Find the product:((3/4)a + (2/3)b) (4a + 3b)

1.	Find the product:((3/4)a + (2/3)b) (4a + 3b)
Answer» ((3/4)a + (2/3)b) (4a + 3b) 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) × (c + d) = a × (c + d) + b × (c + d) = (a × c + a × d) + (b × c + b × d) = ac + ad + bc + bd Let, a= (3/4)a, b=(2/3)b, c= 4a, d= 3b Now, = (3/4)a × (4a + 3b) + (2/3)b × (4a + 3b) = [((3/4)a × 4a) + ((3/4)a + 3b)] – [((2/3)b × 4a) + ((2/3)b × + 3b)] = [3a² + (9/4)ab + (8/3)ab + 2b²] = [3a² + ((27+32)/12)ab + 2b²] = [3a² + (59/12)ab + 2b²]

Answer»

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

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) × (c + d) = a × (c + d) + b × (c + d) = (a × c + a × d) + (b × c + b × d)

= ac + ad + bc + bd

Let,

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

Now,

= (3/4)a × (4a + 3b) + (2/3)b × (4a + 3b)

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

= [3a² + (9/4)ab + (8/3)ab + 2b²]

= [3a² + ((27+32)/12)ab + 2b²]

= [3a² + (59/12)ab + 2b²]

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