The DAV Class 7 Maths Solutions and **DAV Class 7 Maths Chapter 6 Worksheet 2** Solutions of Algebraic Expressions offer comprehensive answers to textbook questions.

## DAV Class 7 Maths Ch 6 WS 2 Solutions

Question 1. Find the product:

(i) (x^{2} + 3xy) (4x)

Answer:

(x^{2} + 3xy) (4x)

= x^{2} × 4x + 3xy × 4x

= 4x^{3} + 12 ×^{2}y

(ii) 9 pqr (2 p^{2}q – 3q^{2}r)

Answer:

9pqr (2p^{2}q – 3q^{2}r)

= 9 pqr × 2 p^{2}q – 9 pqr × 3q^{2}r

= 18p^{3}q^{2}r – 27 pq^{3}r^{2}

(iii) \(\frac{3}{4}\)a^{2}(\(\frac{2}{3}\)b^{2} + 8ab)

Answer:

\(\frac{3}{4}\)a^{2}(\(\frac{2}{3}\)b^{2} + 8ab)

= \(\frac{3}{4}\)a^{2} × \(\frac{2}{3}\) b^{2} + \(\frac{3}{4}\) a^{2} × 8 ab

= \(\frac{1}{2}\)a^{2}b^{2} + 6a^{3}b

Question 2.

Find the following products and then evaluate when x = 2, y = – 1.

(i) (7x + 9y^{2}) (3xy^{2})

Solution:

(i)(7x + 9y^{2}) (3xy^{2})

Put x = 2 and y = -1

= 7x × 3 xy^{2} + 9y^{2} × 3 xy^{2}

= 21x^{2}y^{2} + 27xy^{4}

= 21 (2)^{2} (- 1)^{2} + 27(2) (- 1)^{4}

= 21 × 4 + 27 × 2 × 1

= 84 + 54

= 138

(ii) -11x(2y^{5} – \(\frac{3}{11}\)x^{2}y^{3})

Answer:

-11x(2y^{5} – \(\frac{3}{11}\)x^{2}y^{3})

= -11x × 2y^{5} – 11x × –\(\frac{3}{11}\) x^{2}y^{3}

= -22xy^{5} + 3x^{3}y^{3}

Put x = 2, y = -1

= -22(2)(-1)^{5} + 3(2)^{3}(-1)^{3}

= -22 × 2 × -1 + 3 × 8 × – 1

= 44 – 24

= 20

(iii) 0.5x (2x^{2}y^{2} + 1.5 xy^{3}) = 0.5x × 2x^{2}y^{2} + 0.5x × 1.5xy^{3}

= 1.0 x^{3}y^{2} + 0.75 x^{2}y^{3}

Put x = 2, y = – 1

= 1.0 (2)^{3} (-1)^{2} + 0.75(2)^{2} (-1)^{3}

= 1.0 × 8 × 1-0.75 × 4 × -1

= 8 + 3.00 = 11

Question 3.

Find the product of 8s^{2}(t^{2} – 2st) and verify the result when s = 1 and t = 5.

Answer:

8s^{2} (t^{2} – 2st)

= 8s^{2}t^{2} – 8s^{2} x 2st

= 8s^{2}t^{2} – 16s^{3}t

Verification:

L.H.S. = 8s^{2} (t^{2} – 2st)

Put s = 1 and t = 5 = 8(1)^{2} [(5)^{2} – 2(1) (5)]

= 8 × 1 [25 -10]

= 8 × 15

= 120

R.H.S. = 8s^{2}t^{2} -16 s^{3}t

Put s = 1 and t = 5

= 8(1)^{2}(5)^{2} – 16(1)^{3}(5)

= 8 × 1 × 25 – 16 × 1 × 5

= 200 – 80

= 120

Hence Verified L.H.S. = R.H.S.

Question 4.

Find the product of \(\frac{2}{7}\)x^{2} (7y + 14x) and verify the result when x = 2 and y = 3.

Answer:

\(\frac{2}{7}\) x^{2} (7y + 14x) = \(\frac{2}{7}\) x^{2} × 7y + \(\frac{2}{7}\) x^{2} × 14x

= 2x^{2}y + 4x^{3}

Verification:

L.H.S. = jx^{2}(7y + 14x)

Put x = 2 and y = 3

= \(\frac{2}{7}\)(2)^{2} [7(3) + 14 (2)]

= \(\frac{2}{7}\) × 4 [21 + 48]

= \(\frac{2}{7}\) × 49 = 56

R.H.S. = 2 x^{2}y + 4x^{3}

= 2(2)^{2} (3) + 4 (2)^{3}

= 2 × 4 × 3 + 4 × 8

= 24 + 32

= 56

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

Question 5.

Find the product of 0.1xy (3x + 2y) and verify the result when x = 5 and y = – 1.

Answer:

0.2xy(3x + 2y) = 0.2 xy × 3x + 0.2 xy × 2y

= 0.6 x^{2}y + 0.4 xy^{2}

Verification:

L.H.S. = 0.2 xy (3x + 2y)

Put x = 5 and y = – 1

= 0.2(5)(- 1) [3(5) + 2(-1)] = -1.0 [15 – 2]

= -1 × 13

=-13

= 0.6x^{2}y + 0.4 xy^{2}

= 0.6(5)^{2}(-1) + 0.4(5) (-1)^{2}

= 0.6 × 25 × – 1 + 0.4 × 5 × 1

= – 15 + 2

= – 13

Hence Verified L.H.S. = R.H.S.

Question 6.

Simplify:

(i) ab (a^{2} – b^{2}) + b^{3} (a- 2b)

Answer:

ab(a^{2} – b^{2}) + b^{3} (a – 2b)

= ab × a^{2} – ab × b^{2} + b^{3} a – b^{3} × 2b

= a^{3}b – ab^{3} + ab^{3} – 2b^{4}

= a^{3}b – 2b^{4}

(ii) -6 x(xy + 2y^{2}) – 3y^{2}(2x^{2} + y)

Answer:

– 6x (xy + 2y^{2}) – 3y^{2} (2x^{2} + y)

= – 6x × xy – 6x × 2y^{2} – 3y^{2} × 2x^{2} – 3y^{2} × y

= – 6x^{2}y – 12xy^{2} – 6x^{2}y^{2} – 3y^{3}

(iii) 2x (x – y^{2}) – 3y (xy + 2x) – xy (x + y)

Answer:

2x(x – y^{2}) – 3y (xy + 2x) – xy (x + y)

= 2x × x – 2x × y^{2} – 3y × xy – 3y × 2x – xy × x – xy × y

= 2x^{2} – 2xy^{2} – 3xy^{2} – 6xy – x^{2}y – xy^{2}

= 2x^{2} – 6xy^{2} – x^{2}y – 6xy