## NCERT Solutions for class 7 Maths Chapter 12: Algebraic Expressions## Exercise 12.1
One half of the sum = ½ (x + y)
One fourth of the product = 1/4 × pq = pq/4 Or 1/4 pq
Squared number y = y Squared numbers added = x
Three times the product = 3 × mn = 3mn Number 5 added = 5 + 3mn
Product subtracted from 10 = 10 - yz
Product of numbers a and b = ab Sum subtracted from the product = ab - (a + b)
The factors of (x - 3) are: - X
- -3
The factors of (1 + x + x - 1
- X
- x
^{2}
x - x
- x
y - y y × 1 - y × y = y (1 - y The factors of y - y - y
- -y
^{3}
-y - -1
- y
- y
- y
The factors of 5xy - 5xy
^{2}- 5
- x
- y
- y
- 7x
^{2}y- 7
- x
- x
- y
The factors of - ab + 2b - - ab
- -1
- a
- b
- 2b
^{2}- 2
- b
- b
- - 3a
^{2}- -3
- a
- a
5 is the coefficient of t in the expression 5t. A single number does not have any coefficient. For example, The number 3 does not have any coefficient.
y is the coefficient of t in the expression yt. A single number does not have any coefficient. For example, The number 3 does not have any coefficient.
- An expression that contains only one term is known as the monomials expression.
- An expression that contains two unlike terms is known as the binomial expression.
- An expression that contains three terms is known as trinomials expression.
- The terms with different factors in an expression are called unlike terms.
- The terms with the same factors in an expression are called as like terms.
-7 x = -7, x 5/2 x = 5/2, x Both the expression contains the variable x with numbers as coefficient. Hence, these terms are considered as like terms.
14xy = 14, x, y 4yx = 42, y, x Both the expression contain the variables x and y with numbers as coefficient. Hence, these terms are considered as like terms.
4m 4mp It contains two different expressions. Hence, these terms are considered as unlike terms.
12xz = 12, x, z 12x It contains two different expressions. Hence, these terms are considered as unlike terms.
The like terms in the given series of expression are: - xy - 4yx 8x 7 y, y - 100 x, 3 x - 11yx, 2xy
The like terms in the given series of expression are: - p - 5p 13p 10pq, - 7qp, 78qp 7p, 2405p 8q, - 100q - 23, 41 ## Exercise 12.2
We will combine all the terms with b together for easy calculation. = 21b + 7b - 20b - 32 = 28b - 20b - 32 = 8b - 32
= 7z = 7z
Let's open the brackets. = p - p + q - q - q + p Putting together the p and q terms, = p - p + p + q - q - q = p - q
Let's open the brackets. = 3a - 2b - ab - a + b - ab + 3ab + b - a Putting together the a, b, and ab terms, = 3a - a - a - 2b + b + b - ab - ab + 3ab = a + 0 + ab = a + ab
Putting together the x = - 5x = - 4x Or 8yx
Let's open the brackets. = 3y Putting together the y = 3y = 4y = 4y
= 3mn - 5mn + 8mn - 4mn Let's separate positive and negative numbers for easy calculation. = 3mn + 8mn - 5mn - 4mn = 11mn - 9mn = 2mn
= t - t - z + z - 8tz + 3tz = 0 + 0 + (-5tz) = - 5tz
= - 7mn + 5 + 12mn + 2 + 9mn - 8 - 2mn - 3 Let's separate out the terms. = - 7mn + 12mn + 9mn - 2mn + 5 + 2 - 8 - 3 = 21mn - 9mn + 7 - 11 = 12mn - 4
Let's separate out the terms. = a - a + a + b + b - b - 3 + 3 + 3 = a + b + 3
Let's separate out the x, y, and xy terms. = 14x - 7x + 10y - 10y - 12xy + 8xy + 4xy - 13 + 18 = 7x + 0 - 12xy + 12xy + 5 = 7x + 0 + 0 + 5 = 7x + 5
= 5m - 4m + 2m - 7n + 3n + 2 - 3mn - 5 = 3m - 4n - 3mn - 3
= 4x = 9x
= 3p = 3p = 10p = 0 + 5pq + 5 = 5pq + 5
Let's separate the a, b, and ab terms. = 4a - 4a + 4b - 4b + ab - ab = 0 + 0 + 0 = 0
= x = - x
= y = 6y
= -12xy - 6xy = -18xy
= (a + b) - (a - b) = a + b - a + b = a - a + b + b = 0 + 2b = 2b
= b (5 - a) - a (b - 5) = 5b - ab - ab + 5a = 5a + 5b - 2ab
= 4m = 4m = 4m = 5m
= 5x - 10 - (- x = 5x - 10 + x = x = x
= 3ab - 2a = 3ab - 2a = - 2a = - 7a
= 5p = 5p = 5p = 8p
x A = 2x A = 2x A = 2x A = x Thus,
^{2} + xy + y^{2} to obtain 2x^{2} + 3xy.
2a + 8b + 10 - A = - 3a + 7b + 16 A = 2a + 8b + 10 - (- 3a + 7b + 16) A = 2a + 8b + 10 + 3a - 7b - 16 A = 2a + 3a + 8b - 7b + 10 - 16 A = 5a + b - 6 Thus,
Let the expression to be subtracted be A. 3x A = 3x A = 3x A = 3x A = 4x A = 4x Thus,
^{2} - 4y^{2} + 5xy + 20 to obtain - x^{2} - y^{2} + 6xy + 20.
= [3x - y + 11 - y - 11] - (3x - y - 11) = [3x - y - y + 11 - 11] - (3x - y - 11) = [3x - 2y + 0] - (3x - y - 11) = [3x - 2y] - (3x - y - 11) = 3x - 2y - 3x + y + 11 = 3x - 3x - 2y + y + 11 = 0 - y + 11 = - y + 11
= [4 + 3x + 5 - 4x + 2x = [2x = [2x = 2x = 2x = 0 + 2x + 4 = 2x + 4 ## Exercise 12.3
(2) - 2 = 0
= 3m - 5 = 3 × 2 - 5 = 6 - 5 = 1
= 9 - 5m = 9 - 5 × 2 = 9 - 10 = -1
Substituting the value of m = 2, we get: = 3m = 3 × (2) = 3 × 2 × 2 - 2 × 2 - 7 = 12 - 4 - 7 = 1
= 5m/2 - 4 = (5 × 2)/2 - 4 = 10/2 - 4 = 5 - 4 = 1
= 4p + 7 = 4 × (-2) + 7 = - 8 + 7 = 7 - 8 = -1
= - 3p = - 3 × (- 2) = - 3 × 4 + 4 × (- 2) + 7 = - 12 - 8 + 7 = - 20 + 7 = - 13
= - 2p = - 2 × (- 2) = - 2 × (- 8) - 3 × (4) + 4 × (- 2) + 7 = 16 - 12 - 8 + 7 = 16 + 7 - 12 - 8 = 23 - 20 = 3
= 2x - 7 = 2 × (- 1) - 7 = - 2 - 7 = - 9
= - x + 2 = - (-1) + 2 = 1 + 2 = 3
= x = (-1) = 1 - 2 + 1 = 1 + 1 - 2 = 2 - 2 = 0
= 2x = 2 × (-1) = 2 + 1 - 2 = 1
= a = 2 = 4 + 4 = 8
= a =2 = 4 - 4 + 4 = 4 + 4 - 4 = 8 - 4 = 4
= a = 2 = 4 - 4 = 0
= 2a + 2b = 2 × 0 + 2 × (-1) = 0 - 2 = -2
= 2a = 2 × 0 = 0 + 1 + 1 = 2
= 2a = 2 × 0 = 0 + 0 + 0 = 0
= a = 0 = 0 + 0 + 2 = 2
x + 7 + 4 (x - 5) = x + 7 + 4x - 20 = 5x + 7 - 20 = 5x - 13 Substituting the value of x = 2, we get: = 5x - 13 = 5 × 2 - 13 = 10 - 13 = -3 Or = x + 7 + 4 (x - 5) = 2 + 7 + 4 × (2 - 5) = 2 + 7 + 4 × (-3) = 2 + 7 + (- 12) = 9 - 12 = - 3
= 3x + 6 + 5x - 7 = 3x + 5x + 6 - 7 = 8x - 1 Substituting the value of x = 2, we get: 8x - 1 = 8 × 2 - 1 = 16 - 1 = 15
= 6x + 5x - 10 = 11x - 10 Substituting the value of x = 2, we get: =11x - 10 = 11 × 2 - 10 = 22 - 10 = 12
= 8x - 4 + 3x + 11 = 8x + 3x - 4 + 11 = 11x + 7 Substituting the value of x = 2, we get: = 11x + 7 = 11 × 2 + 7 = 22 + 7 = 29
= 3x - x - 5 + 9 = 2x + 4 Substituting the value of x = 3, we get: = 2 × 3 + 4 = 6 + 4 = 10
= 2 + 4 - 4x = 6 - 4x = - 4x + 6 Substituting the value of x = 3, we get: - 4x + 6 = - 4 × 3 + 6 = - 12 + 6 = - 6
= 3a - 8a + 5 + 1 = -5a + 6 Substituting the value of a = - 1, we get: -5a + 6 = - 5 × (- 1) + 6 = 5 + 6 = 11
= - 3b - 5b + 10 - 4 = - 8b + 6 Substituting the value of b = - 2, we get: - 8b + 6 = - 8 × (- 2) + 6 = 16 + 6 = 22
= 2a + a - 2b - 4 - 5 = 3a - 2b - 9 Substituting the value of a = - 1, b = - 2, we get: = 3a - 2b - 9 = 3 × (- 1) - 2 × (- 2) - 9 = -3 + 4 - 9 = - 8
= z = 10 = 1000 - 3 (0) = 1000 - 0 = 1000
= p = (- 10) = 100 + 20 - 100 = 20
2x : Substituting the value of x = 0, we get: 2 (0) = 0 + 0 - a = 5 - a = 5 a = - 5
= 2a = 2a = 2a Substituting the value of a = 5 and b = - 3, we get: = 2 × (5) = 2 × 25 - 15 + 3 = 50 - 15 + 3 = 35 + 3 = 38 ## Exercise 12.4
5 For n = 5, The number of segments = 5 (5) + 1 = 25 + 1 = 10 For n = 10, The number of segments = 5 (10) + 1 = 50 + 1 = 100 For n = 100, The number of segments = 5 (100) + 1 = 500 + 1 =
5 For n = 5, The number of segments = 3 (5) + 1 = 15 + 1 = 10 For n = 10, The number of segments = 3 (10) + 1 = 30 + 1 = 100 For n = 100, The number of segments = 3 (100) + 1 = 300 + 1 =
5 For n = 5, The number of segments = 5 (5) + 2 = 25 + 2 = 10 The number of segments = 5 (10) + 2 = 50 + 2 = 100 For n = 100, The number of segments = 5 (100) + 2 = 500 + 2 =
Here, we have to find the 100 For n = 100, 2n - 1 = 2 (100) - 1 = 200 - 1 =
Here, we have to find the 5 For n = 5, 3n + 2 = 3 (5) + 2 = 15 + 2 = For n = 10, 3n + 2 = 3 (10) + 2 = 30 + 2 = For n = 100, 3n + 2 = 3 (100) + 2 = 300 + 2 =
Here, we have to find the 5 For n = 5, 4n + 1 = 4 (5) + 1 = 20 + 1 = For n = 10, 4n + 1 = 4 (10) + 1 = 40 + 1 = For n = 100, 4n + 1 = 4 (100) + 1 = 400 + 1 =
Here, we have to find the 5 For n = 5, 7n + 20 = 7 (5) + 20 = 35 + 20 = For n = 10, 7n + 20 = 7 (10) + 20 = 70 + 20 = For n = 100, 7n + 20 = 7 (100) + 20 = 700 + 20 =
Here, we have to find the 5 For n = 5, n = 25 + 1 = For n = 10, n = 100 + 1 = Next Topicclass 7 Maths Chapter 13 |