Algebraic Expressions and Identities

This chapter introduces the fundamental concepts of algebra.

Key Topics:

• Algebraic Expressions: Combinations of variables (like x, y, z) and constants (like 2, -5) connected by mathematical operations (addition, subtraction, multiplication, division).
• Terms: Individual components of an algebraic expression separated by + or – signs.
• Factors: Numbers or variables multiplied together to form a term.
• Coefficients: Numerical factors of a term.
• Like and Unlike Terms: Terms with the same or different literal factors.
• Monomials, Binomials, Trinomials, and Polynomials: Classifications based on the number of terms.
• Addition and Subtraction of Algebraic Expressions: Combining like terms.
• Multiplication of Algebraic Expressions: Using the distributive law.
• Identities: Equations that hold true for all values of the variables involved. Common identities include:
  • (a + b)² = a² + 2ab + b²
  • (a – b)² = a² – 2ab + b²
  • (a + b)(a – b) = a² – b²

Exercise 8.1

1. Add the following:

(i) ab – bc, bc – ca, ca – ab

(ii) a – b + ab, b – c + bc, c – a + ac

(iii) 2p2q2 – 3pq + 4, 5 + 7pq – 3p2q2

(iv) l2 + m2, m2 + n2, n2 + l2, 2lm + 2mn + 2nl

Ans : 

Let’s add the algebraic expressions:

(i) 

Adding the like terms, we get:

• ab – bc + bc – ca + ca – ab
• = 0

(ii) 

Combining like terms:

• a – b + ab + b – c + bc + c – a + ac
• = ab + bc + ac

(iii) 

Combining like terms:

• 2p²q² – 3pq + 4 + 5 + 7pq – 3p²q²
• = -p²q² + 4pq + 9

(iv) l² + m², m² + n², n² + l², 2lm + 2mn + 2nl

Combining like terms:

• l² + m² + m² + n² + n² + l² + 2lm + 2mn + 2nl
• = 2l² + 2m² + 2n² + 2lm + 2mn + 2nl
• = 2(l² + m² + n² + lm + mn + nl)

2. (a) Subtract 4a – 7ab + 3b + 12 from 12a – 9ab + 5b – 3

(b) Subtract 3xy + 5yz – 7zx from 5xy – 2yz – 2zx + 10xyz 

(c) Subtract 4p 2q – 3pq + 5pq2 – 8p + 7q – 10 from 18 – 3p – 11q + 5pq – 2pq2 + 5p 2q

Ans : 

(a) Subtract 4a – 7ab + 3b + 12 from 12a – 9ab + 5b – 3

• We change the signs of the terms in the expression being subtracted and then combine like terms.
• (12a – 9ab + 5b – 3) – (4a – 7ab + 3b + 12)
• = 12a – 9ab + 5b – 3 – 4a + 7ab – 3b – 12
• = (12a – 4a) + (-9ab + 7ab) + (5b – 3b) + (-3 – 12)
• = 8a – 2ab + 2b – 15

(b) Subtract 3xy + 5yz – 7zx from 5xy – 2yz – 2zx + 10xyz

• Following the same process:
• (5xy – 2yz – 2zx + 10xyz) – (3xy + 5yz – 7zx)
• = 5xy – 2yz – 2zx + 10xyz – 3xy – 5yz + 7zx
• = (5xy – 3xy) + (-2yz – 5yz) + (-2zx + 7zx) + 10xyz
• = 2xy – 7yz + 5zx + 10xyz

(c) Subtract 4p²q – 3pq + 5pq² – 8p + 7q – 10 from 18 – 3p – 11q + 5pq – 2pq² + 5p²q

• Similarly:
• (18 – 3p – 11q + 5pq – 2pq² + 5p²q) – (4p²q – 3pq + 5pq² – 8p + 7q – 10)
• = 18 – 3p – 11q + 5pq – 2pq² + 5p²q – 4p²q + 3pq – 5pq² + 8p – 7q + 10
• = (18 + 10) + (-3p + 8p) + (-11q – 7q) + (5p/1qq + 3pq) + (-2pq² – 5pq²) + (5p²q – 4p²q)
• = 28 + 5p – 18q + 8pq – 7pq² + p²q

Exercise 8.2

1. Find the product of the following pairs of monomials.

(i) 4, 7p

(ii) -4p, 7p

(iii) -4p, 7pq

(iv) 4p3, -3p

(v) 4p, 0

Ans : 

(i) 4, 7p

• Product = 4 * 7p = 28p

(ii) -4p, 7p

• Product = (-4p) * (7p) = -28p²

(iii) -4p, 7pq

• Product = (-4p) * (7pq) = -28p²q

(iv) 4p³, -3p

• Product = (4p³) * (-3p) = -12p⁴

(v) 4p, 0

• Product = 4p * 0 = 0

2. Find the areas of rectangles with the following pairs of monomials as their lengths and breadths respectively.

(p, q); (10m, 5n); (20x2, 5y2); (4x, 3x2); (3mn, 4np)

Ans :

(i) Length = p, Breadth = q

• Area = p * q = pq square units

(ii) Length = 10m, Breadth = 5n

• Area = 10m * 5n = 50mn square units

(iii) Length = 20x², Breadth = 5y²

• Area = 20x² * 5y² = 100x²y² square units

(iv) Length = 4x, Breadth = 3x²

• Area = 4x * 3x² = 12x³ square units

(v) Length = 3mn, Breadth = 4np

• Area = 3mn * 4np = 12mn²p square units

3. Complete the table of Products.

Ans : 

First monomial →	2x	-5y	3x²	-4xy	7x²y	-9x²y²
Second monomial ↓
2x	4x²	-10xy	6x³	-8x²y	14x³y	-18x³y²
-5y	-10xy	25y²	-15x²y	20xy²	-35x²y²	45x³y³
3x²	6x³	-15x²y	9x⁴	-12x³y	21x⁴y	-27x⁴y²
-4xy	-8x²y	20xy²	-12x³y	16x²y²	-28x³y²	36x³y³
7x²y	14x³y	-35x²y²	21x⁴y	-28x³y²	49x⁴y²	-63x⁵y³
-9x²y²	-18x³y²	45x³y³	-27x⁴y²	36x³y³	-63x⁵y³	81x⁴y⁴

4. Obtain the volume of rectangular boxes with the following length, breadth and height respectively.

(i) 5a, 3a2, 7a4

(ii) 2p, 4q, 8r

(iii) xy, 2x2y, 2xy2

(iv) a, 2b, 3c

Ans : 

(i) Length = 5a, Breadth = 3a², Height = 7a⁴

• Volume = 5a * 3a² * 7a⁴ = 105a⁷ cubic units

(ii) Length = 2p, Breadth = 4q, Height = 8r

• Volume = 2p * 4q * 8r = 64pqr cubic units

(iii) Length = xy, Breadth = 2x²y, Height = 2xy²

• Volume = xy * 2x²y * 2xy² = 4x⁴y⁴ cubic units

(iv) Length = a, Breadth = 2b, Height = 3c

• Volume = a * 2b * 3c = 6abc cubic units

5. Obtain the product of

(i) xy, yz, zx

(ii) a, -a2, a3

(iii) 2, 4y, 8y2, 16y3

(iv) a, 2b, 3c, 6abc

(v) m, -mn, mnp

Ans : 

(i) xy, yz, zx

• Product = xy * yz * zx = x²y²z²

(ii) a, -a², a³

• Product = a * (-a²) * a³ = -a⁶

(iii) 2, 4y, 8y², 16y³

• Product = 2 * 4y * 8y² * 16y³ = 1024y⁶

(iv) a, 2b, 3c, 6abc

• Product = a * 2b * 3c * 6abc = 36a²b²c²

(v) m, -mn, mnp

• Product = m * (-mn) * mnp = -m³n²p

Exercise 8.3

1. Carry out the multiplication of the expressions in each of the following pairs:

(i) 4p, q + r

(ii) ab, a – b

(iii) a + b, 7a2b2

(iv) a2 – 9, 4a

(v) pq + qr + rp, 0

Ans : 

(i) 4p, q + r

Using the distributive property, we get:

• 4p(q + r) = 4pq + 4pr

(ii) ab, a – b

Using the distributive property, we get:

• ab(a – b) = a²b – ab²

(iii) a + b, 7a²b²

Using the distributive property, we get:

• (a + b) * 7a²b² = 7a³b² + 7ab³

(iv) a² – 9, 4a

Using the distributive property, we get:

• (a² – 9) * 4a = 4a³ – 36a

(v) pq + qr + rp, 0

• (pq + qr + rp) * 0 = 0

2. Complete the table.

S.No.	First Expression	SecondExpression	Product
(i)	a	b + c + d	–
(ii)	x + y – 5	5xy	–
(iii)	p	6p2 – 7p + 5	–
(iv)	4p2q2	p2 – q2	–
(v)	a + b + c	abc	–

Ans : 

S.No.	First Expression	Second Expression	Product
(i)	ab + c + d	–	ab + ac + ad
(ii)	x + y – 5	5xy	5x²y + 5xy² – 25xy
(iii)	p	6p² – 7p + 5	6p³ – 7p² + 5p
(iv)	4p²q²	p² – q²	4p⁴q² – 4p²q⁴ (using difference of squares)
(v)	a + b + c	abc	a²bc + ab²c + abc² (using distributive property)

3. Find the products.

Ans : 

(i) a² x 2a²² x 4a²⁶ = 8a³⁰

(ii) (2/3) xy x (-9/10)x²y² = -3/5 x³y³

(iii) (-10/3)pq³ x (6/5)p³q = -4p⁴q⁴

(iv) x x x² x x³ x x⁴ = x¹⁰

4. (a) Simplify: 3x(4x – 5) + 3 and find its values for (i) x = 3 (ii) x = 1/2.

(b) Simplify: a(a2 + a + 1) + 5 and find its value for (i) a = 0 (ii) a = 1 (iii) a = -1

Ans : 

(a) 

Using the distributive property, we get:

• 3x(4x – 5) + 3 = 12x² – 15x + 3

Finding the values:

• (i) For x = 3:
  • 12(3)² – 15(3) + 3 = 108 – 45 + 3 = 66
• (ii) For x = 1/2:
  • 12(1/2)² – 15(1/2) + 3 = 12(1/4) – 15/2 + 3 = 3 – 15/2 + 3 = -3/2

(b) 

Using the distributive property, we get:

• a(a² + a + 1) + 5 = a³ + a² + a + 5

Finding the values:

• (i) For a = 0:
  • 0³ + 0² + 0 + 5 = 5
• (ii) For a = 1:
  • 1³ + 1² + 1 + 5 = 1 + 1 + 1 + 5 = 8
• (iii) For a = -1:
  • (-1)³ + (-1)² + (-1) + 5 =
  •  -1 + 1 – 1 + 5 
  • = 4

5. (a) Add: p(p – q), q(q – r) and r(r – p)

(b) Add: 2x(z – x – y) and 2y(z – y – x)

(c) Subtract: 3l(l – 4m + 5n) from 4l(10n – 3m + 2l)

(d) Subtract: 3a(a + b + c) – 2b(a – b + c) from 4c(-a + b + c)

Ans : 

(a) 

• p(p – q) + q(q – r) + r(r – p)
• = p² – pq + q² – qr + r² – rp
• = p² + q² + r² – pq – qr – rp

(b) 

• 2x(z – x – y) + 2y(z – y – x)
• = 2xz – 2x² – 2xy + 2yz – 2y² – 2xy
• = -2x² – 2y² + 2xz + 2yz – 4xy

(c) 

• 4l(10n – 3m + 2l) – 3l(l – 4m + 5n)
• = 40ln – 12lm + 8l² – 3l² + 12lm – 15ln
• = 5l² + 25ln

(d)

• 4c(-a + b + c) – [3a(a + b + c) – 2b(a – b + c)]
• = -4ac + 4bc + 4c² – (3a² + 3ab + 3ac – 2ab + 2b² – 2bc)
• = -4ac + 4bc + 4c² – 3a² – 3ab – 3ac + 2ab – 2b² + 2bc
• = -3a² – 2b² + 4c² – ab + 6bc – 7ac

Exercise 8.4

1. Multiply the binomials:

(i) (2x + 5) and (4x – 3)

(ii) (y – 8) and (3y – 4)

(iii) (2.5l – 0.5m) and (2.5l + 0.5m)

(iv) (a + 3b) and (x + 5)

(v) (2pq + 3q²) and (3pq – 2q²)

(vi) (3/4 a² + 3b²) and 4(a² – 2/3 b²)

Ans : 

(i) 

• (2x + 5)(4x – 3) = 2x(4x – 3) + 5(4x – 3) = 8x² – 6x + 20x – 15 = 8x² + 14x – 15

(ii) 

• (y – 8)(3y – 4) = y(3y – 4) – 8(3y – 4) = 3y² – 4y – 24y + 32 = 3y² – 28y + 32

(iii) 

• (2.5l – 0.5m)(2.5l + 0.5m) = (2.5l)² – (0.5m)² (This is a difference of squares) = 6.25l² – 0.25m²

(iv) 

• (a + 3b)(x + 5) = a(x + 5) + 3b(x + 5) = ax + 5a + 3bx + 15b

(v) 

• (2pq + 3q²)(3pq – 2q²) = 2pq(3pq – 2q²) + 3q²(3pq – 2q²) = 6p²q² – 4pq³ + 9pq³ – 6q⁴ = 6p²q² + 5pq³ – 6q⁴

(vi) 

• First, distribute the 4 in the second binomial:
  • 4(a² – 2/3 b²) = 4a² – 8/3 b²
• Then multiply the binomials:
  • (3/4 a² + 3b²)(4a² – 8/3 b²) = (3/4 a²)(4a²) + (3/4 a²)(-8/3 b²) + (3b²)(4a²) + (3b²)(-8/3 b²) = 3a⁴ – 2a²b² + 12a²b² – 8b⁴ = 3a⁴ + 10a²b² – 8b⁴

2. Find the product:

(i) (5 – 2x) (3 + x)

(ii) (x + 7y) (7x – y)

(iii) (a2 + b) (a + b2)

(iv) (p2 – q2)(2p + q)

Ans : 

(i) (5 – 2x)(3 + x)

Using the distributive property, we get:

• (5 – 2x)(3 + x) = 15 + 5x – 6x – 2x² = -2x² – x + 15

(ii) (x + 7y)(7x – y)

• (x + 7y)(7x – y) = 7x² – xy + 49xy – 7y² = 7x² + 48xy – 7y²

(iii) (a² + b)(a + b²)

• (a² + b)(a + b²) = a³ + a²b² + ab + b³

(iv) (p² – q²)(2p + q)

• (p² – q²)(2p + q) = 2p³ + p²q – 2pq² – q³

3. Simplify:

(i) (x2 – 5) (x + 5) + 25

(ii) (a2 + 5)(b3 + 3) + 5

(iii) (t + s2) (t2 – s)

(iv) (a + b) (c – d) + (a – b) (c + d) + 2(ac + bd)

(v) (x + y) (2x + y) + (x + 2y) (x – y)

(vi) (x + y)(x2 – xy + y2)

(vii) (1.5x – 4y)(1.5x + 4y + 3) – 4.5x + 12y

(viii) (a + b + c) (a + b – c)

Ans : 

(i) (x² – 5)(x + 5) + 25

• Using the formula (a + b)(a – b) = a² – b², we get:
  • x³ + 5x – 5x – 25 + 25 = x³

(ii) (a² + 5)(b³ + 3) + 5

• Expanding the brackets, we get:
  • a²b³ + 3a² + 5b³ + 15 + 5
  • a²b³ + 3a² + 5b³ + 20

(iii) (t + s²)(t² – s)

• Expanding the brackets, we get:
  • t³ – st + s²t² – s³

(iv) (a + b)(c – d) + (a – b)(c + d) + 2(ac + bd)

• Expanding the brackets, we get:
  • ac – ad + bc – bd + ac + ad – bc – bd + 2ac + 2bd
  • Combining like terms, we get:
  • 4ac

(v) (x + y)(2x + y) + (x + 2y)(x – y)

• Expanding the brackets, we get:
  • 2x² + xy + 2xy + y² + x² – xy + 2xy – 2y²
  • Combining like terms, we get:
  • 3x² + 4xy – y²

(vi) (x + y)(x² – xy + y²)

• Using the formula (a + b)(a² – ab + b²) = a³ + b³, we get:
  • x³ + y³

(vii) (1.5x – 4y)(1.5x + 4y + 3) – 4.5x + 12y

• Expanding the first bracket, we get:
  • 2.25x² + 6xy – 6xy – 16y² + 4.5x – 12y – 4.5x + 12y
  • Simplifying, we get:
  • 2.25x² – 16y²

(viii) (a + b + c) (a + b – c)

• Expanding the brackets, we get:
  • a² + ab – ac + ab + b² – bc + ac + bc – c²
  • Combining like terms, we get:
  • a² + b² – c² + 2ab