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)
Answer:
(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)
Answer:
(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)
Answer:
(\(\frac{2}{p}\)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)
Answer:
(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)
Answer:
(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)
Answer:
(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
Answer:
(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
Answer:

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

(iii) (m + n) (2m – 3n)
Answer:
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
Answer:
(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)
Answer:
(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)
Answer:
(\(\frac{2}{3}\)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)
Answer:
(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)
Answer:
(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)
Answer:
(x + 2y) (- 3x – y) – (x + y) (x – y) + (x – 2y) (- 2x + y)
= -3x² – xy – 6xy – 2y² – x² + xy – xy + y² – 2x² + xy +xy – 2y²
= -6x² – 2xy – 3y²

(vi) (a² + b²)(a² + b²) – (a² – b²) (a² – b²)
Answer:
(a² + b²) (a² + b²) – (a² – b²) (a² – b²)
= [a²(a² + b²) + b² (a² + b²)] – [a² (a² – b²) – b² (a² – b²)]
= (a⁴ + a²b² + a²b² + b⁴) – (a⁴ – a²b² – a²b² + b⁴)
= (a⁴ + 2a²b² + b⁴) – (a⁴ – 2a²b² + b⁴)
= a⁴ + 2a²b² + b⁴ – a⁴ + 2a²b² – b⁴
= 2a²b² + 2a²b²
= 4a²b²