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Short and Long Question: Algebraic Expressions - Class 7 PDF Download

Q1: Group the like terms together from the following expressions:
-8x2y, 3x, 4y,  -3/2x , 2x2y, -y
Ans:

Group of like terms are:
(i) -8x2y, 2x2y
(ii) 3x, -3/2x
(iii) 4y, -y 

Q2: Identify the pairs of like and unlike terms:
(i) −3/2x, y
(ii) -x, 3x
(iii) −1/2y2x, 3/2xy2
(iv) 1000, -2
Ans:

(i) −3/2x, y → Unlike Terms
(ii) -x, 3x → Like Terms
(iii) −1/2y2x, 3/2xy2 → Like Terms
(iv) 1000, -2 → Like Terms

Q3: Classify the following into monomials, binomial and trinomials.
(i) -6
(ii) -5 + x
(iii) 3/2x – y
(iv) 6x2 + 5x – 3
(v) z2 + 2
Ans:

(i) -6 is monomial
(ii) -5 + x is binomial
(iii) 3/2x – y is binomial
(iv) 6x2 + 5x – 3 is trinomial
(v) z2 + z is binomial

Q4: Identify the constant terms in the following expressions:
(i) -3 + 3/2x
(ii) 3/2 – 5y + y2
(iii) 3x2 + 2y – 1
Ans:

(i) Constant term = -3
(ii) Constant term = 3/2
(iii) Constant term = -1

Q5: Add:
(i) 3x2y, -5x2y, -x2y
(ii) a + b – 3, b + 2a – 1
Ans:

(i) 3x2y, -5x2y, -x2y
= 3x2y + (-5x2y) + (-x2y)
= 3x2y – 5x2y – x2y
= (3 – 5 – 1 )x2y
= -3x2y
(ii) a + b – 3, b + 2a – 1
= (a + b – 3) + (b + 2a – 1)
= a + b – 3 + b + 2a – 1
= a + 2a + b + b – 3 – 1
= 3a + 2b – 4

Q6: Subtract 3x2 – x from 5x – x2.
Ans:
 
(5x – x2) – (3x2 – x)
= 5x – x2 – 3x2 + x
= 5x + x – x2 – 3x2
= 6x – 4x2

Q7: Simplify combining the like terms:
(i) a – (a – b) – b – (b – a)
(ii) x2 – 3x + y– x – 2y2
Ans:

(i) a – (a – b) – b – (b – a)
= a – a + b – b – b + a
= (a – a + a) + (b – b – b)
= a – b
(ii) x2 – 3x + y2 – x – 2y2
= x2 + y2 – 2y2 – 3x – x
= x2y2 – 4x

Q8: Subtract 24xy – 10y – 18x from 30xy + 12y – 14x.
Ans:

(30xy + 12y – 14x) – (24xy – 10y – 18x)
= 30xy + 12y – 14x – 24xy + 10y + 18x

= 30xy – 24xy + 12y + 10y – 14x + 18x
= 6xy + 22y + 4x

Q9: From the sum of 2x2 + 3xy – 5 and 7 + 2xy – x2 subtract 3xy + x2 – 2.
Ans:

Sum of the given term is (2x+ 3xy – 5) + (7 + 2xy – x2)
= 2x2 + 3xy – 5 + 7 + 2xy – x2
= 2x2 – x2 + 3xy + 2xy – 5 + 7
= x2 + 5xy + 2
Now (x2 + 5xy + 2) – (3xy + x2 – 2)
= x2 + 5xy + 2 – 3xy – x2 + 2
= x2 – x2 + 5xy – 3xy + 2 + 2
= 0 + 2xy + 4
= 2xy + 4

Q10: Subtract 3x– 5y – 2 from 5y – 3x+ xy and find the value of the result if x = 2, y = -1.
Ans:

(5y – 3x2 + xy) – (3x2 – 5y – 2)
= 5y – 3x2 + xy – 3x2 + 5y + 2
= -3x2 – 3x2 + 5y + 5y + xy + 2
= -6x2 + 10y + xy + 2
Putting x = 2 and y = -1, we get
-6(2)2 + 10(-1) + (2)(-1) + 2
= -6 × 4 – 10 – 2 + 2
= -24 – 10 – 2 + 2
= -34

Q11: Simplify the following expressions and then find the numerical values for x = -2.
(i) 3(2x – 4) + x2 + 5
(ii) -2(-3x + 5) – 2(x + 4)
Ans:

(i) 3(2x – 4) + x2 + 5
= 6x – 12 + x2 + 5
= x2 + 6x – 7
Putting x = -2, we get
= (-2)2 + 6(-2) – 7
= 4 – 12 – 7
= 4 – 19
= -15
(ii) -2(-3x + 5) – 2(x + 4)
= 6x – 10 – 2x – 8
= 6x – 2x – 10 – 8
= 4x – 18
Putting x = -2, we get
= 4(-2) – 18
= -8 – 18
= -26

Q12: Find the value of t if the value of 3x2 + 5x – 2t equals to 8, when x = -1.
Ans:

3x2 + 5x – 2t = 8 at x = -1
⇒ 3(-1)2 + 5(-1) – 2t = 8
⇒ 3(1) – 5 – 2t = 8
⇒ 3 – 5 – 2t = 8
⇒ -2 – 2t = 8
⇒ 2t = 8 + 2
⇒ -2t = 10
⇒ t = -5
Hence, the required value of t = -5.

Q13: To what expression must 99x3 – 33x2 – 13x – 41 be added to make the sum zero?
Ans:

Given expression:
99x3 – 33x2 – 13x – 41
Negative of the above expression is
-99x3 + 33x2 + 13x + 41
(99x3 – 33x2 – 13x – 41) + (-99x3 + 33x2 + 13x + 41)
= 99x3 – 33x2 – 13x – 41 – 99x3 + 33x2 + 13x + 41
= 0
Hence, the required expression is -99x3 + 33x2 + 13x + 41

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