# RS Aggarwal Solutions for Class 7 Maths Exercise 6D Chapter 6 Algebraic Expressions

RS Aggarwal Solutions for Class 7 Maths Exercise 6D Chapter 6 Algebraic Expressions, Multiplication of two binomials is the topic covered in this exercise of RS Aggarwal Solutions for Class 7 Chapter 6. By using the distributive law of multiplication over addition twice, we may find their product. These RS Aggarwal Solutions for Class 7 Maths Chapter 6 Algebraic Expressions are extremely helpful for the students to clear all their doubts quickly.

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Find each of the following products:

1. (5x + 7) (3x + 4)

Solution:-

Suppose (a + b) and (c + d) are two binomials. By using the distributive law of multiplication over addition twice, we may find their product as given below.

(a + b) Ã— (c + d) = a Ã— (c + d) + b Ã— (c + d) = (a Ã— c + a Ã— d) + (b Ã— c + b Ã— d)

= ac + ad + bc + bd

Let,

a= 5x, b= 7, c= 3x, d= 4

Now,

= 5x Ã— (3x + 4) +7 Ã— (3x + 4)

= [(5x Ã— 3x) + (5x Ã— 4)] + [(7 Ã— 3x) + (7 Ã— 4)]

= [15x2 + 20x + 21x + 28]

= [15x2 + 41x + 28]

2. (4x – 3) (2x + 5)

Solution:-

Suppose (a – b) and (c + d) are two binomials. By using the distributive law of multiplication over addition twice, we may find their product as given below.

(a – b) Ã— (c + d) = a Ã— (c + d) – b Ã— (c + d) = (a Ã— c + a Ã— d) – (b Ã— c + b Ã— d)

= ac + ad – bc – bd

Let,

a= 4x, b= 3, c= 2x, d= 5

Now,

= 4x Ã— (2x + 5) -3 Ã— (2x + 5)

= [(4x Ã— 2x) + (4x Ã— 5)] – [(3 Ã— 2x) + (3 Ã— 5)]

= [8x2 + 20x – 6x – 15]

= [8x2 + 14x – 15]

3. (x – 6) (4x + 9)

Solution:-

Suppose (a – b) and (c + d) are two binomials. By using the distributive law of multiplication over addition twice, we may find their product as given below.

(a – b) Ã— (c + d) = a Ã— (c + d) – b Ã— (c + d) = (a Ã— c + a Ã— d) – (b Ã— c + b Ã— d)

= ac + ad – bc – bd

Let,

a= x, b= 6, c= 24x, d= 9

Now,

= x Ã— (4x + 9) -6 Ã— (4x + 5)

= [(x Ã— 4x) + (x Ã— 9)] – [(6 Ã— 4x) + (6 Ã— 9)]

= [4x2 + 9x – 24x – 54]

= [4x2 – 15x – 54]

4. (5y – 1) (3y – 8)

Solution:-

Suppose (a – b) and (c – d) are two binomials. By using the distributive law of multiplication over addition twice, we may find their product as given below.

(a – b) Ã— (c – d) = a Ã— (c – d) – b Ã— (c – d) = (a Ã— c – a Ã— d) – (b Ã— c – b Ã— d)

= ac – ad – bc + bd

Let,

a= 5y, b= 1, c= 3y, d= 8

Now,

= 5y Ã— (3y – 8) -1 Ã— (3y – 8)

= [(5y Ã— 3y) + (5y Ã— -8)] – [(1 Ã— 3y) + (1 Ã— -8)]

= [15y2 â€“ 40y â€“ 3y + 8]

= [15y2 â€“ 43y + 8]

5. (7x + 2y) (x + 4y)

Solution:-

Suppose (a + b) and (c + d) are two binomials. By using the distributive law of multiplication over addition twice, we may find their product as given below.

(a + b) Ã— (c + d) = a Ã— (c + d) + b Ã— (c + d) = (a Ã— c + a Ã— d) + (b Ã— c + b Ã— d)

= ac + ad + bc + bd

Let,

a= 7x, b= 2y, c= x, d= 4y

Now,

= 7x Ã— (x + 4y) +2y Ã— (x + 4y)

= [(7x Ã— x) + (7x Ã— 4y)] + [(2y Ã— x) + (2y Ã— 4y)]

= [7x2 + 28xy + 2yx + 8y2]

= [7x2 + 30xy + 8y2]

6. (9x + 5y) (4x + 3y)

Solution:-

Suppose (a + b) and (c + d) are two binomials. By using the distributive law of multiplication over addition twice, we may find their product as given below.

(a + b) Ã— (c + d) = a Ã— (c + d) + b Ã— (c + d) = (a Ã— c + a Ã— d) + (b Ã— c + b Ã— d)

= ac + ad + bc + bd

Let,

a= 9x, b= 5y, c= 4x, d= 3y

Now,

= 9x Ã— (4x + 3y) +5y Ã— (4x + 3y)

= [(9x Ã— 4x) + (9x Ã— 3y)] + [(5y Ã— 4x) + (5y Ã— 3y)]

= [36x2 + 27xy + 20yx + 15y2]

= [36x2 + 47xy + 15y2]

7. (3m â€“ 4n) (2m â€“ 3n)

Solution:-

Suppose (a – b) and (c – d) are two binomials. By using the distributive law of multiplication over addition twice, we may find their product as given below.

(a – b) Ã— (c – d) = a Ã— (c – d) – b Ã— (c – d) = (a Ã— c – a Ã— d) – (b Ã— c – b Ã— d)

= ac – ad – bc + bd

Let,

a= 3m, b= 4n, c= 2m, d= 3n

Now,

= 3m Ã— (2m â€“ 3n) -4n Ã— (2m â€“ 3n)

= [(3m Ã— 2m) + (3m Ã— -3n)] – [(4n Ã— 2m) + (4n Ã— -3n)]

= [6m2 â€“ 9mn â€“ 8mn + 12n2]

= [6m2 â€“ 17mn + 12n2]

8. (0.8x â€“ 0.5y) (1.5x â€“ 3y)

Solution:-

Suppose (a – b) and (c – d) are two binomials. By using the distributive law of multiplication over addition twice, we may find their product as given below.

(a – b) Ã— (c – d) = a Ã— (c – d) – b Ã— (c – d) = (a Ã— c – a Ã— d) – (b Ã— c – b Ã— d)

= ac – ad – bc + bd

Let,

a= 0.8x, b= 0.5y, c= 1.5x, d= 3y

Now,

= 0.8x Ã— (1.5x â€“ 3y) – 0.5y Ã— (1.5x â€“ 3y)

= [(0.8x Ã— 1.5x) + (0.8x Ã— -3y)] – [(0.5y Ã— 1.5x) + (0.5y Ã— -3y)]

= [1.2x2 â€“ 2.4xy â€“ 0.75yx + 1.5y2]

= [1.2x2 â€“ 3.15xy + 1.5y2]

9. ((1/5)x + 2y) ((2/3)x â€“ y)

Solution:-

Suppose (a + b) and (c – d) are two binomials. By using the distributive law of multiplication over addition twice, we may find their product as given below.

(a + b) Ã— (c – d) = a Ã— (c – d) + b Ã— (c – d) = (a Ã— c – a Ã— d) + (b Ã— c – b Ã— d)

= ac – ad + bc – bd

Let,

a= (1/5)x, b= 2y, c= (2/3)x, d= y

Now,

= (1/5)x Ã— ((2/3)x â€“ y) + 2y Ã— ((2/3)x â€“ y)

= [((1/5)x Ã— (2/3)x) + ((1/5)x Ã— -y)] + [(2y Ã— (2/3)x) + (2y Ã— -y)]

= [(2/15)x2 – (1/5)xy + (4/3)yx – 2y2)]

= [(2/15)x2 + (17/15) xy – 2y2]

10. ((2/5)x â€“ (1/2)y) (10x â€“ 8y)

Solution:-

Suppose (a – b) and (c – d) are two binomials. By using the distributive law of multiplication over addition twice, we may find their product as given below.

(a – b) Ã— (c – d) = a Ã— (c – d) – b Ã— (c – d) = (a Ã— c – a Ã— d) – (b Ã— c – b Ã— d)

= ac – ad – bc + bd

Let,

a= (2/5)x, b=(1/2)y, c= 10x, d= 8y

Now,

= (2/5)x Ã— (10x â€“ 8y) – (1/2)y Ã— (10x â€“ 8y)

= [((2/5)x Ã— 10x) + ((2/5)x Ã— -8y)] – [((1/2)y Ã— 10x) + ((1/2)y Ã— -8y)]

= [4x2 â€“ (16/5)xy â€“ 5yx + 4y2]

= [4x2 â€“ (41/5)xy + 4y2]

11. ((3/4)a + (2/3)b) (4a + 3b)

Solution:-

Suppose (a – b) and (c – d) are two binomials. By using the distributive law of multiplication over addition twice, we may find their product as given below.

(a + b) Ã— (c + d) = a Ã— (c + d) + b Ã— (c + d) = (a Ã— c + a Ã— d) + (b Ã— c + b Ã— d)

= ac + ad + bc + bd

Let,

a= (3/4)a, b=(2/3)b, c= 4a, d= 3b

Now,

= (3/4)a Ã— (4a + 3b) + (2/3)b Ã— (4a + 3b)

= [((3/4)a Ã— 4a) + ((3/4)a + 3b)] – [((2/3)b Ã— 4a) + ((2/3)b Ã— + 3b)]

= [3a2 + (9/4)ab + (8/3)ab + 2b2]

= [3a2 + ((27+32)/12)ab + 2b2]

= [3a2 + (59/12)ab + 2b2]

12. (x2 â€“ a2) (x â€“ a)

Solution:-

Suppose (a – b) and (c – d) are two binomials. By using the distributive law of multiplication over addition twice, we may find their product as given below.

(a – b) Ã— (c – d) = a Ã— (c – d) – b Ã— (c – d) = (a Ã— c – a Ã— d) – (b Ã— c – b Ã— d)

= ac – ad – bc + bd

Let,

a= x2, b= a2, c= x, d= a

Now,

= x2 Ã— (x â€“ a) – a2 Ã— (x â€“ a)

= [(x2 Ã— x) + (x2 Ã— -a)] – [(a2 Ã— x) + (a2 Ã— -a)]

= [x3 â€“ x2a â€“ a2x + a3]

13. (3p2 + q2) (2p2 â€“ 3q2)

Solution:-

Suppose (a + b) and (c – d) are two binomials. By using the distributive law of multiplication over addition twice, we may find their product as given below.

(a + b) Ã— (c – d) = a Ã— (c – d) + b Ã— (c – d) = (a Ã— c – a Ã— d) + (b Ã— c – b Ã— d)

= ac – ad + bc – bd

Let,

a= 3p2, b= q2, c= 2p2, d= 3q2

Now,

= 3p2Ã— (2p2 â€“ 3q2) + q2 Ã— (2p2 â€“ 3q2)

= [(3p2Ã— 2p2) + (3p2Ã— -3q2)] + [(q2 Ã— 2p2) + (q2 Ã— -3q2)]

= [6p4 â€“ 9p2q2 + 2q2p2 â€“ 3q4)]

= [6p4 â€“ 7p2q2 â€“ 3q4]

14. (2x2 â€“ 5y2) (x2 + 3y2)

Solution:-

Suppose (a – b) and (c + d) are two binomials. By using the distributive law of multiplication over addition twice, we may find their product as given below.

(a – b) Ã— (c + d) = a Ã— (c + d) – b Ã— (c + d) = (a Ã— c + a Ã— d) – (b Ã— c + b Ã— d)

= ac + ad – bc – bd

Let,

a= 2x2, b= 5y2, c= x2, d= 3y2

Now,

= 2x2 Ã— (x2 + 3y2) â€“ 5y2 Ã— (x2 + 3y2)

= [(2x2 Ã— x2) + (2x2 Ã— 3y2)] – [(5y2 Ã— x2) + (5y2 Ã— 3y2)]

= [2x4 + 6x2y2 â€“ 5y2x2 â€“ 15y4]

= [2x4 + x2y2 â€“ 15y4]

15. (x3 â€“ y3) (x2 + y2)

Solution:-

Suppose (a – b) and (c + d) are two binomials. By using the distributive law of multiplication over addition twice, we may find their product as given below.

(a – b) Ã— (c + d) = a Ã— (c + d) – b Ã— (c + d) = (a Ã— c + a Ã— d) – (b Ã— c + b Ã— d)

= ac + ad – bc – bd

Let,

a= x3, b= y3, c= x2, d= y2

Now,

= x3 Ã— (x2 + y2) â€“ y3 Ã— (x2 + y2)

= [(x3 Ã— x2) + (x3 Ã— y2)] – [(y3 Ã— x2) + (y3 Ã— y2)]

= [x5 + x3y2 â€“ y3x2 â€“ y5]

### Access other exercises of RS Aggarwal Solutions For Class 7 Chapter 6 – Algebraic Expressions

Exercise 6C Solutions