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.