The DAV Class 7 Maths Solutions and DAV Class 7 Maths Chapter 6 Worksheet 4 Solutions of Algebraic Expressions offer comprehensive answers to textbook questions.
DAV Class 7 Maths Ch 6 WS 4 Solutions
Question 1.
Find the following products:
(i) (x + 3) (x2 + 2x- 1)
Answer:
(x + 3) (x2 + 2x – 1) = x(x2 + 2x – 1) + 3 (x2 + 2x – 1)
= x3 + 2x2 -x + 3x2 + 6x – 3
= x3 + 5x2 + 5x – 3
(ii) (7y – 2) (5y2 – 3y + 2)
Answer:
(7y – 2) (5y2 – 3y + 2) = 7y(5y2 – 3y + 2) – 2 (5y2 -3y + 2)
= 35y3 – 21 y2 + 14y – 10y2 + 6y – 4
= 35y3 – 31y2 + 20y – 4
(iii) (p3 + 3p + q) (9p + 2q)
Answer:
(p3 + 3p + q) (9p + 2q) = p3(9p + 2q) + 3p(9p + 2q) + q(9p + 2q)
= 9 p4 + 2 p3q + 27 p2 + 6pq + 9 pq + 2 q2
= 9 p4 + 2 p3q + 27p2 + 15 pq + 2 q2
(iv) (- 2x2 + xy – y2) (3x + 4y)
Answer:
(- 2x2 + xy – y2) (3x + 4y) = – 2x2(3x + 4y) + xy (3x + 4y) – y2 (3x + 4y)
= – 6x3 – 8x2y + 3x2y + 4xy2 – 3xy2 – 4y3
= -6x3 – 5x2y + xy2 – 4y3
(v) (\(\frac{2}{5}\)a + \(\frac{1}{7}\)b) (3a + 4b – 2)
Answer:
(vi) (0.1a – 0.2c) (a + c + ac)
Answer:
(0.1 a – 0.2 c) (a + c + ac) = 0.1 a(a + c + ac) -0.2 c(a + c + ac)
= 0.1a2 + 0.1ac + 0.1a2C – 0.2ac – 0.2c2 – 0.2ac2
= 0.1a2 + 0.3ac + 0.1a2c – 0.2c2 – 0.2ac2
Question 2.
Simplify the following and verify the result for the given values:
(i) (x2 – 4xy + y2) (x – 2y); x = 3 and y = 2
Answer:
(x2 – 4xy + y2) (x – 2y); x = 3 and y = 2
= x2(x – 2y) – 4 xy (x – 2y) + y2(x – 2y)
= x3 – 2x2y – 4x2y + 8xy2 + xy2 – 2y3
= x3 – 6 x2y + 9 xy2 – 2y3
= (x2 – 4xy + y2) (x – 2y)
Verification:
L.H.S = [(3)2 – 4(3) (2) + (2)2] [3 – 2(2)]
= [9 – 24 + 4] [3 – 4]
= [13 – 24] [- 1] = -11 x -1 = 11
R.H.S = x3 – 6x2 y + 9xy2 – 2y3
= (3)3 – 6(3)2 (2) + 9(3)(2)2 – 2(2)3
= 27 – 108 + 108 – 16
= 27 – 16
= 11
Hence verified L.H.S. = R.H.S.
(ii) (7x2y – 3z2) (x + y + z); x = 1, y = 1, z = -1
Answer:
(7x2 y – 3z2 ) (x + y + z); x = 1, y = 1, z = – 1
= 7x2 y (x + y + z) – 3z2 (x + y + z)
= 7x3 y + 7x2 y2 + 7x2 yz – 3xz2 – 3yz2 – 3z3
Verification:
L.H.S. = (7x2 y – 3z2 ) (x + y + z)
Putting x = 1, y = 1, z = – 1
= [7x(1)2 x(1)-3x(-1)2] [1 + 1 – 1]
= (7 – 3) (1)
= 4 x 1 = 4
R.H.S. = 7x3 y + 7x2 y2 + 7x2 yz – 3xz2 – 3yz2 – 3z3
= 7(1)3 (1) + 7 (1)2 (1)2 + 7 (1)2 (1) (- 1) – 3(1) (- 1)2 – 3(1)(- 1)2 – 3(- 1)3
= 7 + 7- 7- 3 – 3 + 3 = 4
Hence verified L.H.S. = R.H.S.
(iii) (4a2 – 6ab + 9b2) (2a + 3b ); a = 2, b = 1
Answer:
(4a2 – 6ab + 9b2) (2a + 3b); a = 2, b = 1
= 4a2(2a + 3b) – 6ab(2a + 3b) + 9b2(2a + 3b)
= 8a3 + 12a2b – 12a2b – 18ab2 + 18ab2 + 27b3
= 8a3 + 27b3
Verification
L.H.S. = [4 × (2)2 – 6(2)(1) + 9(1)2](2 × 2 + 3 × 1)
= (16 – 12 + 9) (4 + 3)
= 13 × 7 = 91
R.H.S. 8a3 + 27b3 = 8(2)3 + 27(1)3
= 64 + 27 = 91
Hence verified L.H.S. = R.H.S.
(iv) (m2 – 10m + 25) (m – 5);m = -2
Answer:
(m2 – 10m + 25) (m – 5); m = -2
= m2(m – 5) – 10m (m – 5) + 25 (m – 5)
Verification
L.H.S. = (m2 – 10m + 25) (m – 5)
Putting m = – 2
= [(-2)2 – 10 × (-2) + 25] [-2 – 5]
= (4 + 20 + 25) (-7)
= 49 × – 7
= -343
R.H.S. = m3 – 15m2 + 75m – 125
Putting m = – 2
= (-2)3 – 15(-2)2 + 75(-2) – 125
= – 8 – 60 – 150 – 125
= -343
Hence verified L.H.S. = R.H.S.
(v) (p2 + q2 + r2) (pq + qr): p = 2,q = -3,r = 1
Answer:
(p2 + q2 + r2) (pq + qr); p = 2,q = -3,r = l
= p2(pq + qr) + q2 (pq + qr) + r2 (pq + qr)
= p3q + p2qr + pq3 + q3r + pr2q + qr3
Verification:
L.H.S. = (p2 + q2 + r2) (pq + qr)
Putting p = 2, q = -3, r = 1
= [(2)2 + (- 3)2 + (1)2] [2 × – 3 + (- 3) (1)]
= (4 + 9 + 1) (- 6 – 3)
= 14 × – 9
= -126
R.H.S. = p3q + p2qr + pq3 + q3r + pr2q + qr3
Putting p = 2, q = -3, r = 1
= (2)3 (- 3) + (2)2 (- 3) (1) + 2(- 3)3 + (- 3)3 (1) + (2) (1)2 (- 3) + (- 3) (1)3
= – 24 – 12 – 54 – 27 – 6 – 3
= – 126
Hence verified L.H.S. = R.H.S.
(vi) (\(\frac{5}{4}\)x2 – \(\frac{3}{2}\)xy)(x + y + y2); x = 2,y = 2
Answer:
= \(\frac{5}{4}\)x2(x + y + y2) – \(\frac{3}{2}\)xy(x + y + y2)
= \(\frac{5}{4}\)x3 + \(\frac{5}{4}\)x2y + \(\frac{5}{4}\)x2y2 – \(\frac{3}{2}\)x2y – \(\frac{3}{2}\)xy2 – \(\frac{3}{2}\)xy3
= \(\frac{5}{4}\)x3 – \(\frac{1}{4}\)x2y + \(\frac{5}{4}\)x2y2 – \(\frac{3}{2}\)xy2 – \(\frac{3}{2}\)xy3
Verification:
L.H.S = (\(\frac{5}{4}\)x2 – \(\frac{3}{2}\)xy)(x + y + y2)
Putting x = 2, y = 2
= [\(\frac{5}{4}\)(2)2 – \(\frac{3}{2}\)(2)(2))(2 + 2 + (2)2)]
= [\(\frac{5}{4}\) × 4 – 6] [4 + 4]
= (5 – 6)(8)
= (-1)(8) = -8
R.H.S = \(\frac{5}{4}\)x3 – \(\frac{1}{4}\)x2y + \(\frac{5}{4}\)x2y2 – \(\frac{3}{2}\)xy2 – \(\frac{3}{2}\)xy3
Putting x = 2, y = 2
= \(\frac{5}{4}\)(2)3 – \(\frac{1}{4}\)(2)2(2) + \(\frac{5}{4}\)(2)2(2)2 – \(\frac{3}{2}\)(2)(2)2 – \(\frac{3}{2}\)(2)(2)3
= \(\frac{5}{4}\) × 8 – \(\frac{1}{4}\) × 4 × 2 + \(\frac{5}{4}\) × 4 × 4 – \(\frac{3}{2}\) × 2 × 4 – \(\frac{3}{2}\) × 2 × 8
= 10 – 2 + 20 – 12 – 24
= 30 – 38
= – 8
Hence verified L.H.S. = R.H.S.