Algebraic Expressions

Question 1.
Find the product of the following pairs:

• (i) 6, 7k
• (ii) - 31, - 2m
• (iii) -5t²- 3t²
• (iv) 6n, 3m
• (v) - 5p², - 2p

Solution:
The product of 6, 7k = 6 × 7k = 42k

• ii) The product of - 3l, - 2m = (- 3l) × (- 2m) = 6/m
• iii) The product of - 5t², - 3t²= (- 5t²) × (- 3t²) = 15t⁴
• iv) The product of 6n, 3m = 6n × 3m = 18mn
• v) The product of - 5p², - 2p = (- 5p²) × (- 2p) = 10p³

Question 2.
Complete the table of the products.

Solution:

Question 3.
Find the volumes of rectangular boxes with given length, breadth and height in the following table.

Solution:

Question 4.
Find the product of the following monomials

(i) xy, x²y , xy, x
(ii) a, b, ab, a³ b, ab³
(iii) kl, lm, km, klm
(iv) pq ,pqr, r
(v) - 3a, 4ab, - 6c, d

Solution:
i) The product of xy, x²y, xy, x = xy × x²y × xy × x
= x⁵ × y³= x⁵y³

ii) The product of a, b, ab, a³b, ab³= a × b × ab × a³b × ab³
= a⁶× b⁶= a⁶b⁶

iii) The product of kl, lm, km, klm = kl × lm × km × klm
k³× l³ × m³=k³l³m³

iv) The product of pq, pqr, r = pq × pqr × r
= p² × q²× r²- p²q²r²

v) The product of - 3a, 4ab, - 6c, d = (- 3a) × 4ab × (- 6c) x d
= + 72a²× b × c × d
= 72a²bcd

Question 5.
If A = xy,B = yz and C = zx, then find ABC=

Solution:
ABC = xy × yz × zx = x²y²z²

Question 7.
Write some monomials of your own and find their products.

Solution:
The product of,some monomials is given below :
i) abc × a²bc = a³b²c²
ii) xy × x²z × yz²= x³y²z³
iii) p × q × r = p³q³r³

Question 1.
Complete the table:

Solution

Question 2.
Simplify: 4y(3y + 4)

Solution:
4y(3y + 4) = 4y × 3y + 4y × 4
= 12y²+ 6y

Question 3.
Simplify x(2x²- 7x + 3) and find the values of it for (i) x = 1 and (ii) x = 0

Solution:
x(2x²- 7x + 3)
= x × 2x²- x × 7x + x × 3
= 2x³- 7x²+ 3x
= 3 × 3 - 1 × 3 - 3 × 1 + 1 × 1
=9 - 3 - 3 + 1 = 4
∴(a - b)²= 4sq.units
[∵ (3 - 1)²= 2²= 4]

(ii)

∴ (a - b)²= a²- 2ab + b²
Area of ABCD + Area of CYZS
= a²- 2ab + b²
area of ABCD - area of BXYC - area of DCST + area of CYZS
= 5 × 5 - 2 × 5 - 2 × 5 + 2 × 2
= 25 - 10 - 10 + 4
= 9 sq.units
[∵ (5 - 2)²= (3)²= 9]

Question 4.
Add the product: a(a - b), b(b - c), c(c - a)

Solution:
a(a - b) + b(b - c) + c(c - a)
=a × a - a × b + b × b - b × c + c × c - c × a
=a²- ab + b²- bc + c²- ca
=a²+ b²+ c²- ab - bc - ca

Question 5.
Add the product: x(x + y - r), y(x - y+r), z(x - y - z)

Solution:
x(x + y - r) +y(x - y + r) + z(x - y - z)
= x² + xy - xr + xy - y² + yr + zx - yz - z²
= x² - y² - z² + 2xy - xr + yr + zx - yz

Question 6.
Subtract the product of 2x(5x - y) from product of 3x(x+2y)

Solution:
3x(x + 2y) - 2x(5x - y)
=(3x × x + 3x × 2y)-(2x × 5x - 2x × y)
= 3x² + 6xy - (10x² - 2xy)
= 3x² + 6xy- 10x² + 2xy
= 8xy - 7x²

Question 7.
Subtract 3k(5k - l + 3rn) from 6k(2k + 3l - 2rn)

Solution:
6k(2k + 3l - 2m) - 3k(5k - l + 3m)
= 12k²+ 18kl - 12km - 15k² + 3kl - 9km
= -3k² + 21kl - 21km

Question 8.
Simplify: a²(a - b + c) + b²(a + b - c) - c²(a - b - c)

Solution:
a²(a - b + c) + b²(a + b - c) - c²(a - b - c)
= a³ - a²b + a²c + ab² + b³ - b²c - ac² + bc² + c³
= a³ + b³ + c³ - a²b + a²c + ab² - b²c - ac² - bc²

Question 1.
Multiply the binomials:

(i) 2a - 9 and 3a + 4
(ii) x - 2y and 2x - y
(iii) kl + lm and k - l
(iv) m² - n² and m + n
Solution:
i) 2a - 9 and 3a + 4
(2a - 9) (3a + 4) = 2a (3a + 4) - 9(3a + 4)
= 6a² + 8a - 27a - 36
= 6a² - 19a - 36
ii) x - 2y and 2x - y
(x - 2y) (2x - y) = x(2x - y) - 2y(2x - y)
= 2x² - xy - 4xy + 2y²
= 2x² - 5xy + 2y²

iii) kl + lm and k - l
(kl + lm) (k - l) = kl(k - l) + lm(k - l)
= k²l - l²k + klm - l²m
iv) m² - n² and m + n
(m² - n²) (m + n) = m²(m + n) - n²(m + n)
= m³ + m²n - n²m - n³

Question 2.
Find the product:

(i) (x + y)(2x - 5y + 3xy)
(ii) (mn - kl + km) (kl - lm)
(iii) (a - 2b + 3c)(ab² - a²b)
(iv) (p³ + q³)(p - 5q+6r)
Solution:
i) (x + y) (2x - 5y + 3xy)
= x(2x - 5y + 3xy) + y(2x - 5y + 3xy)
= 2x² - 5xy + 3x²y + 2xy - 5y² + 3xy²
= 2x² - 5y² - 3xy + 3x²y + 3xy²
ii) (mn - kl + km) (kl - lm)
= kl(mn - kl + km) - lm(mn - kl + km)
= klmn - k2l2 + k²lm - lm²n + kl²m - klm²

iii) (a - 2b + 3c) (ab² - a²b) = a(ab² - a²b) - 2b(ab² - a²b) + 3c(ab²- a²b) = a²b² - a³b - 2ab³ + 2a²b² + 3ab²c - 3a²bc = 3a²b² - a³b - 2ab³ + 3ab²c - 3a²bc iv) (p³ + q³) (p - 5q + 6r) = p³(p - 5q + 6r) + q³(p - 5q + 6r) = p⁴ - 5p³q + 6p³r + pq³ - 5q⁴ + 6rq³ = p⁴ - 5q⁴ - 5p³q + 6p³r + pq³ + 6rq³

Question 3.
Simplify the following:

(i) (x-2y) (y - 3x) + (x+y) (x-3y) - (y - 3x) (4x - 5y)
(ii) (m + n) (m² - mn + n²)
(iii) (a - 2b + 5c) (a - b) - (a - b - c) (2a + 3c) + (6a + b) (2c - 3a - 5b)
(iv) (pq-qr-i-pr) (pq-i-qr) - (pr-i-pq) (p-i-q - r)
Solution:
i) (x - 2y) (y - 3x) + (x + y) (x - 3y) - (y - 3x) (4x - 5y)
= (y - 3x) [x - 2y - (4x - 5y)] + (x + y)(x - 3y)
= (y - 3x) [x - 2y - 4x + 5y] + (x + y) (x - 3y)
= (y - 3x) (3y - 3x) + (x + y) (x - 3y)
= y(3y - 3x) - 3x(3y - 3x) + x(x - 3y) + y(x - 3y)
= 3y² - 3xy - 9xy + 9x² + x² - 3xy + xy - 3y²
= 10x² - 14xy

ii) (m + n) (m²- mn + n²)
= m(m² - mn + n²) + n(m² - mn + n²)
= m³ - m²n + n²m + nm² - mn² + n³
= m³ + n³

iii) (a - 2b + 5c) (a - b) - (a - b - c) (2a + 3c) + (6a + b) (2c - 3a - 5b)
= a(a - 2b + 5c) - b(a - 2b + 5c) - 2a(a - b - c) - 3c(a - b - c) + 6a(2c - 3a - 5b) + b(2c - 3a - 5b)
= a² - 2ab + 5ac - ab + 2b² - 5bc - 2a² + 2ab + 2ac - 3ac + 3bc + 3c² + 12ac - 18a² - 30ab + 2bc - 3ab - 5b²
= - 19a² - 3b² - 34ab + 16ac + 3c2

iv) (pq - qr + pr) (pq + qr) - (pr + pq) (p + q - r)
= pq(pq - qr + pr) + qr(pq - qr + pr) - pr(p + q - r) - pq(p + q - r)
= p²q² - pq²r + p²qr + pq²r - q²r² + pqr² - p²r - pqr + pr² - p²q - pq² + pqr
= p²q² - q²r² + p²qr + pqr² - p²r + pr² - p²q - pq ²

Question 1.
Select a suitable identity and find the following products

(i) (3k + 4l)(3k + 4l)
(ii) (ax² + by²)(ax² + by²)
(iii) (7d - 9e)(7d - 9e)
(iv) (m² - n²)(m² + n²)
(v) (3t + 9s) (3t - 9s)
(vi) (kl - mn) (kl + mn)
(vii) (6x + 5)(6x + 6)
(viii) (2b - a)(2b +c)

Solution:
(3k + 4l) (3k 4l) = (3k + 4l)² is in the form of (a + b)².
=(3k)² + 2 × 3k × 4l+ (4l)² [ (a+ b)² = a² + 2ab + b²
= 3k × 3k + 24kl + 4l × 4l
= 9k2 + 24kl + 16l2
ii) (ax² + by²) (ax² + by²) = (ax² + by²)² is in the form of (a + b)².
= (ax²)² + 2 × ax² × by² + (by²)² [ ∵ (a + b)² = a² + 2ab + b²]
= ax² × ax² + 2abx²y² + by² × by²
= a²x⁴ + 2ab x²y² + b²y⁴

iii) (7d - 9e) (7d - 9e)
= (7d - 9e)² is in the form of (a - b)².
= (7d)² - 2 × 7d × 9e + (9e)² [ ∵ (a - b)² = a² - 2ab + b²]
= 7d × 7d - 126de + 9e × 9e
= 49d² - 126de + 81e²
iv) (m² - n²) (m² + n²) is in the form of (a + b) (a - b).
∴ (a + b) (a - b) = a² - b²
∴ (m² + n²) (m² - n²) = (m²)² - (n²)² = m⁴ - n⁴
v) (3t + 9s) (3t - 9s) = (3t)² - (9s)² [ ∵ (a + b) (a - b) = a² - b² ]
= 3t × 3t - 9s × 9s
= 9t² - 81s²

vi) (kl - mn) (kl + mn) = (kl)² - (mn)² [ ∵(a + b) (a - b) = a² - b² ]
= kl × kl - mn × mn
= k²l² - m²n²
vii) (6x + 5) (6x + 6) is in the form of
(ax + b) (ax + c).
(ax + b) (ax + c) = a²x² + ax(b + c) + bc
(6x + 5) (6x + 6) = (6)²x² + 6x (5 + 6) + 5 × 6
= 36x² + 6x × 11 + 30
= 36x² + 66x + 30

viii) (2b - a) (2b + c) is in the form of (ax - b) (ax + c).
(ax - b) (ax + c) = a²x² + ax(c - b) - cb
(2b - a) (2b + c) = (2)²(b)² + 2b (c - a) - ca
= 4b2 + 2bc - 2ab - ca

Question 2.
Evaluate the following by using suitable identities:

(i) 304²
(ii) 509²
(iii) 992²
(iv) 799²
(v) 304 × 296
(vi) 83 × 77
(vii) 109 × 108
(viii) 204 × 206

Solution:
i) 304² = (300 + 4)² is in the form of (a + b)².
∵ (a+b)² = a² + 2ab + b²
a = 300, b = 4
(300 + 4)² = (300)² + 2 × 300 × 4 + (4)²
= 300 × 300+ 2400 + 4 × 4
= 90,000 + 2400 + 16
= 92,416

ii) 509² = (500 + 9)²
a = 500, b = 9
= (500)² + 2 × 500 × 9 + (9)²
[ ∵ (a + b)² = a² + 2ab + b²]
= 500 × 500 + 9000 + 9 × 9
= 2,50,000 + 9000 + 81
= 2,59,081

iii) 992² = (1000 - 8)²
a = 1000, b = 8
= (1000)² - 2 × 1000 × 8 + (8)² [∵ (a-b)² = a² - 2ab + b²]
= 1000 × 1000 - 16,000 + 8 × 8
= 10,00,000 - 16000 + 64
= 10,00,064 - 1600
= 9,98,464

iv) 799² = (800 - 1)²
a = 800, b = 1
= (800)² - 2 × 800 × 1 + (1)²
= 800 × 800 - 1600 + 1
= 6,40,000 - 1600 + 1
= 6,40,001 - 1600
= 6,38,401
v) 304 × 296 = (300 + 4) (300 - 4) is in the form of (a + b) (a - b).
(a + b) (a - b) = a² - b²
∴ (300 + 4) (300 - 4) = (300)² - (4)2
= 300 × 300 - 4 × 4
= 90,000 - 16
= 89,984

vi) 83 × 77 = (80 + 3) (80 - 3)
= (80)² - (3)² [ ∵ (a + b) (a - b) = a² - b²]
= 80 × 80 - 3 × 3
= 6400 - 9
= 6391
vii) 109 × 108 = (100 + 9) (100 + 8)
= (100)² + (9 + 8)100 + 9 × 8
= 10,000 + 1700 + 72
= 11,772
viii) 204 × 206 = (205 - 1) (205 + 1)
= (205)² - (1)² [∵ (a + b)(a-b) = a² - b²]
= 205 × 205 - 1 × 1
= 42,025 -1
= 42,024

Question 1.
Verify the identity (a + b)² ? a² + 2ab + b² geometrically by taking

(i) a = 2 units, b = 4 units
(ii) a = 3 units, b = 1 unit
(iii) a = 5 units, b = 2 unit

Solution:
(i)

= 4 × 4 + 4 × 2 + 2 × 2 + 4 × 2
= 16+ 8 + 4 + 8 = 36 sq.units
[∵ (2 + 4)² = 6² = 36]

(ii)

Area of a square AEGI
= area of square ABCD + area of rectangle CDEF + area of square CFGH + area of rectangle BIHC.
= 3 × 3 + 3 × 1 + 1 × 1+3 × 1
= 9 + 3 + 1 + 3
= 16 sq. units
[∵ (3 + 1)2 = 4² = 16]

(iii)

= 5 × 5 + 2 × 5 + 2 × 2 + 5 × 2 = 25 + 10 + 4 + 10 = 49 sq.units [∵ (5 + 2)² = 7² = 49]

Question 2.
Verify the identity (a - b)²≡ a² - 2ab+ b² geometrically by taking

(i) a = 3 units, b= 1 unit
(ii) a = 5 units, b = 2 units

Solution:
(i)

Area of AIFE + Area of FGCH = (a - b)² = a² - 2ab + b² [area of AIFE - area of IBGF - area of EFHD + area of FGCH]
= 3 × 3 - 1 × 3 - 3 × 1 + 1 × 1
= 9 - 3 - 3 + 1 = 4
∴ (a - b)² = 4 sq. units
[∵ (3 - 1 )² = 2² = 4]

(ii)

∴ (a - b)² = a² - 2ab + b²
Area of ABCD + Area of CYZS = a² - 2ab + b²
area of ABCD - area of BXYC - area of DCST + area of CYZS
=5 × 5 - 2 × 5 - 2 × 5 + 2 × 2
= 25 - 10 - 10 + 4
= 9 sq.units
[∵ (5 - 2)² = (3)² = 9]

Question 3.
Verify the identity(a + b)(a - b) ≡ a² - b² geometrically by taking

(i) a = 3 units, b = 2 units
(ii) a = 2 units, b = 1 unit

Solution:
i)

a² - b² = Area of Fig I. + Area of Fig II
= a(a - b) + b(a - b)
= (a - b) (a + b)
= 3 × 3 - 2 × 2
a² - b² = 9 - 4= 5sq . units
[ ∵ 3² - 2² = 9 - 4 = 5]

(ii)

a² - b² = Area of Fig I. + Area of Fig II = a(a - b) + b(a - b) = (a + b) (a - b) =(2 + 1)(2 - 1) = 3 × 1 = 3 a² - b² = 3 sq. units [∵ (2² - 1²) = 4 - 1 = 3]