CBSE Class 7 Mathematics Algebraic Expression Worksheet Set 03

Algebraic Expressions

Question I. Fill in the blanks :
a. 4xy + 2xy is a __________________
(bionomial, trinomial, monomial)
b. The product of twice of ‘a’ and thrice of ‘b’ _________________
c. The constant term of 12 + 19a3b2 is _________________
d. The like terms of a2b, -b2a, -5ba2, 5ab , 3a3b, 4ab3 are __________
e. The value which satisfies an equation is called its _______________
f. Algebraic terms that have the same algebraic factors are called _______________.
g. The numerical factor of a term is called _______________.
h. A _______________ is product of factors.
i. Any expression with one or more terms is called a _______________.
j. A combination of constants and variables connected by the signs of the fundamental operations is called ____________
Answer:
a. monomial (Since \( 4xy + 2xy = 6xy \), it simplifies to a single term)
b. \( 6ab \) (Multiply \( 2a \) by \( 3b \) to get \( 2 \times 3 \times a \times b = 6ab \))
c. 12 (The term without any variables is the constant term)
d. \( a^2b \) and \( -5ba^2 \) (Both terms have the identical variable factors \( a^2b \))
e. solution (or root)
f. like terms
g. numerical coefficient (or coefficient)
h. term
i. polynomial
j. algebraic expression
In simple words: Add like terms together first before naming an expression. Numbers without letters are called constants.

Exam Tip: Always simplify terms by combining like terms before classifying an expression as a monomial, binomial, or trinomial.

Question II. a. (p-q) – (p+q) = _______________
Answer:
Remove the brackets and change the signs inside the second bracket:
\[ (p - q) - (p + q) = p - q - p - q \]
Group like terms:
\[ = (p - p) + (-q - q) = 0 - 2q = -2q \]
Hence, \( (p - q) - (p + q) = \mathbf{-2q} \).
In simple words: The letter p cancels out completely, leaving minus 2q.

Exam Tip: Be careful to change both signs inside the second bracket when subtracting.

Question II. b. _______________ should be subtracted from 3x3- 1 to get x3
Answer:
Let the required expression be \( E \).
According to the statement:
\[ (3x^3 - 1) - E = x^3 \]
Rearrange to solve for \( E \):
\[ E = (3x^3 - 1) - x^3 \]
\[ E = (3x^3 - x^3) - 1 = 2x^3 - 1 \]
Therefore, \( 2x^3 - 1 \) should be subtracted.
In simple words: Subtract the final result from the starting expression to find what was taken away.

Exam Tip: Check your answer by subtracting \( 2x^3 - 1 \) from \( 3x^3 - 1 \) to see if you get \( x^3 \).

Question II. c. The sum of -4pqr, 5pqr and 7pqr is _______________
Answer:
Add the like terms by adding their numerical coefficients:
\[ (-4pqr) + 5pqr + 7pqr = (-4 + 5 + 7)pqr \]
\[ = (1 + 7)pqr = 8pqr \]
The sum is \( 8pqr \).
In simple words: All three parts have the same letters, so just add minus 4, 5, and 7 together.

Exam Tip: Keep the variable part unchanged and add only the numbers in front.

Question II. d. The (x+2) + (x-2) = _______________
Answer:
Add the two binomials by removing the brackets:
\[ (x + 2) + (x - 2) = x + 2 + x - 2 \]
Group like terms:
\[ = (x + x) + (2 - 2) = 2x + 0 = 2x \]
Hence, \( (x + 2) + (x - 2) = \mathbf{2x} \).
In simple words: Plus 2 and minus 2 make zero, leaving two x terms added together.

Exam Tip: Opposite constant numbers cancel each other out to zero.

Question III. a. Find the sum of 2x2 - 3y2, 9x2 + 6y2, -3x2 -5y2
Answer:
Write down the sum of the given expressions:
\[ (2x^2 - 3y^2) + (9x^2 + 6y^2) + (-3x^2 - 5y^2) \]
Combine like terms together:
- For \( x^2 \) terms: \( 2x^2 + 9x^2 - 3x^2 = (2 + 9 - 3)x^2 = 8x^2 \)
- For \( y^2 \) terms: \( -3y^2 + 6y^2 - 5y^2 = (-3 + 6 - 5)y^2 = -2y^2 \)

Sum = \( 8x^2 - 2y^2 \)
In simple words: Put the matching x-squared terms together and the y-squared terms together, then add their numbers.

Exam Tip: Group like terms inside brackets before combining coefficients to avoid sign errors.

Question III. b. Subtract (a2 + b2 + 2ab) from (a2 + b2 – 2ab)
Answer:
Write the second expression first and subtract the first expression:
\[ (a^2 + b^2 - 2ab) - (a^2 + b^2 + 2ab) \]
Change the signs of all terms inside the subtracted bracket:
\[ = a^2 + b^2 - 2ab - a^2 - b^2 - 2ab \]
Group like terms:
\[ = (a^2 - a^2) + (b^2 - b^2) + (-2ab - 2ab) \]
\[ = 0 + 0 - 4ab = -4ab \]
In simple words: The squared terms cancel out to zero, leaving minus 2ab and minus 2ab, which makes minus 4ab.

Exam Tip: The expression written after "from" must always be placed first in the subtraction setup.

Question III. c. Write the term containing a2 and find its coefficient
i) 16a2 – 4b2 +10
ii) 22b2 - a2
Answer:
i) In \( 16a^2 - 4b^2 + 10 \):
- Term containing \( a^2 \) is \( 16a^2 \)
- Coefficient of \( a^2 \) is 16

ii) In \( 22b^2 - a^2 \):
- Term containing \( a^2 \) is \( -a^2 \)
- Coefficient of \( a^2 \) is -1
In simple words: Pick the term with a-squared. The number multiplying it is its coefficient.

Exam Tip: When a term has a minus sign with no visible number like \( -a^2 \), its coefficient is \( -1 \).

Question III. d. Simplify 2 (x2 + 2xy) + 5 – xy – y2
Answer:
Expand the bracket by multiplying each inside term by 2:
\[ 2(x^2 + 2xy) = 2x^2 + 4xy \]
Substitute this back into the full expression:
\[ 2x^2 + 4xy + 5 - xy - y^2 \]
Combine like terms:
- Terms with \( xy \): \( 4xy - xy = 3xy \)

Rearrange the terms:
\[ = 2x^2 + 3xy - y^2 + 5 \]
In simple words: Multiply out the bracket first, then combine the matching xy terms.

Exam Tip: Remember to multiply both terms inside the bracket by the number outside.

Question III. e. Find the value of the given expressions when a=0, b=-1, c=1
i) a3 – b3
ii) a2 + 2ab +b3
iii) 3ab + 3ac + c2
iv) \frac{5a}{10} - 4b
Answer:
Given values: \( a = 0 \), \( b = -1 \), and \( c = 1 \).

i) For \( a^3 - b^3 \):
\[ = (0)^3 - (-1)^3 = 0 - (-1) = 0 + 1 = 1 \]

ii) For \( a^2 + 2ab + b^3 \):
\[ = (0)^2 + 2(0)(-1) + (-1)^3 = 0 + 0 + (-1) = -1 \]

iii) For \( 3ab + 3ac + c^2 \):
\[ = 3(0)(-1) + 3(0)(1) + (1)^2 = 0 + 0 + 1 = 1 \]

iv) For \( \frac{5a}{10} - 4b \):
\[ = \frac{5(0)}{10} - 4(-1) = 0 - (-4) = 0 + 4 = 4 \]
In simple words: Put zero, minus 1, and 1 into each problem. Multiplying any number by zero gives zero.

Exam Tip: Any negative number raised to an odd power remains negative, so \( (-1)^3 = -1 \).