# DAV Class 7 Maths Chapter 6 Worksheet 3 Solutions

The DAV Class 7 Maths Solutions and DAV Class 7 Maths Chapter 6 Worksheet 3 Solutions of Algebraic Expressions offer comprehensive answers to textbook questions.

## DAV Class 7 Maths Ch 6 WS 3 Solutions

Question 1.
Find the product of the following binomials:
(i) (5x + 3) (2x + 4)
(5x + 3) (2x + 4) = 5x (2x + 4) + 3 (2x + 4)
= 5x × 2x + 5x × 4 + 3 × 2x + 3 × 12
= 10x2 + 20x + 6x + 36
= 10x2 + 26x + 36

(ii) (7p – 3q) (2p + 5q)
(7p – 3q) (2p + 5q)
= 7p (2p + 5q) – 3q (2p + 5q)
= 7p × 2p + 7p × 5q – 3q × 2p – 3q × 5q
= 14p2 + 35pq – 6pq – 15q2
= 14p2 + 29pq – 15q2

(iii) ($$\frac{2}{p}$$a + b) [a2 – $$\frac{1}{5}$$b2)
($$\frac{2}{p}$$a + b) [a2 – $$\frac{1}{5}$$b2)
= $$\frac{2}{p}$$a(a2 – $$\frac{1}{5}$$b2) + b(a2 – $$\frac{1}{5}$$b2)
= $$\frac{2}{p}$$a × a2 + $$\frac{2}{p}$$a × $$\frac{-1}{5}$$b2 + ba2 – $$\frac{1}{5}$$b2 × b
= $$\frac{2 a^3}{p}-\frac{2 a b^2}{5 p}$$ + a2b – $$\frac{-1}{5}$$b3

(iv) (m2n -5)(6 – mn2)
(m2n – 5) (6 – mn2)
= m2n (6 – mn2) – 5 (6 – mn2)
= 6m2n – m2n x mn2 -5 × 6 – 5x – mn2
= 6m2n – m3n3 – 30 + 5mn2

(v) (4x2 + 7y3) (3xy – 2y2)
(4x2 + 7y3) (3xy – 2y2)
= 4x2 (3xy – 2y2) + 7y3 (3xy – 2y2)
= 4x2 × 3xy – 4x2 × 2y2 + 7y3 × 3xy – 7y3 × 2y2
= 12x3y – 8x2y2 + 21xy4 – 14y5

(vi) (1.1x + 2.7y) (1.1x – 2.7y)
(1.1x + 2.7y) (1.1 x – 2.7y)
= 1.1x (1.1x – 2.7y) + 2.7y (1.1x – 2.7y)
= 1.1 x × 1.1 x – 1.1 x × 2.7y + 2.7y × 1.1x – 2.7y × 2.7 y
= 1.21x2 + 2.97xy + 2.97xy – 7.29y2
= 1.21x2 – 7.29 y2

Question 2.
Multiply the following binomials and verify the residt for the given values.
(i) (2x2 – 5y) (5x + 2y2); x = 2, y = -1
(2x2 – 5y) (5x + 2y2)
= 2x2 × 5x + 2x2 × 2y2 – 5y × 5x – 5y × 2y2
= 10x3 + 4x2y2 – 25xy – 10y3

Verifications:
L.H.S = (2x2 – 5y)(5x + 2y2)

Put x = 2, y = -1
= [2 x (2)2 – 5 x – 1] [5 x 2 + 2 (- 1)2]
= [2 x 4 + 5] [10 + 2 x 1]
= [8 + 5] [10 + 2]
= 13 x 12 = 156
= 10x3 + 4x2y2 – 25 xy – 10y3
= 10(2)3 + 4(2)2 (- 1)2 – 25(2) (- 1) – 10 (- 1)3
= 10 × 8 + 4 × 4 × 1 + 50 – 10 × -1
= 80 + 16 + 50 + 10
= 156
L.H.S. = R.H.S.
Hence verified L.H.S = R.H.S

(ii) ([-$$\frac{1}{4}$$a + $$\frac{1}{5}$$b]([$$\frac{1}{4}$$a + $$\frac{1}{5}$$b); a = 8,b = 5

Hence verified L.H.S = R.H.S

(iii) (m + n) (2m – 3n)
m (2m – 3n) + n (2m – 3n)
2m2 – 3mn + 2mn – 3n2 = 2m2 – mn – 3n2
So, (m + n)(2m – 3n) = 2m2 -mn- 3 n2
Now, to verify the result, put m = -2, n = 0 in LHS and RHS.
LHS = (-2 + 0) [2 x (-2)-3(0)]
= (-2) (-4 – 0) = (-2) (-4) = 8
RHS = 2m2 -mn- 3n2
= 2(-2)2 – (-2) (0) – 3(0)2
= 2 × 4 – 0 – 0
= 8
LHS = RHS
Hence, verified.

(iv) (0.1p + 0.2q) (0.2q – 0.1p); p = 10, q = 5
(0.1 p + 0.2q) (0.2q – 0.1 p) = 0.1 p (0.2q – 0.1 p) + 0.2q (0.2q – 0.1 p)
= 0.1 p × 0.2q – 0.1 p × 0.1 p + 0.2q × 0.2q – 0.2q × 0.1p
= 0.02pq – 0.01 p2 + 0.04q2 – 0.02 pq
= – 0.01 p2 + 0.04q2

Verification:
L.H.S. = (0.1p + 0.2q) (0.2q – 0.1 p)
Putting p = 10, q = 5
= (0.1 × 10 + 0.2 × 5) (0.2 × 5 – 0.1 × 10)
= (1 + 1) (1 – 1)
= 2 × 0
= 0
R.H.S. = -0.01 (10)2 + 0.04 × (5)2
= – 0.01 × 100 + 0.04 × 25
= -1 + 1
= 0
Hence verified L.H.S. = R.H.S.

Question 3.
Simplify the following:
(i) (2x – 3y)(3x + y) + (x + 2y)(x – y)
(2x – 3y)(3x + y) + (x + 2y)(x – y)
= 2x (3x + y) – 3y (3x + y) + x(x – y) + 2y (x – y)
= 2x × 3x + 2x × y – 3y × 3x – 3y × y + x × x – x × y + 2y × x – 2y × y
= 6x2 + 2xy – 9xy – 3y2 + x2 – xy + 2xy – 2y2
= 7x2 – 6xy – by2

(ii) ($$\frac{2}{3}$$x + 4)($$\frac{3}{2}$$x + 6) – ($$\frac{1}{7}$$x – 1) ($$\frac{1}{7}$$x + 1)
($$\frac{2}{3}$$x + 4)($$\frac{3}{2}$$x + 6) – ($$\frac{1}{7}$$x – 1) ($$\frac{1}{7}$$x + 1)
= ($$\frac{2}{3}$$x × $$\frac{3}{2}$$x × $$\frac{1}{7}$$x × $$\frac{1}{7}$$x ×6 + 4 × $$\frac{3}{2}$$x + 4 × 6) – ($$\frac{1}{7}$$x × $$\frac{1}{7}$$x + $$\frac{1}{7}$$x – $$\frac{1}{7}$$x – 1)
= x2 + 4x + 6x + 24 – $$\frac{1}{49}$$x2 + 1
= x2 + 10x + 24 – $$\frac{1}{49}$$x2 + 1
= $$\frac{48}{49}$$x2 + 10x + 25

(iii) (7abc + b2) (7 -bc) + c (2b3 – 9ab)
(7abc + b2) (7 – bc) + c (2b3 – 9ab)
= 7abc (7 – bc) + b2 (7 – bc) + c × 2b3 – 9ab × c
= 7abc × 7 – 7 abc × bc + b2 × 7 – b2 × bc + 2b3c – 9abc
= 49abc – 7ab2c + 7b2 – b3c + 2b3c – 9abc
= 40abc – 7ab2c2 + 7b2 + b3c

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

(v) (x + 2y) (- 3x – y) – (x + y) (x – y) + (x – 2y) (- 2x + y)