Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 3 Algebra Ex 3.1

Students can Download Maths Chapter 3 Algebra Ex 3.1 Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 3 Algebra Ex 3.1

8th Maths Algebra Exercise 3.1 Question 1.
Multiply a monomial by a monomial.
(i) 6x, 4
(ii) -3x, 7y
(iii) -2m², (-5m)³
(iv) a³, – 4a²b
(v) 2p²q³, -9pq²
Solution:
(i) 6x × 4 = (6 × 4) (x) = 24x
(ii) -3x × 7y = (-3 × 7) (x × y) = -21xy
(iii) (-2m²) × (-5m)³ = -2m² × (-)³ (5³ (m)³) = -2m² × (-125m³)
= (-) × (-)(2 × 125)(m² × m³) = + 250m5 = 250 m
(iv) a³ × (-4a² b) = (-4) × (a³ × a²) × (b) = -4a⁵b
(v) (2p²q³) × (-9pq²) = (+) × (-) × (2 × 9) (p² × p(q³ × q²)) = -18p³q⁵

Maths 8th Samacheer Kalvi Question 2.
Complete the table

Solution:

8th Standard Maths 3.1 Solution Question 3.
Find the product of the terms.
(i) -2mn, (2m)², -3mn
(ii) 3x²y, -3xy³, x²y²
Solution:
(i) (-2mn) × (2m)² × (-3mn) = (-2mn) × 2² m² × (-3mn) = (-2mn) × 4m² × (-3mn)
= (-) (+)(-) (2 × 4 × 3) (m × m² × m) (n × n)
= + 24 m⁴ n²

(ii) (3²y) × (-31xy³) × (x²y²) = (+) × (-) × (+) × (3 × 3 × 1) (x² × x × x²) x (y × y³ × y²)
= -9x⁵y⁶

Samacheer Kalvi Guru 8th Maths Question 4.
If l = 4pq², b = -3p 2q, h = 2p³q³ then, find the value of 1 × b × h.
Solution:
Given l = 4pq²
b = -3p²q
h = 2p³q³
l × b × h = (4pq²) × (-3p² q) × (2p³q³)
= (+) (-) (+) (4 × 3 × 2) (p × p² × p³) (q² × q × q³)
= -24p⁶q⁶

Samacheer Kalvi.Guru 8th Maths Question 5.
Expand
(i) 5x (2y – 3)
(ii) -2p (5p² – 3p + 7)
(iii) 3mn (m³n³ – 5m²n + 7mn²)
(iv) x² (x + y + z) y² (x + y + z) + z² (x – y – z)
Solution:
(i) 5x(2y – 3) = (5x) (2y) – (5x) (3)
= (5 × 2) (x × y) – (5 × 3) x
= 10xy – 15x

(ii) -2p (5p² – 3p + 7) = (-2p) (5p²) + (-2p) (-3p) + (-2p) (7)
= [(-) (+) (2 × 5) (p × p²)] + [(-) (+) (2 × 3) (p × q)] + (-) (+) (2 × 7) p
= -10p³ + 6p² – 14p

(iii) 3mn(m³n³ – 5m²n + 7mn²)
= (3mn) (m³n³) + (3mn) (-5m²n) + (3mn)(7mn²)
= (3) (m × m³) (n × n³) + (+) (-) (3 × 5) (m × m²) (n × n) + (3 × 7) (m × m)(n × n²)
= 3m⁴n⁴ – 15m³ n² + 21m²n³

(iv) x² (x + y + z) + y² (x + y + z) + z² (x – y – z)
= (x² × x) + (x² × y) + (x² × z) + (y² × x) + (y² × y) + (y² × z) + (z² × x) + z² (-y) + z² (-z)
= x³ + x²y + x²z + xy² + y³ + y²z + xz² – yz² – z³
= x³ + y³ – z² + x²y + x²z + xy² + zy² + xz² – yz²

Samacheer Kalvi Guru Maths 8th Question 6.
Find the product of
(i) (2x + 3)(2x – 4)
(ii) (y² – 4) (2y² + 3y)
(iii) (m² – m) (5m²n² – n²)
(iv) 3(x – 5) × 2(x – 1)
Solution:
(2x) (2x – 4) + 3 (2x – 4) = (2x) (2x – 4) + 3 (2x – 4)
= (2x × 2x) – 4 (2x) + 3(2x) – 3 (4)
= 4x² – 8x + 6x – 12
= 4x² + (- 8 + 6)x – 12
= 4x² – 2x – 12

(ii) (y² -4) (2y² + 3y) = y² (2y² + 3y) – 4 (2y² + 37)
= y²(2y²) + y²(3y) – 4(2y²) -4 (3y)
= 2y⁴ + 3y³ – 8y² – 12y

(iii) (m² – n) (5m²n² – n²) = m² (5m²n² – n²) – n (5m²n² – n²)
= m² (5m²n²) + m² (-n²) – n (5m²n²) + (-) (-) n (n²)
= 5m⁴n² – m²n² – 5m²n³ + n³

(iv) 3(x – 5) × 2(x – 1) = (3 × 2) (x – 5) (x – 1)
= 6 × [x (x – 1) – 5 (x- 1)]
= 6 [x.x – x . 1 – 5.x + (-1) (-) 5 1]
= 6 [x² – x – 5x + 5] = 6 [x² + (-1 – 5)x + 5]
= 6 [x² – 6x + 5] = 6x² – 36x + 30

Samacheer Kalvi 8 Maths Question 7.
Find the missing term.
(i) 6xy – × ______ = -12x³y
(ii) ________ × (-15m²n³p) = 45m³n³p²
(iii) 2y(5x²y – ___ + 3 ___) = 10x²y² – 2xy + 6y³
Solution:
(i) 6xy – × (-2x²) = -12x³y
(ii) -3mp × (-15m²n³p) = 45m³n³p²
(iii) 2y(5x²y – x + 3 y²) = 10x²y² – 2xy + 6y³

Samacheer Kalvi 8th Maths Book Solutions Question 8.
Match the following

(A) iv, v, ii, i, iii
(B) v, iv, iii, ii, i
(C) iv, v, ii, iii, i
(D) iv, v, ii, iii, i
Solution:
(a) iv
(b) v
(c) ii
(d) iii
(e) i

Samacheer Kalvi 8 Maths Book Question 9.
A car moves at a uniform speed of (x + 30) km/hr. Find the distance covered by the car in (y + 2)hours. (Hint: distance = speed × time).
Solution:
Sppeed of the car = (x + 30) km / hr.
Time = (y + 2) hours
Distance = Speed × time = (x + 30) (y + 2) = x(y + 2) + 30 (y + 2) = x (y + 2) + 30 (y + 2)
= (x) (y) + (x) (2) + (30) (y) + (30) (2)
= xy + 2x + 30y + 60
Distance covered = (xy + 2x + 30y + 60) km

Objective Type Questions

Samacheer Kalvi Class 8 Maths Solutions Question 10.
The product of 7p³ and (2p²)² is
(A) 14p²
(B) 28p⁷
(C) 9p⁷
(D) 11p¹²
Solution:
(B) 28p⁷

Samacheerkalvi.Guru 8th Maths Question 11.
The missing terms in the product -3m³ n × 9(- -) = ____ m⁴n³ are
(A) mn², 27
(B) m²n, 27
(C) m²n², -27
(D) mn², -27
Solution
(A) mn² ,27

Samacheer Kalvi Guru 8th Maths Guide Question 12.
If the area of a square is 36x⁴y² then, its side is ______ .
(A) 6x⁴y²
(B) 8x²y²
(C) 6x²y
(D) -6x²y
Solution:
(C) 6x²y

Samacheer Kalvi 8th Maths Question 13.
If the area of a rectangle is 48m²n³ and whose length is 8mn2 then, its breadth is ____ .
(A) 6 mn
(B) 8m²n
(C) 7m²n²
(D) 6m²n²
Solution:
(A) 6mn

Maths Class 8 Samacheer Kalvi Question 14.
If the area of a rectangular land is (a² – b²) sq.units whose breadth is (a – b) then, its length is _____
(A) a – b
(B) a + b
(C) a² – b
(D) (a + b)²
Solution:
(B) a + b