: $x+5-5=12-5$, so $x=7$. Check: $7+5=12$. ✓

3. Solving one-step equations

A one-step equation needs a single inverse operation to free the variable. The inverse of $+$ is $-$, and the inverse of $\times$ is $\div$.

Worked examples — one step each

(i) $x-4=9\Rightarrow x=9+4=13$ (add $4$ to both sides).

(ii) $3x=21\Rightarrow x=\dfrac{21}{3}=7$ (divide both sides by $3$).

(iii) $\dfrac{y}{5}=4\Rightarrow y=4\times5=20$ (multiply both sides by $5$).

4. Solving two-step equations

Most equations need two steps. Undo addition/subtraction first, then undo multiplication/division (the reverse of the usual order of operations).

Worked example — $2x+3=11$

Step 1: subtract $3$ from both sides → $2x=8$.

Step 2: divide both sides by $2$ → $x=4$.

Check: $2(4)+3=8+3=11$. ✓

Worked example — $\dfrac{m}{4}-2=3$

Step 1: add $2$ to both sides → $\dfrac{m}{4}=5$.

Step 2: multiply both sides by $4$ → $m=20$.

Check: $\dfrac{20}{4}-2=5-2=3$. ✓

5. The transposition shortcut

Transposing means moving a term from one side of the $=$ to the other, while changing its sign. It is just a quick way of doing the same operation to both sides.

When a term crosses the $=$ sign: $+$ becomes $-$, $-$ becomes $+$, $\times$ becomes $\div$, and $\div$ becomes $\times$.
Worked example — transposing

Solve $5x-7=18$. Transpose $-7$ to the right (it becomes $+7$): $5x=18+7=25$. Transpose the $\times5$ (it becomes $\div5$): $x=\dfrac{25}{5}=5$. Check: $5(5)-7=18$. ✓

6. Variables on both sides

When the variable appears on both sides, gather all variable terms on one side and all numbers on the other by transposing.

Worked example — $4x+5=2x+13$

Transpose $2x$ to the left: $4x-2x+5=13\Rightarrow 2x+5=13$. Transpose $5$: $2x=8$. Divide by $2$: $x=4$. Check: LHS $=4(4)+5=21$; RHS $=2(4)+13=21$. ✓

7. Equations with brackets

First open the brackets using the distributive rule $a(b+c)=ab+ac$, then solve as usual.

Worked example — $3(x-2)=12$

Open brackets: $3x-6=12$. Transpose $-6$: $3x=18$. Divide by $3$: $x=6$. (Or simply divide both sides by $3$ first: $x-2=4\Rightarrow x=6$.) Check: $3(6-2)=3\times4=12$. ✓

8. Turning words into equations

This is the heart of the chapter. Follow four steps:

  • Step 1 — Let: choose a variable for the unknown (e.g. let the number be $x$).
  • Step 2 — Translate: rewrite the sentence as an equation.
  • Step 3 — Solve: find the variable.
  • Step 4 — Check & answer: verify and write the answer in words.
Words Maths
a number increased by 5$x+5$
7 less than a number$x-7$
thrice a number$3x$
half of a number$\dfrac{x}{2}$
is / equals$=$

9. Word problems — fully solved

Worked example — think of a number

"Three times a number, increased by $4$, is $25$. Find the number." Let the number be $x$. Equation: $3x+4=25$. Solve: $3x=21\Rightarrow x=7$. Check: $3(7)+4=25$. ✓ The number is $7$.

Worked example — ages

"Rahul is $5$ years older than Sita. The sum of their ages is $27$. Find their ages." Let Sita be $x$ years; Rahul is $x+5$. Equation: $x+(x+5)=27\Rightarrow 2x+5=27\Rightarrow 2x=22\Rightarrow x=11$. So Sita is $11$ and Rahul is $16$. Check: $11+16=27$. ✓

Worked example — consecutive numbers

"The sum of two consecutive numbers is $47$." Let them be $x$ and $x+1$. Then $x+(x+1)=47\Rightarrow 2x+1=47\Rightarrow 2x=46\Rightarrow x=23$. The numbers are $23$ and $24$.

Worked example — money

"Anya has ₹$x$. She spends ₹$30$ and is left with ₹$70$." Equation: $x-30=70\Rightarrow x=100$. She had ₹$100$.

Worked example — perimeter

"The length of a rectangle is $3$ cm more than its breadth, and its perimeter is $26$ cm." Let breadth $=b$; length $=b+3$. Perimeter $=2(\text{length}+\text{breadth})$, so $2(b+3+b)=26\Rightarrow 2(2b+3)=26\Rightarrow 4b+6=26\Rightarrow 4b=20\Rightarrow b=5$. Breadth $=5$ cm, length $=8$ cm.

10. Checking your solution

Always substitute the answer back into the original equation. If LHS $=$ RHS, your solution is correct. This single habit catches almost every careless error.

Worked example — verifying

For $5x-7=18$ we found $x=5$. Substitute: LHS $=5(5)-7=25-7=18=$ RHS. ✓ The solution is verified.

11. Common mistakes to avoid

  • Forgetting to change the sign when transposing — a $+$ must become $-$ on crossing the $=$.
  • Operating on only one side — the balance breaks if you do not treat both sides alike.
  • Dividing by the coefficient before moving the constant in a two-step equation — undo $+/-$ first.
  • Not opening brackets correctly: $3(x-2)=3x-6$, not $3x-2$.
  • Skipping the check — a quick substitution confirms the answer.

12. Quick revision checklist

  • Equation = two equal expressions; solution = value making LHS $=$ RHS.
  • Do the same operation to both sides (balance rule).
  • Transpose: move a term across $=$ and flip its sign ($+\leftrightarrow-$, $\times\leftrightarrow\div$).
  • Two-step: undo $+/-$ first, then $\times/\div$.
  • Word problems: Let → Translate → Solve → Check.
  • Always verify by substitution.
Practice MCQs
1. The solution of $x+7=15$ is:
  1. $22$
  2. $8$
  3. $7$
  4. $15$
Answer: (B) Subtract $7$: $x=15-7=8$.
2. The solution of $4x=36$ is:
  1. $9$
  2. $32$
  3. $40$
  4. $144$
Answer: (A) Divide by $4$: $x=9$.
3. To solve $2x+3=11$, the first step is to:
  1. divide both sides by $2$
  2. subtract $3$ from both sides
  3. multiply both sides by $2$
  4. add $3$ to both sides
Answer: (B) Undo addition/subtraction before multiplication/division.
4. The solution of $\dfrac{y}{3}=6$ is:
  1. $2$
  2. $9$
  3. $18$
  4. $3$
Answer: (C) Multiply both sides by $3$: $y=18$.
5. When a $+5$ is transposed across the $=$ sign, it becomes:
  1. $+5$
  2. $-5$
  3. $\times5$
  4. $\div5$
Answer: (B) Transposing changes the sign: $+5\to-5$.
6. The solution of $5x-7=18$ is:
  1. $3$
  2. $4$
  3. $5$
  4. $11$
Answer: (C) $5x=25\Rightarrow x=5$.
7. Solve $4x+5=2x+13$.
  1. $x=2$
  2. $x=3$
  3. $x=4$
  4. $x=9$
Answer: (C) $2x=8\Rightarrow x=4$.
8. "Three times a number plus $4$ equals $25$" becomes:
  1. $3x-4=25$
  2. $3+x=25$
  3. $3x+4=25$
  4. $x+4=25$
Answer: (C) Thrice the number is $3x$, increased by $4$, equals $25$.
9. Opening the brackets, $3(x-2)$ equals:
  1. $3x-2$
  2. $3x-6$
  3. $x-6$
  4. $3x+6$
Answer: (B) Distribute: $3\times x-3\times2=3x-6$.
10. The sum of two consecutive numbers is $47$. The smaller number is:
  1. $22$
  2. $23$
  3. $24$
  4. $25$
Answer: (B) $x+(x+1)=47\Rightarrow x=23$.
11. A number, when divided by $4$ and then reduced by $2$, gives $3$. The number is:
  1. $12$
  2. $16$
  3. $20$
  4. $24$
Answer: (C) $\dfrac{x}{4}-2=3\Rightarrow\dfrac{x}{4}=5\Rightarrow x=20$.
12. The value of the variable that makes an equation true is called its:
  1. coefficient
  2. solution (root)
  3. constant
  4. expression
Answer: (B) It is the solution, also called the root.
13. Rahul is $5$ years older than Sita and their ages add to $27$. Sita's age is:
  1. $10$
  2. $11$
  3. $16$
  4. $22$
Answer: (B) $x+(x+5)=27\Rightarrow 2x=22\Rightarrow x=11$.
14. To keep an equation balanced, any operation you do to the LHS you must also do to the:
  1. variable only
  2. RHS
  3. constant only
  4. coefficient only
Answer: (B) Both sides must be treated identically.
15. The solution of $\dfrac{x}{2}+3=8$ is:
  1. $5$
  2. $10$
  3. $16$
  4. $22$
Answer: (B) $\dfrac{x}{2}=5\Rightarrow x=10$.
Important Questions
Q1. Solve $2x+3=11$ showing each step, and verify your answer. (3 marks)
Answer: Subtract $3$ from both sides: $2x=8$. Divide both sides by $2$: $x=4$. Verify: $2(4)+3=8+3=11=$ RHS. ✓
Q2. Explain what "transposing" means and use it to solve $5x-7=18$. (3 marks)
Answer: Transposing means moving a term across the $=$ sign and changing its sign. Transpose $-7$: $5x=18+7=25$. Transpose $\times5$ as $\div5$: $x=\dfrac{25}{5}=5$. Check: $5(5)-7=18$. ✓
Q3. The sum of three consecutive numbers is $72$. Form an equation and find the numbers. (3 marks)
Answer: Let them be $x$, $x+1$, $x+2$. Then $x+(x+1)+(x+2)=72\Rightarrow 3x+3=72\Rightarrow 3x=69\Rightarrow x=23$. The numbers are $23$, $24$ and $25$.
Q4. The length of a rectangle is $3$ cm more than its breadth and its perimeter is $26$ cm. Find its dimensions. (3 marks)
Answer: Let breadth $=b$, length $=b+3$. Perimeter $=2(b+3+b)=26\Rightarrow 4b+6=26\Rightarrow 4b=20\Rightarrow b=5$. Breadth $=5$ cm, length $=8$ cm.
Q5. Solve $4x+5=2x+13$ and explain why the variable terms are gathered on one side. (2 marks)
Answer: Transpose $2x$ to the left: $4x-2x+5=13\Rightarrow 2x+5=13\Rightarrow 2x=8\Rightarrow x=4$. We gather variable terms together so the unknown appears once, making it easy to isolate.
Q6. A number multiplied by $3$ and then increased by $7$ gives $34$. Find the number. (2 marks)
Answer: Let the number be $x$. Then $3x+7=34\Rightarrow 3x=27\Rightarrow x=9$. The number is $9$.
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Large Numbers Around Us · Arithmetic Expressions · A Peek Beyond the Point · Expressions using Letter-Numbers · Parallel and Intersecting Lines · Number Play · A Tale of Three Intersecting Lines · Working with Fractions
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