NCERT Solutions for Class 7 Maths Chapter 12 - Algebraic Expressions
Unlock the Power of Algebraic Expressions withour quest to facilitate your learning, we proudly offer the PDF solutions for NCERT Class 7 Maths Chapter 12 - "Algebraic Expressions." These solutions serve as a valuable resource, especially when students encounter challenges while working on questions from this chapter.Our solutions for Exercise 12.1 focus on providing a clear understanding of fundamental concepts in algebra. This exercise covers topics related to the formation of expressions, understanding terms within expressions, distinguishing between like and unlike terms, and exploring monomials, binomials, trinomials, and polynomials.
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The NCERT Solutions for Class 7 Maths Chapter 12 - Algebraic Expressions are tailored to help the students master the concepts that are key to success in their classrooms. The solutions given in the PDF are developed by experts and correlate with the CBSE syllabus of 2023-2024. These solutions provide thorough explanations with a step-by-step approach to solving problems. Students can easily get a hold of the subject and learn the basics with a deeper understanding. Additionally, they can practice better, be confident, and perform well in their examinations with the support of this PDF.
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Students can access the NCERT Solutions for Class 7 Maths Chapter 12 - Algebraic Expressions. Curated by experts according to the CBSE syllabus for 2023–2024, these step-by-step solutions make Maths much easier to understand and learn for the students. These solutions can be used in practice by students to attain skills in solving problems, reinforce important learning objectives, and be well-prepared for tests.
Exercise 12.1
Get the algebraic expressions in the following cases using variables, constants and arithmetic operations.
(i) Subtraction of z from y.
(ii) One-half of the sum of numbers x and y.
(iii) The number z multiplied by itself.
(iv) One-fourth of the product of numbers p and q.
(v) Numbers x and y both squared and added.
(vi) Number 5 added to three times the product of numbers m and n.
(vii) Product of numbers y and z subtracted from 10.
(viii) Sum of numbers a and b subtracted from their product.
(i) = Y – z
(ii) = ½ (x + y)
= (x + y)/2
(iii) = z × z
= z2
(iv) = ¼ (p × q)
= pq/4
(v) = x2 + y2
(vi) = 3mn + 5
(vii) = 10 – (y × z)
= 10 – yz
(viii) = (a × b) – (a + b)
= ab – (a + b)
(i) Identify the terms and their factors in the following expressions
Show the terms and factors by tree diagrams.
(a) x – 3
(b) 1 + x + x2
(c) y – y3
(d) 5xy2 + 7x2y
(e) – ab + 2b2 – 3a2
(ii) Identify terms and factors in the expressions given below:
(a) – 4x + 5 (b) – 4x + 5y (c) 5y + 3y2 (d) xy + 2x2y2
(e) pq + q (f) 1.2 ab – 2.4 b + 3.6 a (g) ¾ x + ¼
(h) 0.1 p2 + 0.2 q2
(i)(a) Expression: x – 3
Terms: x, -3
Factors: x; -3
(b) Expression: 1 + x + x2
Terms: 1, x, x2
Factors: 1; x; x,x
(c) Expression: y – y3
Terms: y, -y3
Factors: y; -y, -y, -y
(d) Expression: 5xy2 + 7x2y
Terms: 5xy2, 7x2y
Factors: 5, x, y, y; 7, x, x, y
(e) Expression: -ab + 2b2 – 3a2
Terms: -ab, 2b2, -3a2
Factors: -a, b; 2, b, b; -3, a, a
(ii) Expressions is defined as, numbers, symbols and operators (such as +. – , × and ÷) grouped together that show the value of something.
In algebra a term is either a single number or variable, or numbers and variables multiplied together. Terms are separated by + or – signs or sometimes by division.
Factors is defined as, numbers we can multiply together to get another number.
|
Sl.No. |
Expression |
Terms |
Factors |
|
(a) |
– 4x + 5 |
-4x 5 |
-4, x 5 |
|
(b) |
– 4x + 5y |
-4x 5y |
-4, x 5, y |
|
(c) |
5y + 3y2 |
5y 3y2 |
5, y 3, y, y |
|
(d) |
xy + 2x2y2 |
xy 2x2y2 |
x, y 2, x, x, y, y |
|
(e) |
pq + q |
pq q |
P, q Q |
|
(f) |
1.2 ab – 2.4 b + 3.6 a |
1.2ab -2.4b 3.6a |
1.2, a, b -2.4, b 3.6, a |
|
(g) |
¾ x + ¼ |
¾ x ¼ |
¾, x ¼ |
|
(h) |
0.1 p2 + 0.2 q2 |
0.1p2 0.2q2 |
0.1, p, p 0.2, q, q |
State whether a given pair of terms is of like or unlike terms.
(i) 1, 100
(ii) –7x, (5/2)x
(iii) – 29x, – 29y
(iv) 14xy, 42yx
(v) 4m2p, 4mp2
(vi) 12xz, 12x2z2
(i) Like term.
When terms have the same algebraic factors, they are like terms.
(ii) Like term.
When terms have the same algebraic factors, they are like terms.
(iii) Unlike terms.
The terms have different algebraic factors, they are unlike terms.
(iv) Like term.
When terms have the same algebraic factors, they are like terms.
(v) Unlike terms.
When terms have different algebraic factors, they are unlike terms.
(vi) Unlike terms.
When terms have different algebraic factors, they are unlike terms.
Identify like terms in the following:
(a) – xy2, – 4yx2, 8x2, 2xy2, 7y, – 11x2, – 100x, – 11yx, 20x2y, – 6x2, y, 2xy, 3x
(b) 10pq, 7p, 8q, – p2q2, – 7qp, – 100q, – 23, 12q2p2, – 5p2, 41, 2405p, 78qp,13p2q, qp2, 701p2
(a) When terms have the same algebraic factors, they are like terms.
They are,
– xy2, 2xy2
– 4yx2, 20x2y
8x2, – 11x2, – 6x2
7y, y
– 100x, 3x
– 11yx, 2xy
(b) When terms have the same algebraic factors, they are like terms.
They are,
10pq, – 7qp, 78qp
7p, 2405p
8q, – 100q
– p2q2, 12q2p2
– 23, 41
– 5p2, 701p2
13p2q, qp2
Classify into monomials, binomials and trinomials.
(i) 4y – 7z
(ii) y2
(iii) x + y – xy
(iv) 100
(v) ab – a – b
(vi) 5 – 3t
(vii) 4p2q – 4pq2
(viii) 7mn
(ix) z2 – 3z + 8
(x) a2 + b2
(xi) z2 + z
(xii) 1 + x + x2
(i) Binomial.
An expression which contains two unlike terms is called a binomial.
(ii) Monomial.
An expression with only one term is called a monomial.
(iii) Trinomial.
An expression which contains three terms is called a trinomial.
(iv) Monomial.
An expression with only one term is called a monomial.
(v) Trinomial.
An expression which contains three terms is called a trinomial.
(vi) Binomial.
An expression which contains two unlike terms is called a binomial.
(vii) Binomial.
An expression which contains two unlike terms is called a binomial.
(viii) Monomial.
An expression with only one term is called a monomial.
(ix) Trinomial.
An expression which contains three terms is called a trinomial.
(x) Binomial.
An expression which contains two unlike terms is called a binomial.
(xi) Binomial.
An expression which contains two unlike terms is called a binomial.
(xii) Trinomial.
An expression which contains three terms is called a trinomial.
Identify the numerical coefficients of terms (other than constants) in the following expressions:
(i) 5 – 3t2
(ii) 1 + t + t2 + t3
(iii) x + 2xy + 3y
(iv) 100m + 1000n
(v) – p2q2 + 7pq
(vi) 1.2 a + 0.8 b
(vii) 3.14 r2
(viii) 2 (l + b)
(ix) 0.1 y + 0.01 y2
Expressions are defined as, numbers, symbols and operators (such as +. – , × and ÷) grouped together that show the value of something.
In algebra, a term is either a single number or variable, or numbers and variables multiplied together. Terms are separated by + or – signs or sometimes by division.
A coefficient is a number used to multiply a variable (2x means 2 times x, so 2 is a coefficient) Variables on their own (without a number next to them) actually have a coefficient of 1 (x is really 1x)
|
Sl.No. |
Expression |
Terms |
Coefficients |
|
(i) |
5 – 3t2 |
– 3t2 |
-3 |
|
(ii) |
1 + t + t2 + t3 |
t t2 t3 |
1 1 1 |
|
(iii) |
x + 2xy + 3y |
x 2xy 3y |
1 2 3 |
|
(iv) |
100m + 1000n |
100m 1000n |
100 1000 |
|
(v) |
– p2q2 + 7pq |
-p2q2 7pq |
-1 7 |
|
(vi) |
1.2 a + 0.8 b |
1.2a 0.8b |
1.2 0.8 |
|
(vii) |
3.14 r2 |
3.142 |
3.14 |
|
(viii) |
2 (l + b) |
2l 2b |
2 2 |
|
(ix) |
0.1 y + 0.01 y2 |
0.1y 0.01y2 |
0.1 0.01 |
(a) Identify terms which contain x and give the coefficient of x.
(i) y2x + y
(ii) 13y2 – 8yx
(iii) x + y + 2
(iv) 5 + z + zx
(v) 1 + x + xy
(vi) 12xy2 + 25
(vii) 7x + xy2
(b) Identify terms which contain y2 and give the coefficient of y2.
(i) 8 – xy2
(ii) 5y2 + 7x
(iii) 2x2y – 15xy2 + 7y2
(a)
|
Sl.No. |
Expression |
Terms |
Coefficient of x |
|
(i) |
y2x + y |
y2x |
y2 |
|
(ii) |
13y2 – 8yx |
– 8yx |
-8y |
|
(iii) |
x + y + 2 |
x |
1 |
|
(iv) |
5 + z + zx |
x zx |
1 z |
|
(v) |
1 + x + xy |
xy |
y |
|
(vi) |
12xy2 + 25 |
12xy2 |
12y2 |
|
(vii) |
7x + xy2 |
7x xy2 |
7 y2 |
(b)
|
Sl.No. |
Expression |
Terms |
Coefficient of y2 |
|
(i) |
8 – xy2 |
– xy2 |
– x |
|
(ii) |
5y2 + 7x |
5y2 |
5 |
|
(iii) |
2x2y – 15xy2 + 7y2 |
– 15xy2 7y2 |
– 15x 7 |
Exercise 12.2
(a) What should be added to x2 + xy + y2 to obtain 2x2 + 3xy?
(b) What should be subtracted from 2a + 8b + 10 to get – 3a + 7b + 16?
(a) Let us assume p be the required term
Then,
p + (x2 + xy + y2) = 2x2 + 3xy
p = (2x2 + 3xy) – (x2 + xy + y2)
p = 2x2 + 3xy – x2 – xy – y2
p = 2x2 – x2 + 3xy – xy – y2
p = x2 + 2xy – y2
(b) Let us assume x be the required term
Then,
2a + 8b + 10 – x = -3a + 7b + 16
x = (2a + 8b + 10) – (-3a + 7b + 16)
x = 2a + 8b + 10 + 3a – 7b – 16
x = 2a + 3a + 8b – 7b + 10 – 16
x = 5a + b – 6
What should be taken away from 3x2 – 4y2 + 5xy + 20 to obtain – x2 – y2 + 6xy + 20?
Let us assume a be the required term
Then,
3x2 – 4y2 + 5xy + 20 – a = -x2 – y2 + 6xy + 20
a = 3x2 – 4y2 + 5xy + 20 – (-x2 – y2 + 6xy + 20)
a = 3x2 – 4y2 + 5xy + 20 + x2 + y2 – 6xy – 20
a = 3x2 + x2 – 4y2 + y2 + 5xy – 6xy + 20 – 20
a = 4x2 – 3y2 – xy
Simplify combining like terms:
(i) 21b – 32 + 7b – 20b
(ii) – z2 + 13z2 – 5z + 7z3 – 15z
(iii) p – (p – q) – q – (q – p)
(iv) 3a – 2b – ab – (a – b + ab) + 3ab + b – a
(v) 5x2y – 5x2 + 3yx2 – 3y2 + x2 – y2 + 8xy2 – 3y2
(vi) (3y2 + 5y – 4) – (8y – y2 – 4)
(i) When terms have the same algebraic factors, they are like terms.
Then,
= (21b + 7b – 20b) – 32
= b (21 + 7 – 20) – 32
= b (28 – 20) – 32
= b (8) – 32
= 8b – 32
(ii) When terms have the same algebraic factors, they are like terms.
Then,
= 7z3 + (-z2 + 13z2) + (-5z – 15z)
= 7z3 + z2 (-1 + 13) + z (-5 – 15)
= 7z3 + z2 (12) + z (-20)
= 7z3 + 12z2 – 20z
(iii) When terms have the same algebraic factors, they are like terms.
Then,
= p – p + q – q – q + p
= p – q
(iv) When terms have the same algebraic factors, they are like terms.
Then,
= 3a – 2b – ab – a + b – ab + 3ab + b – a
= 3a – a – a – 2b + b + b – ab – ab + 3ab
= a (1 – 1- 1) + b (-2 + 1 + 1) + ab (-1 -1 + 3)
= a (1 – 2) + b (-2 + 2) + ab (-2 + 3)
= a (1) + b (0) + ab (1)
= a + ab
(v) When terms have the same algebraic factors, they are like terms.
Then,
= 5x2y + 3yx2 – 5x2 + x2 – 3y2 – y2 – 3y2
= x2y (5 + 3) + x2 (- 5 + 1) + y2 (-3 – 1 -3) + 8xy2
= x2y (8) + x2 (-4) + y2 (-7) + 8xy2
= 8x2y – 4x2 – 7y2 + 8xy2
(vi) When terms have the same algebraic factors, they are like terms.
Then,
= 3y2 + 5y – 4 – 8y + y2 + 4
= 3y2 + y2 + 5y – 8y – 4 + 4
= y2 (3 + 1) + y (5 – 8) + (-4 + 4)
= y2 (4) + y (-3) + (0)
= 4y2 – 3y
Add:
(i) 3mn, – 5mn, 8mn, – 4mn
(ii) t – 8tz, 3tz – z, z – t
(iii) – 7mn + 5, 12mn + 2, 9mn – 8, – 2mn – 3
(iv) a + b – 3, b – a + 3, a – b + 3
(v) 14x + 10y – 12xy – 13, 18 – 7x – 10y + 8xy, 4xy
(vi) 5m – 7n, 3n – 4m + 2, 2m – 3mn – 5
(vii) 4x2y, – 3xy2, –5xy2, 5x2y
(viii) 3p2q2 – 4pq + 5, – 10 p2q2, 15 + 9pq + 7p2q2
(ix) ab – 4a, 4b – ab, 4a – 4b
(x) x2 – y2 – 1, y2 – 1 – x2, 1 – x2 – y2
(i) When terms have the same algebraic factors, they are like terms.
Then, we have to add the like terms
= 3mn + (-5mn) + 8mn + (- 4mn)
= 3mn – 5mn + 8mn – 4mn
= mn (3 – 5 + 8 – 4)
= mn (11 – 9)
= mn (2)
= 2mn
(ii) When terms have the same algebraic factors, they are like terms.
Then, we have to add the like terms
= t – 8tz + (3tz – z) + (z – t)
= t – 8tz + 3tz – z + z – t
= t – t – 8tz + 3tz – z + z
= t (1 – 1) + tz (- 8 + 3) + z (-1 + 1)
= t (0) + tz (- 5) + z (0)
= – 5tz
(iii) When terms have the same algebraic factors, they are like terms.
Then, we have to add the like terms
= – 7mn + 5 + 12mn + 2 + (9mn – 8) + (- 2mn – 3)
= – 7mn + 5 + 12mn + 2 + 9mn – 8 – 2mn – 3
= – 7mn + 12mn + 9mn – 2mn + 5 + 2 – 8 – 3
= mn (-7 + 12 + 9 – 2) + (5 + 2 – 8 – 3)
= mn (- 9 + 21) + (7 – 11)
= mn (12) – 4
= 12mn – 4
(iv) When terms have the same algebraic factors, they are like terms.
Then, we have to add the like terms
= a + b – 3 + (b – a + 3) + (a – b + 3)
= a + b – 3 + b – a + 3 + a – b + 3
= a – a + a + b + b – b – 3 + 3 + 3
= a (1 – 1 + 1) + b (1 + 1 – 1) + (-3 + 3 + 3)
= a (2 -1) + b (2 -1) + (-3 + 6)
= a (1) + b (1) + (3)
= a + b + 3
(v) When terms have the same algebraic factors, they are like terms.
Then, we have to add the like terms
= 14x + 10y – 12xy – 13 + (18 – 7x – 10y + 8xy) + 4xy
= 14x + 10y – 12xy – 13 + 18 – 7x – 10y + 8xy + 4xy
= 14x – 7x + 10y– 10y – 12xy + 8xy + 4xy – 13 + 18
= x (14 – 7) + y (10 – 10) + xy(-12 + 8 + 4) + (-13 + 18)
= x (7) + y (0) + xy(0) + (5)
= 7x + 5
(vi) When terms have the same algebraic factors, they are like terms.
Then, we have to add the like terms
= 5m – 7n + (3n – 4m + 2) + (2m – 3mn – 5)
= 5m – 7n + 3n – 4m + 2 + 2m – 3mn – 5
= 5m – 4m + 2m – 7n + 3n – 3mn + 2 – 5
= m (5 – 4 + 2) + n (-7 + 3) – 3mn + (2 – 5)
= m (3) + n (-4) – 3mn + (-3)
= 3m – 4n – 3mn – 3
(vii) When terms have the same algebraic factors, they are like terms.
Then, we have to add the like terms
= 4x2y + (-3xy2) + (-5xy2) + 5x2y
= 4x2y + 5x2y – 3xy2 – 5xy2
= x2y (4 + 5) + xy2 (-3 – 5)
= x2y (9) + xy2 (- 8)
= 9x2y – 8xy2
(viii) When terms have the same algebraic factors, they are like terms.
Then, we have to add the like terms
= 3p2q2 – 4pq + 5 + (- 10p2q2) + 15 + 9pq + 7p2q2
= 3p2q2 – 10p2q2 + 7p2q2 – 4pq + 9pq + 5 + 15
= p2q2 (3 -10 + 7) + pq (-4 + 9) + (5 + 15)
= p2q2 (0) + pq (5) + 20
= 5pq + 20
(ix) When terms have the same algebraic factors, they are like terms.
Then, we have to add the like terms
= ab – 4a + (4b – ab) + (4a – 4b)
= ab – 4a + 4b – ab + 4a – 4b
= ab – ab – 4a + 4a + 4b – 4b
= ab (1 -1) + a (4 – 4) + b (4 – 4)
= ab (0) + a (0) + b (0)
= 0
(x) When terms have the same algebraic factors, they are like terms.
Then, we have to add the like terms
= x2 – y2 – 1 + (y2 – 1 – x2) + (1 – x2 – y2)
= x2 – y2 – 1 + y2 – 1 – x2 + 1 – x2 – y2
= x2 – x2 – x2 – y2 + y2 – y2 – 1 – 1 + 1
= x2 (1 – 1- 1) + y2 (-1 + 1 – 1) + (-1 -1 + 1)
= x2 (1 – 2) + y2 (-2 +1) + (-2 + 1)
= x2 (-1) + y2 (-1) + (-1)
= -x2 – y2 -1
Subtract:
(i) –5y2 from y2
(ii) 6xy from –12xy
(iii) (a – b) from (a + b)
(iv) a (b – 5) from b (5 – a)
(v) –m2 + 5mn from 4m2 – 3mn + 8
(vi) – x2 + 10x – 5 from 5x – 10
(vii) 5a2 – 7ab + 5b2 from 3ab – 2a2 – 2b2
(viii) 4pq – 5q2 – 3p2 from 5p2 + 3q2 – pq
(i) When terms have the same algebraic factors, they are like terms.
Then, we have to subtract the like terms
= y2 – (-5y2)
= y2 + 5y2
= 6y2
(ii) When terms have the same algebraic factors, they are like terms.
Then, we have to subtract the like terms
= -12xy – 6xy
= – 18xy
(iii) When terms have the same algebraic factors, they are like terms.
Then, we have to subtract the like terms
= (a + b) – (a – b)
= a + b – a + b
= a – a + b + b
= a (1 – 1) + b (1 + 1)
= a (0) + b (2)
= 2b
(iv) When terms have the same algebraic factors, they are like terms.
Then, we have to subtract the like terms
= b (5 -a) – a (b – 5)
= 5b – ab – ab + 5a
= 5b + ab (-1 -1) + 5a
= 5a + 5b – 2ab
(v) When terms have the same algebraic factors, they are like terms.
Then, we have to subtract the like terms
= 4m2 – 3mn + 8 – (- m2 + 5mn)
= 4m2 – 3mn + 8 + m2 – 5mn
= 4m2 + m2 – 3mn – 5mn + 8
= 5m2 – 8mn + 8
(vi) When terms have the same algebraic factors, they are like terms.
Then, we have to subtract the like terms
= 5x – 10 – (-x2 + 10x – 5)
= 5x – 10 + x2 – 10x + 5
= x2 + 5x – 10x – 10 + 5
= x2 – 5x – 5
(vii) When terms have the same algebraic factors, they are like terms.
Then, we have to subtract the like terms
= 3ab – 2a2 – 2b2 – (5a2 – 7ab + 5b2)
= 3ab – 2a2 – 2b2 – 5a2 + 7ab – 5b2
= 3ab + 7ab – 2a2 – 5a2 – 2b2 – 5b2
= 10ab – 7a2 – 7b2
(viii) When terms have the same algebraic factors, they are like terms.
Then, we have to subtract the like terms
= 5p2 + 3q2 – pq – (4pq – 5q2 – 3p2)
= 5p2 + 3q2 – pq – 4pq + 5q2 + 3p2
= 5p2 + 3p2 + 3q2 + 5q2 – pq – 4pq
= 8p2 + 8q2 – 5pq
(a) From the sum of 3x – y + 11 and – y – 11, subtract 3x – y – 11.
(b) From the sum of 4 + 3x and 5 – 4x + 2x2, subtract the sum of 3x2 – 5x and
–x2 + 2x + 5.
(a) First we have to find out the sum of 3x – y + 11 and – y – 11
= 3x – y + 11 + (-y – 11)
= 3x – y + 11 – y – 11
= 3x – y – y + 11 – 11
= 3x – 2y
Now, subtract 3x – y – 11 from 3x – 2y
= 3x – 2y – (3x – y – 11)
= 3x – 2y – 3x + y + 11
= 3x – 3x – 2y + y + 11
= -y + 11
(b) First we have to find out the sum of 4 + 3x and 5 – 4x + 2x2
= 4 + 3x + (5 – 4x + 2x2)
= 4 + 3x + 5 – 4x + 2x2
= 4 + 5 + 3x – 4x + 2x2
= 9 – x + 2x2
= 2x2 – x + 9 … [equation 1]
Then, we have to find out the sum of 3x2 – 5x and – x2 + 2x + 5
= 3x2 – 5x + (-x2 + 2x + 5)
= 3x2 – 5x – x2 + 2x + 5
= 3x2 – x2 – 5x + 2x + 5
= 2x2 – 3x + 5 … [equation 2]
Now, we have to subtract equation (2) from equation (1)
= 2x2 – x + 9 – (2x2 – 3x + 5)
= 2x2 – x + 9 – 2x2 + 3x – 5
= 2x2 – 2x2 – x + 3x + 9 – 5
= 2x + 4
Exercise 12.3
If p = – 2, find the value of:
(i) 4p + 7
(ii) – 3p2 + 4p + 7
(iii) – 2p3 – 3p2 + 4p + 7
(i) From the question it is given that p = -2
Then, substitute the value of p in the question
= (4 × (-2)) + 7
= -8 + 7
= -1
(ii) From the question it is given that p = -2
Then, substitute the value of p in the question
= (-3 × (-2)2) + (4 × (-2)) + 7
= (-3 × 4) + (-8) + 7
= -12 – 8 + 7
= -20 + 7
= -13
(iii) From the question it is given that p = -2
Then, substitute the value of p in the question
= (-2 × (-2)3) – (3 × (-2)2) + (4 × (-2)) + 7
= (-2 × -8) – (3 × 4) + (-8) + 7
= 16 – 12 – 8 + 7
= 23 – 20
= 3
If a = 2, b = – 2, find the value of:
(i) a2 + b2
(ii) a2 + ab + b2
(iii) a2 – b2
(i) From the question it is given that a = 2, b = -2
Then, substitute the value of a and b in the question
= (2)2 + (-2)2
= 4 + 4
= 8
(ii) From the question it is given that a = 2, b = -2
Then, substitute the value of a and b in the question
= 22 + (2 × -2) + (-2)2
= 4 + (-4) + (4)
= 4 – 4 + 4
= 4
(iii) From the question it is given that a = 2, b = -2
Then, substitute the value of a and b in the question
= 22 – (-2)2
= 4 – (4)
= 4 – 4
= 0
When a = 0, b = – 1, find the value of the given expressions:
(i) 2a + 2b
(ii) 2a2 + b2 + 1
(iii) 2a2b + 2ab2 + ab
(iv) a2 + ab + 2
(i) From the question it is given that a = 0, b = -1
Then, substitute the value of a and b in the question
= (2 × 0) + (2 × -1)
= 0 – 2
= -2
(ii) From the question it is given that a = 0, b = -1
Then, substitute the value of a and b in the question
= (2 × 02) + (-1)2 + 1
= 0 + 1 + 1
= 2
(iii) From the question it is given that a = 0, b = -1
Then, substitute the value of a and b in the question
= (2 × 02 × -1) + (2 × 0 × (-1)2) + (0 × -1)
= 0 + 0 +0
= 0
(iv) From the question it is given that a = 0, b = -1
Then, substitute the value of a and b in the question
= (02) + (0 × (-1)) + 2
= 0 + 0 + 2
= 2
Find the value of the following expressions, when x = –1:
(i) 2x – 7
(ii) – x + 2
(iii) x2 + 2x + 1
(iv) 2x2 – x – 2
(i) From the question it is given that x = -1
Then, substitute the value of x in the question
= (2 × -1) – 7
= – 2 – 7
= – 9
(ii) From the question it is given that x = -1
Then, substitute the value of x in the question
= – (-1) + 2
= 1 + 2
= 3
(iii) From the question it is given that x = -1
Then, substitute the value of x in the question
= (-1)2 + (2 × -1) + 1
= 1 – 2 + 1
= 2 – 2
= 0
(iv) From the question it is given that x = -1
Then, substitute the value of x in the question
= (2 × (-1)2) – (-1) – 2
= (2 × 1) + 1 – 2
= 2 + 1 – 2
= 3 – 2
= 1
If m = 2, find the value of:
(i) m – 2
(ii) 3m – 5
(iii) 9 – 5m
(iv) 3m2 – 2m – 7
(v) (5m/2) – 4
(i) From the question it is given that m = 2
Then, substitute the value of m in the question
= 2 -2
= 0
(ii) From the question it is given that m = 2
Then, substitute the value of m in the question
= (3 × 2) – 5
= 6 – 5
= 1
(iii) From the question it is given that m = 2
Then, substitute the value of m in the question
= 9 – (5 × 2)
= 9 – 10
= – 1
(iv) From the question it is given that m = 2
Then, substitute the value of m in the question
= (3 × 22) – (2 × 2) – 7
= (3 × 4) – (4) – 7
= 12 – 4 -7
= 12 – 11
= 1
(v) From the question it is given that m = 2
Then, substitute the value of m in the question
= ((5 × 2)/2) – 4
= (10/2) – 4
= 5 – 4
= 1
Simplify the expression and find its value when a = 5 and b = – 3.
2(a2 + ab) + 3 – ab
From the question it is given that a = 5 and b = -3
We have,
= 2a2 + 2ab + 3 – ab
= 2a2 + ab + 3
Then, substitute the value of a and b in the equation
= (2 × 52) + (5 × (-3)) + 3
= (2 × 25) + (-15) + 3
= 50 – 15 + 3
= 53 – 15
= 38
What should be the value of a if the value of 2x2 + x – a equals to 5, when x = 0?
From the question it is given that x = 0
We have,
2x2 + x – a = 5
a = 2x2 + x – 5
Then, substitute the value of x in the equation
a = (2 × 02) + 0 – 5
a = 0 + 0 – 5
a = -5
(i) If z = 10, find the value of z3 – 3(z – 10).
(ii) If p = – 10, find the value of p2 – 2p – 100
(i) From the question it is given that z = 10
We have,
= z3 – 3z + 30
Then, substitute the value of z in the equation
= (10)3 – (3 × 10) + 30
= 1000 – 30 + 30
= 1000
(ii) From the question it is given that p = -10
We have,
= p2 – 2p – 100
Then, substitute the value of p in the equation
= (-10)2 – (2 × (-10)) – 100
= 100 + 20 – 100
= 20
Simplify these expressions and find their values if x = 3, a = – 1, b = – 2.
(i) 3x – 5 – x + 9
(ii) 2 – 8x + 4x + 4
(iii) 3a + 5 – 8a + 1
(iv) 10 – 3b – 4 – 5b
(v) 2a – 2b – 4 – 5 + a
(i) From the question it is given that x = 3
We have,
= 3x – x – 5 + 9
= 2x + 4
Then, substitute the value of x in the equation
= (2 × 3) + 4
= 6 + 4
= 10
(ii) From the question it is given that x = 3
We have,
= 2 + 4 – 8x + 4x
= 6 – 4x
Then, substitute the value of x in the equation
= 6 – (4 × 3)
= 6 – 12
= – 6
(iii) From the question it is given that a = -1
We have,
= 3a – 8a + 5 + 1
= – 5a + 6
Then, substitute the value of a in the equation
= – (5 × (-1)) + 6
= – (-5) + 6
= 5 + 6
= 11
(iv) From the question it is given that b = -2
We have,
= 10 – 4 – 3b – 5b
= 6 – 8b
Then, substitute the value of b in the equation
= 6 – (8 × (-2))
= 6 – (-16)
= 6 + 16
= 22
(v) From the question it is given that a = -1, b = -2
We have,
= 2a + a – 2b – 4 – 5
= 3a – 2b – 9
Then, substitute the value of a and b in the equation
= (3 × (-1)) – (2 × (-2)) – 9
= -3 – (-4) – 9
= – 3 + 4 – 9
= -12 + 4
= -8
Simplify the expressions and find the value if x is equal to 2
(i) x + 7 + 4 (x – 5)
(ii) 3 (x + 2) + 5x – 7
(iii) 6x + 5 (x – 2)
(iv) 4(2x – 1) + 3x + 11
(i) From the question it is given that x = 2
We have,
= x + 7 + 4x – 20
= 5x + 7 – 20
Then, substitute the value of x in the equation
= (5 × 2) + 7 – 20
= 10 + 7 – 20
= 17 – 20
= – 3
(ii) From the question it is given that x = 2
We have,
= 3x + 6 + 5x – 7
= 8x – 1
Then, substitute the value of x in the equation
= (8 × 2) – 1
= 16 – 1
= 15
(iii) From the question it is given that x = 2
We have,
= 6x + 5x – 10
= 11x – 10
Then, substitute the value of x in the equation
= (11 × 2) – 10
= 22 – 10
= 12
(iv) From the question it is given that x = 2
We have,
= 8x – 4 + 3x + 11
= 11x + 7
Then, substitute the value of x in the equation
= (11 × 2) + 7
= 22 + 7
= 29
Exercise 12.4
Observe the patterns of digits made from line segments of equal length. You will find such segmented digits on the display of electronic watches or calculators.
If the number of digits formed is taken to be n, the number of segments required to form n digits is given by the algebraic expression appearing on the right of each pattern. How many segments are required to form 5, 10, 100 digits of the kind
(a) From the question it is given that the numbers of segments required to form n digits of the kind
Then,
The number of segments required to form 5 digits = ((5 × 5) + 1)
= (25 + 1)
= 26
The number of segments required to form 10 digits = ((5 × 10) + 1)
= (50 + 1)
= 51
The number of segments required to form 100 digits = ((5 × 100) + 1)
= (500 + 1)
= 501
(b) From the question it is given that the numbers of segments required to form n digits of the kind
Then,
The number of segments required to form 5 digits = ((3 × 5) + 1)
= (15 + 1)
= 16
The number of segments required to form 10 digits = ((3 × 10) + 1)
= (30 + 1)
= 31
The number of segments required to form 100 digits = ((3 × 100) + 1)
= (300 + 1)
= 301
(c) From the question it is given that the numbers of segments required to form n digits of the kind
Then,
The number of segments required to form 5 digits = ((5 × 5) + 2)
= (25 + 2)
= 27
The number of segments required to form 10 digits = ((5 × 10) + 2)
= (50 + 2)
= 52
The number of segments required to form 100 digits = ((5 × 100) + 1)
= (500 + 2)
= 502
Use the given algebraic expression to complete the table of number patterns.
|
S. No. |
Expression |
Terms |
|||||||||
|
1st |
2nd |
3rd |
4th |
5th |
… |
10th |
… |
100th |
… |
||
|
(i) |
2n – 1 |
1 |
3 |
5 |
7 |
9 |
– |
19 |
– |
– |
– |
|
(ii) |
3n + 2 |
5 |
8 |
11 |
14 |
– |
– |
– |
– |
– |
– |
|
(iii) |
4n + 1 |
5 |
9 |
13 |
17 |
– |
– |
– |
– |
– |
– |
|
(iv) |
7n + 20 |
27 |
34 |
41 |
48 |
– |
– |
– |
– |
– |
– |
|
(v) |
n2 + 1 |
2 |
5 |
10 |
17 |
– |
– |
– |
– |
10001 |
– |
(i) From the table (2n – 1)
Then, 100th term =?
Where n = 100
= (2 × 100) – 1
= 200 – 1
= 199
(ii) From the table (3n + 2)
5th term =?
Where n = 5
= (3 × 5) + 2
= 15 + 2
= 17
Then, 10th term =?
Where n = 10
= (3 × 10) + 2
= 30 + 2
= 32
Then, 100th term =?
Where n = 100
= (3 × 100) + 2
= 300 + 2
= 302
(iii) From the table (4n + 1)
5th term =?
Where n = 5
= (4 × 5) + 1
= 20 + 1
= 21
Then, 10th term =?
Where n = 10
= (4 × 10) + 1
= 40 + 1
= 41
Then, 100th term =?
Where n = 100
= (4 × 100) + 1
= 400 + 1
= 401
(iv) From the table (7n + 20)
5th term =?
Where n = 5
= (7 × 5) + 20
= 35 + 20
= 55
Then, 10th term =?
Where n = 10
= (7 × 10) + 20
= 70 + 20
= 90
Then, 100th term =?
Where n = 100
= (7 × 100) + 20
= 700 + 20
= 720
(v) From the table (n2 + 1)
5th term =?
Where n = 5
= (52) + 1
= 25+ 1
= 26
Then, 10th term =?
Where n = 10
= (102) + 1
= 100 + 1
= 101
So the table is completed below.
|
S. No. |
Expression |
Terms |
|||||||||
|
1st |
2nd |
3rd |
4th |
5th |
… |
10th |
… |
100th |
… |
||
|
(i) |
2n – 1 |
1 |
3 |
5 |
7 |
9 |
– |
19 |
– |
199 |
– |
|
(ii) |
3n + 2 |
5 |
8 |
11 |
14 |
17 |
– |
32 |
– |
302 |
– |
|
(iii) |
4n + 1 |
5 |
9 |
13 |
17 |
21 |
– |
41 |
– |
401 |
– |
|
(iv) |
7n + 20 |
27 |
34 |
41 |
48 |
55 |
– |
90 |
– |
720 |
– |
|
(v) |
n2 + 1 |
2 |
5 |
10 |
17 |
26 |
– |
101 |
– |
10001 |
– |
Frequently Asked Questions
The NCERT solution for Class 7 Chapter 12: Algebraic Expressions is important as it provides a structured approach to learning, ensuring that students develop a strong understanding of foundational concepts early in their academic journey. By mastering these basics, students can build confidence and readiness for tackling more difficult concepts in their further education.
Yes, the NCERT solution for Class 7 Chapter 12: Algebraic Expressions is quite useful for students in preparing for their exams. The solutions are simple, clear, and concise allowing students to understand them better. Algebraic Expressionsally, they can solve the practice questions and exercises that allow them to get exam-ready in no time.
You can get all the NCERT solutions for Class 7 Maths Chapter 12 from the official website of the Orchids International School. These solutions are tailored by subject matter experts and are very easy to understand.
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Students can utilize the NCERT solution for Class 7 Maths Chapter 12 effectively by practicing the solutions regularly. Solve the exercises and practice questions given in the solution.