How to Use This Page
1. Solve each problem on your own first — attempt every question before checking solutions. 2. Compare your method — if your answer differs, rework the problem step by step using the method shown. 3. Focus on the steps, not just the final answer — examiners award marks for correct working. 4. Practise the word problems — they form the bulk of the exam and require clear identification of Given, Formula, and Answer.
Key Concepts (Quick Recap)
- Algebraic expression: A combination of constants and variables connected by +, −, ×, ÷. Example: 3x² − 5y + 7.
- Term: A single part of an expression separated by + or −. In 3x² − 5y + 7, the terms are 3x², −5y, and 7.
- Like terms: Terms with the same variable raised to the same power. Examples: 7x and −3x are like terms; 5xy and −2xy are like terms.
- Unlike terms: Terms with different variables or different powers. Examples: 3x and 5y; 2x and 4x².
- Coefficient: The numerical factor of a term. In −7xy, the coefficient is −7.
- Constant: A term with no variable. In 3x + 5, the constant is 5.
- Addition of expressions: Group like terms, add their coefficients.
- Subtraction of expressions: Change the sign of every term in the expression being subtracted, then add.
- Simple equation: A statement that two expressions are equal. Solve by applying the same operation to both sides.
Exercise Solutions
Exercise 3.1
Q1. Fill in the blanks
(i) The variable in the expression 16x − 7 is _____
In 16x − 7, the only letter (unknown) is x.
Answer: x
(ii) The constant term of the expression 2y − 6 is _____
The constant term is the term without a variable.
2y has variable y, so the constant is −6.
Answer: −6
(iii) In the expression 25m + 14M, the type of the terms are _____ terms
25m has variable m (lowercase) and 14M has variable M (uppercase).
Since m ≠ M, these terms have different variables.
Answer: unlike terms
(iv) The number of terms in the expression 3ab + 4c − 9 is _____
The terms are: 3ab, 4c, −9
Count = 3
Answer: three
(v) The numerical coefficient of the term −xy is _____
−xy = (−1) × xy
Answer: −1
Q2. Say True or False
(i) x + (−x) = 0
x + (−x) = x − x = 0
Answer: True
(ii) The coefficient of ab in the term 15abc is 15.
15abc = 15c × ab
The coefficient of ab is 15c, not 15.
Answer: False
(iii) 2pq and −7qp are like terms.
pq and qp are the same variable product.
So 2pq and −7qp have the same variables with the same powers.
Answer: True
(iv) When y = −1, the value of the expression 2y − 1 is 3.
Substitute y = −1:
2(−1) − 1 = −2 − 1 = −3
−3 ≠ 3
Answer: False
Q3. Find the numerical coefficient of each of the following terms: −3yx, 12k, y, 121bc, −x, 9pq, 2ab
| Term | Numerical Coefficient |
|-------|----------------------|
| −3yx | −3 |
| 12k | 12 |
| y | 1 |
| 121bc | 121 |
| −x | −1 |
| 9pq | 9 |
| 2ab | 2 |
Q4. Write the variables, constants and terms of the following expressions
(i) 18 + x − y
Variables: x, y
Constants: 18
Terms: 18, x, −y
(ii) 7p − 4q + 5
Variables: p, q
Constants: 5
Terms: 7p, −4q, 5
(iii) 29x + 13y
Variables: x, y
Constants: None
Terms: 29x, 13y
(iv) b + 2
Variables: b
Constants: 2
Terms: b, 2
Q5. Identify the like terms among the following: 7x, 5y, −8x, 12y, 6z, z, −12x, −9y, 11z
Group by variable:
x-terms: 7x, −8x, −12x
y-terms: 5y, 12y, −9y
z-terms: 6z, z, 11z
These are the three groups of like terms.
Q6. If x = 2 and y = 3, find the value of the following expressions
(i) 2x − 3y
Substitute x = 2, y = 3:
2(2) − 3(3) = 4 − 9 = −5
Answer: −5
(ii) x + y
Substitute x = 2, y = 3:
2 + 3 = 5
Answer: 5
(iii) 4y − x
Substitute x = 2, y = 3:
4(3) − 2 = 12 − 2 = 10
Answer: 10
(iv) x + 1 − y
Substitute x = 2, y = 3:
2 + 1 − 3 = 3 − 3 = 0
Answer: 0
Objective Type Questions
Q1. An algebraic statement equivalent to "Three times the sum of x and y" is
Options: (i) 3(x + y) (ii) 3 + x + y (iii) 3x + y (iv) 3 + xy
"Three times the sum of x and y" = 3 × (x + y) = 3(x + y)
Answer: (i) 3(x + y)
Q2. The numerical coefficient of −7mn is
Options: (i) 7 (ii) −7 (iii) p (iv) −p
−7mn has numerical coefficient −7.
Answer: (ii) −7
Q3. Choose the pair of like terms
Options: (i) 7p, 7x (ii) 7r, 7x (iii) −4x, 4 (iv) −4x, 7x
Only (iv) has the same variable x: −4x and 7x are like terms.
Answer: (iv) −4x, 7x
Q4. The value of 7a − 4b when a = 3, b = 2 is
Options: (i) 21 (ii) 13 (iii) 8 (iv) 32
7(3) − 4(2) = 21 − 8 = 13
Answer: (ii) 13
Exercise 3.2
Q1. Fill in the blanks
(i) The addition of −7b and 2b is _____
−7b + 2b = (−7 + 2)b = −5b
Answer: −5b
(ii) The subtraction of 5m from −3m is _____
−3m − 5m = (−3 − 5)m = −8m
Answer: −8m
(iii) The additive inverse of −37xyz is _____
The additive inverse changes the sign:
−(−37xyz) = 37xyz
Answer: 37xyz
Q2. True or False
(i) The expressions 8x + 3y and 7x + 2y cannot be added.
They CAN be added by grouping like terms:
(8x + 7x) + (3y + 2y) = 15x + 5y
Answer: False
(ii) If x is a natural number, then x + 1 is its predecessor.
The predecessor of x is x − 1, not x + 1.
x + 1 is the successor.
Answer: False
(iii) Sum of a − b + c and −a + b − c is zero.
(a − b + c) + (−a + b − c)
= (a − a) + (−b + b) + (c − c)
= 0 + 0 + 0 = 0
Answer: True
Q3. Add
(i) 8x, 3x
8x + 3x = (8 + 3)x = 11x
Answer: 11x
(ii) 7mn, 5mn
7mn + 5mn = (7 + 5)mn = 12mn
Answer: 12mn
(iii) −9y, 11y, 2y
−9y + 11y + 2y = (−9 + 11 + 2)y = (2 + 2)y = 4y
Answer: 4y
Q4. Subtract
(i) 4k from 12k
12k − 4k = (12 − 4)k = 8k
Answer: 8k
(ii) 15q from 25q
25q − 15q = (25 − 15)q = 10q
Answer: 10q
(iii) 7xyz from 17xyz
17xyz − 7xyz = (17 − 7)xyz = 10xyz
Answer: 10xyz
Q5. Find the sum
(i) 7p + 6q, 5p − q, q + 16p
(7p + 6q) + (5p − q) + (q + 16p)
= 7p + 6q + 5p − q + q + 16p
Group p-terms: 7p + 5p + 16p = 28p
Group q-terms: 6q − q + q = 6q
Answer: 28p + 6q
(ii) a + 5b + 7c, 2a + 10b + 9c
(a + 5b + 7c) + (2a + 10b + 9c)
a-terms: a + 2a = 3a
b-terms: 5b + 10b = 15b
c-terms: 7c + 9c = 16c
Answer: 3a + 15b + 16c
(iii) mn + t, 2mn − 2t, −3t + 3mn
(mn + t) + (2mn − 2t) + (−3t + 3mn)
mn-terms: mn + 2mn + 3mn = 6mn
t-terms: t − 2t − 3t = (1 − 2 − 3)t = −4t
Answer: 6mn − 4t
(iv) u + v, u − v, 2u + 5v, 2u − 5v
(u + v) + (u − v) + (2u + 5v) + (2u − 5v)
u-terms: u + u + 2u + 2u = 6u
v-terms: v − v + 5v − 5v = 0
Answer: 6u
(v) 5xyz − 3xy, 3zxy − 5yx
Note: zxy = xyz and yx = xy
(5xyz − 3xy) + (3xyz − 5xy)
xyz-terms: 5xyz + 3xyz = 8xyz
xy-terms: −3xy − 5xy = −8xy
Answer: 8xyz − 8xy
Q6. Subtract
(i) 13x + 12y − 5 from 27x + 5y − 43
(27x + 5y − 43) − (13x + 12y − 5)
= 27x + 5y − 43 − 13x − 12y + 5
x-terms: 27x − 13x = 14x
y-terms: 5y − 12y = −7y
constants: −43 + 5 = −38
Answer: 14x − 7y − 38
(ii) 3p + 5 from p − 2q + 7
(p − 2q + 7) − (3p + 5)
= p − 2q + 7 − 3p − 5
p-terms: p − 3p = −2p
q-terms: −2q
constants: 7 − 5 = 2
Answer: −2p − 2q + 2
(iii) m + n from 3m − 7n
(3m − 7n) − (m + n)
= 3m − 7n − m − n
m-terms: 3m − m = 2m
n-terms: −7n − n = −8n
Answer: 2m − 8n
(iv) 2y + z from 6z − 5y
(6z − 5y) − (2y + z)
= 6z − 5y − 2y − z
z-terms: 6z − z = 5z
y-terms: −5y − 2y = −7y
Answer: 5z − 7y
Q7. Simplify
(i) (x + y − z) + (3x − 5y + 7z) − (14x + 7y − 6z)
Step 1: Remove brackets and distribute the minus sign:
= x + y − z + 3x − 5y + 7z − 14x − 7y + 6z
Step 2: Group like terms:
x-terms: x + 3x − 14x = (1 + 3 − 14)x = −10x
y-terms: y − 5y − 7y = (1 − 5 − 7)y = −11y
z-terms: −z + 7z + 6z = (−1 + 7 + 6)z = 12z
Answer: −10x − 11y + 12z
(ii) p + p + 2 + p + 3 + p − 4 − p − 5 + p + 10
Group p-terms: p + p + p − p + p = (1 + 1 + 1 − 1 + 1)p = 3p
Group constants: 2 + 3 − 4 − 5 + 10 = 6
Answer: 3p + 6
(iii) n + (m + 1) + (n + 2) + (m + 3) + (n + 4) + (m + 5)
Remove brackets:
= n + m + 1 + n + 2 + m + 3 + n + 4 + m + 5
n-terms: n + n + n = 3n
m-terms: m + m + m = 3m
constants: 1 + 2 + 3 + 4 + 5 = 15
Answer: 3m + 3n + 15
Objective Type Questions
Q8. The addition of 3mn, −5mn, 8mn and −4mn is
Options: (i) mn (ii) −mn (iii) 2mn (iv) 3mn
3mn + (−5mn) + 8mn + (−4mn)
= (3 − 5 + 8 − 4)mn
= (3 + 8) − (5 + 4) = 11 − 9 = 2
= 2mn
Answer: (iii) 2mn
Q9. When we subtract a from −a, we get _____
Options: (i) a (ii) 2a (iii) −2a (iv) −a
−a − a = −2a
Answer: (iii) −2a
Q10. In an expression, we can add or subtract only _____
Options: (i) like terms (ii) unlike terms (iii) all terms (iv) None
Only like terms (same variable, same power) can be combined.
Answer: (i) like terms
Exercise 3.3
Q1. Fill in the blanks
(i) An expression equated to another expression is called _____
When two expressions are set equal to each other, it forms an equation.
Answer: an equation
(ii) If a = 5, the value of 2a + 5 is _____
Substitute a = 5:
2(5) + 5 = 10 + 5 = 15
Answer: 15
(iii) The sum of twice and four times of the variable x is _____
Twice x = 2x
Four times x = 4x
Sum = 2x + 4x = 6x
Answer: 6x
Q2. Say True or False
(i) Every algebraic expression is an equation.
An equation requires two expressions set equal (e.g., 2x = 10).
An algebraic expression like 3x + 5 is NOT an equation.
Answer: False
(ii) The expression 7x + 1 cannot be reduced without knowing the value of x.
7x + 1 has no like terms to combine, so it cannot be simplified further.
Answer: True
(iii) To add two like terms, its coefficients can be added.
Like terms share the same variable, so we only add coefficients.
Example: 5x + 3x = (5 + 3)x = 8x
Answer: True
Q3. Solve
(i) x + 5 = 8
x + 5 = 8
Subtract 5 from both sides:
x + 5 − 5 = 8 − 5
x = 3
Answer: x = 3
(ii) p − 3 = 7
p − 3 = 7
Add 3 to both sides:
p − 3 + 3 = 7 + 3
p = 10
Answer: p = 10
(iii) 2x = 30
2x = 30
Divide both sides by 2:
2x / 2 = 30 / 2
x = 15
Answer: x = 15
(iv) m/6 = 5
m / 6 = 5
Multiply both sides by 6:
(m / 6) × 6 = 5 × 6
m = 30
Answer: m = 30
(v) 7x + 10 = 80
7x + 10 = 80
Subtract 10 from both sides:
7x + 10 − 10 = 80 − 10
7x = 70
Divide both sides by 7:
7x / 7 = 70 / 7
x = 10
Answer: x = 10
Q4. What should be added to 3x + 6y to get 5x + 8y?
We need: (3x + 6y) + ? = 5x + 8y
So ? = (5x + 8y) − (3x + 6y)
= 5x + 8y − 3x − 6y
= (5 − 3)x + (8 − 6)y
= 2x + 2y
Answer: 2x + 2y should be added.
Q5. Nine added to thrice a whole number gives 45. Find the number.
Given: Nine added to thrice a whole number = 45
Let the number be x.
Thrice the number = 3x
Equation: 3x + 9 = 45
Subtract 9 from both sides:
3x + 9 − 9 = 45 − 9
3x = 36
Divide both sides by 3:
3x / 3 = 36 / 3
x = 12
Answer: The whole number is 12.
Q6. Find the two consecutive odd numbers whose sum is 200.
Let the two consecutive odd numbers be x and x + 2.
Given: x + (x + 2) = 200
x + x + 2 = 200
2x + 2 = 200
Subtract 2 from both sides:
2x = 198
Divide both sides by 2:
x = 99
The other number = 99 + 2 = 101
Answer: The two consecutive odd numbers are 99 and 101.
Q7. The taxi charges in a city comprise a fixed charge of Rs. 100 for 5 kms and Rs. 16 per km for every additional km. If the amount paid at the end of the trip was Rs. 740, find the distance traveled.
Given:
Fixed charge for first 5 km = Rs. 100
Additional charge = Rs. 16 per km beyond 5 km
Total amount paid = Rs. 740
Let total distance = x km
Additional distance = (x − 5) km
Equation:
100 + 16(x − 5) = 740
Subtract 100 from both sides:
16(x − 5) = 640
Divide both sides by 16:
x − 5 = 40
Add 5 to both sides:
x = 45
Answer: Total distance traveled = 45 km
Objective Type Questions
Q8. The generalization of the number pattern 3, 6, 9, 12, ... is
Options: (i) n (ii) 2n (iii) 3n (iv) 4n
Each term is a multiple of 3:
3 = 3 × 1, 6 = 3 × 2, 9 = 3 × 3, 12 = 3 × 4
General form: 3n
Answer: (iii) 3n
Q9. The solution of 3x + 5 = x + 9 is
Options: (i) 2 (ii) 3 (iii) 5 (iv) 4
3x + 5 = x + 9
Subtract x from both sides:
3x − x + 5 = 9
2x + 5 = 9
Subtract 5 from both sides:
2x = 4
Divide by 2:
x = 2
Answer: (i) 2
Q10. The equation y + 1 = 0 is true only when y is
Options: (i) 0 (ii) −1 (iii) 1 (iv) −2
y + 1 = 0
y = −1
Answer: (ii) −1
Exercise 3.4
Miscellaneous Practice Problems
Q1. Subtract −3ab − 8 from 3ab + 8. Also subtract 3ab + 8 from −3ab − 8.
Answer
Part (i): Subtract −3ab − 8 from 3ab + 8
(3ab + 8) − (−3ab − 8)
= 3ab + 8 + 3ab + 8
= (3 + 3)ab + (8 + 8)
= 6ab + 16
Answer: 6ab + 16
Part (ii): Subtract 3ab + 8 from −3ab − 8
(−3ab − 8) − (3ab + 8)
= −3ab − 8 − 3ab − 8
= (−3 − 3)ab + (−8 − 8)
= −6ab − 16
Answer: −6ab − 16
Q2. Find the perimeter of a triangle whose sides are x + 3y, 2x + y, x − y.
Perimeter = Sum of all three sides
= (x + 3y) + (2x + y) + (x − y)
= x + 3y + 2x + y + x − y
x-terms: x + 2x + x = 4x
y-terms: 3y + y − y = 3y
Answer: Perimeter = 4x + 3y
Q3. Thrice a number when increased by 5 gives 44. Find the number.
Given: Thrice a number increased by 5 = 44
Let the number be x.
Equation: 3x + 5 = 44
Subtract 5 from both sides:
3x + 5 − 5 = 44 − 5
3x = 39
Divide both sides by 3:
3x / 3 = 39 / 3
x = 13
Answer: The number is 13.
Q4. How much smaller is 2ab + 4b − c than 5ab − 3b + 2c?
Difference = (5ab − 3b + 2c) − (2ab + 4b − c)
= 5ab − 3b + 2c − 2ab − 4b + c
ab-terms: 5ab − 2ab = 3ab
b-terms: −3b − 4b = −7b
c-terms: 2c + c = 3c
Answer: 2ab + 4b − c is smaller by 3ab − 7b + 3c
Q5. Six times a number subtracted from 40 gives −8. Find the number.
Given: 40 minus six times a number = −8
Let the number be x.
Equation: 40 − 6x = −8
Subtract 40 from both sides:
−6x = −8 − 40
−6x = −48
Divide both sides by −6:
−6x / (−6) = −48 / (−6)
x = 8
Answer: The number is 8.
Challenge Problems
Q6. From the sum of 5x + 7y − 12 and 3x − 5y + 2, subtract the sum of 2x − 7y − 1 and −6x + 3y + 9.
Step 1: Find the first sum.
(5x + 7y − 12) + (3x − 5y + 2)
= (5 + 3)x + (7 − 5)y + (−12 + 2)
= 8x + 2y − 10
Step 2: Find the second sum.
(2x − 7y − 1) + (−6x + 3y + 9)
= (2 − 6)x + (−7 + 3)y + (−1 + 9)
= −4x − 4y + 8
Step 3: Subtract the second sum from the first.
(8x + 2y − 10) − (−4x − 4y + 8)
= 8x + 2y − 10 + 4x + 4y − 8
= (8 + 4)x + (2 + 4)y + (−10 − 8)
= 12x + 6y − 18
Answer: 12x + 6y − 18
Q7. Find the expression to be added with 5a − 3b + 2c to get a − 4b − 2c.
We need: (5a − 3b + 2c) + ? = a − 4b − 2c
So ? = (a − 4b − 2c) − (5a − 3b + 2c)
= a − 4b − 2c − 5a + 3b − 2c
a-terms: a − 5a = −4a
b-terms: −4b + 3b = −b
c-terms: −2c − 2c = −4c
Answer: −4a − b − 4c should be added.
Q8. What should be subtracted from 2m + 8n + 10 to get −3m + 7n + 16?
We need: (2m + 8n + 10) − ? = −3m + 7n + 16
So ? = (2m + 8n + 10) − (−3m + 7n + 16)
= 2m + 8n + 10 + 3m − 7n − 16
m-terms: 2m + 3m = 5m
n-terms: 8n − 7n = n
constants: 10 − 16 = −6
Answer: 5m + n − 6 should be subtracted.
Q9. Give an algebraic equation: "The difference between the area and perimeter of a rectangle is 20."
Let length = l, breadth = b
Area of rectangle = l × b = lb
Perimeter of rectangle = 2(l + b)
Difference = Area − Perimeter = 20
Answer: lb − 2(l + b) = 20
Q10. Add: 2a + b + 3c and a + 1/3 b + 2/5 c
(2a + b + 3c) + (a + 1/3 b + 2/5 c)
a-terms: 2a + a = 3a
b-terms: b + 1/3 b = (3/3 + 1/3)b = 4/3 b
c-terms: 3c + 2/5 c = (15/5 + 2/5)c = 17/5 c
Answer: 3a + 4/3 b + 17/5 c
Common Mistakes
1. Combining unlike terms: Never add 3x and 5y together. Only like terms (same variable, same power) can be combined. 3x + 5y stays as 3x + 5y.
2. Sign errors in subtraction: When subtracting an expression, every term changes sign. Subtracting (2x − 3) means adding (−2x + 3), not (−2x − 3).
3. Forgetting the coefficient of 1 or −1: In the term "x", the coefficient is 1 (not 0). In "−y", the coefficient is −1. Always account for these.
4. Wrong equation-solving order: Always isolate the variable step by step. First remove constants (add/subtract), then remove coefficients (multiply/divide). Do not skip steps.
5. Confusing successor and predecessor: x + 1 is the successor of x; x − 1 is the predecessor. Mixing these up loses marks in objective questions.
6. Dropping negative signs mid-calculation: In problems like 40 − 6x = −8, students sometimes write −6x = −48 and then forget the second negative, getting x = −8 instead of x = 8. Track every sign carefully.
7. Not distributing the subtraction sign to all terms: (a + b) − (c + d) = a + b − c − d. A common error is writing only a + b − c + d, which misses the sign change on d.
Exam Checklist
- [ ] I can identify variables, constants, coefficients, and terms in any algebraic expression.
- [ ] I can distinguish like terms from unlike terms.
- [ ] I can add algebraic expressions by grouping like terms and adding coefficients.
- [ ] I can subtract algebraic expressions by changing all signs of the subtrahend and then adding.
- [ ] I can simplify expressions involving both addition and subtraction.
- [ ] I can solve simple equations using the balance method (same operation on both sides).
- [ ] I can translate word problems into algebraic equations and solve them.
- [ ] I can find what should be added or subtracted to obtain a given expression.
- [ ] I correctly handle negative signs throughout multi-step problems.
- [ ] I write the final answer clearly with proper units for word problems.
