Algebraic Expressions (Exercise 1) with Solutions

1. Classify the algebraic expressions as monomials, binomials and trinomials.

a. 6x2 + 3

b. 5x4

c. 2a2 + 6a +3

d. xy + 3y2

e. x3y4z3

Ans:

a. Binomial

b. Monomial

c. Trinomial

d. Binomial

e. Monomial

Must Read: Up-Hill Stanza-Wise Summary

2. Add the following algebraic expressions.

a. 3x, 5x, (-9x) and 2x

b. xy2 + 3x + 5, 4xy2 + (-8x)

c. 9x2 – xy + 8y2, (-4x2) + 3xy – 2y2 and 5x2 + 2xy +7y2

Ans:

a. = 3x + 5x + (-9x) + 2x

= 10x + (-9x)

= 10x – 9x

= x

b. = [xy2 + 3x + 5] + [4xy2 + (-8x)]

= [xy2 + 3x + 5] + [4xy2 – 8x]

= xy2 + 3x +5 + 4xy2 – 8x

= xy2 + 4xy2 + 3x – 8x + 5

= 5xy2 – 5x + 5

c. = [9x2 – xy + 8y2] + [(-4x2) + 3xy – 2y2] + [5x2 + 2xy +7y2]

= [9x2 – xy + 8y2] + [-4x2 + 3xy – 2y2] + [5x2 + 2xy +7y2]

= 9x2 – xy + 8y2 – 4x2 + 3xy – 2y2 + 5x2 + 2xy + 7y2

= 9x2 – 4x2 + 5x2 – xy + 3xy + 2xy + 8y2 – 2y2 + 7y2

= 10x2 + 4xy + 13y2

3. Subtract the following expressions.

a. 6pqr from 8pqr

b. 2a2bc + 6ab – 5 from -4a2bc + 5ab + 6

c. 6a4 – 9a3 – 4a from 7a4 + 6a3 – 13a

Ans:

a. = 8pqr – 6pqr

= 2pqr

b. = (-4a2bc + 5ab + 6) – (2a2bc + 6ab – 5)

= -4a2bc + 5ab + 6 – 2a2bc – 6ab + 5

= -4a2bc – 2a2bc + 5ab – 6ab + 6 + 5

= -6a2bc – ab + 11

c. = (7a4 + 6a3 – 13a) – (6a4 – 9a3 – 4a)

= 7a4 + 6a3 – 13a – 6a4 + 9a3 + 4a

= 7a4 – 6a4 + 6a3 + 9a3 – 13a + 4a

= a4 + 15a3 – 9a

4. Subtract the sum of 2.5a2b + 5ab + 4 and 5.1a2b + 7ab – 8 from 6.5a2b + 7ab – 10

Ans: = 6.5a2b + 7ab – 10 – (2.5a2b + 5ab + 4 + 5.1a2b + 7ab – 8)

= 6.5a2b + 7ab – 10 – (2.5a2b + 5.1a2b + 5ab + 7ab + 4 – 8)

= 6.5a2b + 7ab – 10 – (7.6a2b + 12ab – 4)

= 6.5a2b + 7ab – 10 – 7.6a2b – 12ab + 4

= 6.5a2b – 7.6a2b + 7ab – 12ab – 10 + 4

= -1.1a2b – 5ab – 6

5. Two adjacent sides of a rectangle are 3a2 – 6b2 and 2a2 + 8b2. What will be the perimeter?

Ans: Length of the rectangle: 3a2 – 6b2

Breadth of the rectangle: 2a2 + 8b2

Perimeter of rectangle: 2(l+b)

Therefore,

= 2(3a2 – 6b2 + 2a2 + 8b2)

= 2(3a2 + 2a2 – 6b2 + 8b2)

= 2(5a2 + 2b2)

= 10a2 + 4b2

6. What should be added to 6a2 – 3b2 to get 4a2 – 5ab + 4b2?

Ans: To get the number we will need to subtract 6a2 – 3b2 from 4a2 – 5ab + 4b2. Therefore,

= 4a2 – 5ab + 4b2 – (6a2 – 3b2)

= 4a2 – 5ab + 4b2 – 6a2 + 3b2

= 4a2 – 6a2 – 5ab + 4b2 + 3b2

= -2a2 – 5ab + 7b2

7. What should be subtracted from 19a3 + 6a2b2 to get 4a3 – 4a2b2 + 6?

Ans: To get the number, we will subtract 4a3 – 4a2b2 + 6 from 19a3 + 6a2b2. Therefore,

= 19a3 + 6a2b2 – (4a3 – 4a2b2 + 6)

= 19a3 + 6a2b2 – 4a3 + 4a2b2 – 6

= 19a3 – 4a3 + 6a2b2 + 4a2b2 – 6

= 15a3 + 10a2b2 – 6

So, this was Exercise 1 of Algebraic Expressions. Stay tuned for more exercises.