NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions are provided here. Students can check for the solutions whenever they are facing difficulty, while solving the questions from NCERT Solutions for Class 7 Maths Chapter 12. These solutions for Chapter 12 are available in PDF format so that students can download and learn offline as well. This book is one of the top materials when it comes to providing a question bank to practice. The topics covered in the chapter are as follows.
- How are Expressions Formed
- Terms of An Expression
- Like and Unlike Terms
- Monomials, Binomials, Trinomials and Polynomials
- Addition and Subtraction of Algebraic Expressions
- Finding The Value of An Expression
- Using Algebraic Expressions â€“ Formulas and Rules
Download the PDF of NCERT Solutions For Class 7 Maths Chapter 12 Algebraic Expressions
Â
Access Exercises of NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions
Access answers to Maths NCERT Solutions For Class 7 Chapter 12 â€“ Algebraic Expressions
Exercise 12.1 Page: 234
1. Get the algebraic expressions in the following cases using variables, constants and arithmetic operations.
(i) Subtraction of z from y.
Solution:-
= Y â€“ z
(ii) One-half of the sum of numbers x and y.
Solution:-
= Â½ (x + y)
= (x + y)/2
(iii) The number z multiplied by itself.
Solution:-
= z Ã— z
= z^{2}
(iv) One-fourth of the product of numbers p and q.
Solution:-
= Â¼ (p Ã— q)
= pq/4
(v) Numbers x and y both squared and added.
Solution:-
= x^{2 }+ y^{2}
(vi) Number 5 added to three times the product of numbers m and n.
Solution:-
= 3mn + 5
(vii) Product of numbers y and z subtracted from 10.
Solution:-
= 10 â€“ (y Ã— z)
= 10 â€“ yz
(viii) Sum of numbers a and b subtracted from their product.
Solution:-
= (a Ã— b) â€“ (a + b)
= ab â€“ (a + b)
2. (i) Identify the terms and their factors in the following expressions
Show the terms and factors by tree diagrams.
(a) x â€“ 3
Solution:-
Expression: x â€“ 3
Terms: x, -3
Factors: x; -3
(b) 1 + x + x^{2}
Solution:-
Expression: 1 + x + x^{2}
Terms: 1, x, x^{2}
Factors: 1; x; x,x
(c) y â€“ y^{3}
Solution:-
Expression: y â€“ y^{3}
Terms: y, -y^{3}
Factors: y; -y, -y, -y
(d) 5xy^{2} + 7x^{2}y
Solution:-
Expression: 5xy^{2} + 7x^{2}y
Terms: 5xy^{2}, 7x^{2}y
Factors: 5, x, y, y; 7, x, x, y
(e) â€“ ab + 2b^{2} â€“ 3a^{2}
Solution:-
Expression: -ab + 2b^{2} â€“ 3a^{2}
Terms: -ab, 2b^{2}, -3a^{2}
Factors: -a, b; 2, b, b; -3, a, a
(ii) Identify terms and factors in the expressions given below:
(a) â€“ 4x + 5 (b) â€“ 4x + 5y (c) 5y + 3y^{2} (d) xy + 2x^{2}y^{2}
(e) pq + q (f) 1.2 ab â€“ 2.4 b + 3.6 a (g) Â¾ x + Â¼
(h) 0.1 p^{2} + 0.2 q^{2}
Solution:-
Expressions is defined as, numbers, symbols and operators (such as +. â€“ , Ã— and Ã·) 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 â€“ signs or sometimes by division.
Factors is defined as, numbers we can multiply together to get another number.
Sl.No. | Expression | Terms | Factors |
(a) | â€“ 4x + 5 | -4x
5 |
-4, x
5 |
(b) | â€“ 4x + 5y | -4x
5y |
-4, x
5, y |
(c) | 5y + 3y^{2} | 5y
3y^{2} |
5, y
3, y, y |
(d) | xy + 2x^{2}y^{2} | xy
2x^{2}y^{2} |
x, y
2, x, x, y, y |
(e) | pq + q | pq
q |
P, q
Q |
(f) | 1.2 ab â€“ 2.4 b + 3.6 a | 1.2ab
-2.4b 3.6a |
1.2, a, b
-2.4, b 3.6, a |
(g) | Â¾ x + Â¼ | Â¾ x
Â¼ |
Â¾, x
Â¼ |
(h) | 0.1 p^{2} + 0.2 q^{2} | 0.1p^{2} 0.2q^{2} |
0.1, p, p
0.2, q, q |
3. Identify the numerical coefficients of terms (other than constants) in the following expressions:
(i) 5 â€“ 3t^{2} (ii) 1 + t + t^{2} + t^{3} (iii) x + 2xy + 3y (iv) 100m + 1000n (v) â€“ p^{2}q^{2} + 7pq (vi) 1.2 a + 0.8 b (vii) 3.14 r^{2} (viii) 2 (l + b)
(ix) 0.1 y + 0.01 y^{2}
Solution:-
Expressions is defined as, numbers, symbols and operators (such as +. â€“ , Ã— and Ã·) 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 â€“ signs or sometimes by division.
A coefficient is a number used to multiply a variable (2xÂ meansÂ 2 times x, so 2 is aÂ coefficient) Variables on their own (without a number next to them) actually have aÂ coefficientÂ of 1 (x is really 1x)
Sl.No. | Expression | Terms | Coefficients |
(i) | 5 â€“ 3t^{2} | â€“ 3t^{2} | -3 |
(ii) | 1 + t + t^{2} + t^{3} | t
t^{2} t^{3} |
1
1 1 |
(iii) | x + 2xy + 3y | x
2xy 3y |
1
2 3 |
(iv) | 100m + 1000n | 100m
1000n |
100
1000 |
(v) | â€“ p^{2}q^{2} + 7pq | -p^{2}q^{2} 7pq |
-1
7 |
(vi) | 1.2 a + 0.8 b | 1.2a
0.8b |
1.2
0.8 |
(vii) | 3.14 r^{2} | 3.14^{2} | 3.14 |
(viii) | 2 (l + b) | 2l
2b |
2
2 |
(ix) | 0.1 y + 0.01 y^{2} | 0.1y
0.01y^{2} |
0.1
0.01 |
4. (a) Identify terms which contain x and give the coefficient of x.
(i) y^{2}x + y (ii) 13y^{2} â€“ 8yx (iii) x + y + 2
(iv) 5 + z + zx (v) 1 + x + xy (vi) 12xy^{2} + 25
(vii) 7x + xy^{2}
Solution:-
Sl.No. | Expression | Terms | Coefficient of x |
(i) | y^{2}x + y | y^{2}x | y^{2} |
(ii) | 13y^{2} â€“ 8yx | â€“ 8yx | -8y |
(iii) | x + y + 2 | x | 1 |
(iv) | 5 + z + zx | x
zx |
1
z |
(v) | 1 + x + xy | xy | y |
(vi) | 12xy^{2} + 25 | 12xy^{2} | 12y^{2} |
(vii) | 7x + xy^{2} | 7x
xy^{2} |
7
y^{2} |
(b) Identify terms which contain y^{2} and give the coefficient of y^{2}.
(i) 8 â€“ xy^{2} (ii) 5y^{2} + 7x (iii) 2x^{2}y â€“ 15xy^{2} + 7y^{2}
Solution:-
Sl.No. | Expression | Terms | Coefficient ofÂ y^{2} |
(i) | 8 â€“ xy^{2} | â€“ xy^{2} | â€“ x |
(ii) | 5y^{2} + 7x | 5y^{2} | 5 |
(iii) | 2x^{2}y â€“ 15xy^{2} + 7y^{2} | â€“ 15xy^{2} 7y^{2} |
â€“ 15x
7 |
5. Classify into monomials, binomials and trinomials.
(i) 4y â€“ 7z
Solution:-
Binomial.
An expression which contains two unlike terms is called a binomial.
(ii) y^{2}
Solution:-
Monomial.
An expression with only one term is called a monomial.
(iii) x + y â€“ xy
Solution:-
Trinomial.
An expression which contains three terms is called a trinomial.
(iv) 100
Solution:-
Monomial.
An expression with only one term is called a monomial.
(v) ab â€“ a â€“ b
Solution:-
Trinomial.
An expression which contains three terms is called a trinomial.
(vi) 5 â€“ 3t
Solution:-
Binomial.
An expression which contains two unlike terms is called a binomial.
Â
(vii) 4p^{2}q â€“ 4pq^{2}
Solution:-
Binomial.
An expression which contains two unlike terms is called a binomial.
Â
(viii) 7mn
Solution:-
Monomial.
An expression with only one term is called a monomial.
(ix) z^{2} â€“ 3z + 8
Solution:-
Trinomial.
An expression which contains three terms is called a trinomial.
(x) a^{2} + b^{2}
Solution:-
Binomial.
An expression which contains two unlike terms is called a binomial.
(xi) z^{2} + z
Solution:-
Binomial.
An expression which contains two unlike terms is called a binomial.
(xii) 1 + x + x^{2}
Solution:-
Trinomial.
An expression which contains three terms is called a trinomial.
6. State whether a given pair of terms is of like or unlike terms.
(i) 1, 100
Solution:-
Like term.
When term have the same algebraic factors, they are like terms.
Â
(ii) â€“7x, (5/2)x
Solution:-
Like term.
When term have the same algebraic factors, they are like terms.
Â
(iii) â€“ 29x, â€“ 29y
Solution:-
Unlike terms.
The terms have different algebraic factors, they are unlike terms.
(iv) 14xy, 42yx
Solution:-
Like term.
When term have the same algebraic factors, they are like terms.
Â
(v) 4m^{2}p, 4mp^{2}
Solution:-
Unlike terms.
The terms have different algebraic factors, they are unlike terms.
(vi) 12xz, 12x^{2}z^{2}
Solution:-
Unlike terms.
The terms have different algebraic factors, they are unlike terms.
7. Identify like terms in the following:
(a) â€“ xy^{2}, â€“ 4yx^{2}, 8x^{2}, 2xy^{2}, 7y, â€“ 11x^{2}, â€“ 100x, â€“ 11yx, 20x^{2}y, â€“ 6x^{2}, y, 2xy, 3x
Solution:-
When term have the same algebraic factors, they are like terms.
They are,
â€“ xy^{2}, 2xy^{2}
â€“ 4yx^{2}, 20x^{2}y
8x^{2}, â€“ 11x^{2}, â€“ 6x^{2}
7y, y
â€“ 100x, 3x
â€“ 11yx, 2xy
(b) 10pq, 7p, 8q, â€“ p^{2}q^{2}, â€“ 7qp, â€“ 100q, â€“ 23, 12q^{2}p^{2}, â€“ 5p^{2}, 41, 2405p, 78qp,
13p^{2}q, qp^{2}, 701p^{2}
Solution:-
When term have the same algebraic factors, they are like terms.
They are,
10pq, â€“ 7qp, 78qp
7p, 2405p
8q, â€“ 100q
â€“ p^{2}q^{2}, 12q^{2}p^{2}
â€“ 23, 41
â€“ 5p^{2}, 701p^{2}
13p^{2}q, qp^{2}
Exercise 12.2 Page: 239
1. Simplify combining like terms:
(i) 21b â€“ 32 + 7b â€“ 20b
Solution:-
When term have the same algebraic factors, they are like terms.
Then,
= (21b + 7b â€“ 20b) â€“ 32
= b (21 + 7 â€“ 20) â€“ 32
= b (28 â€“ 20) â€“ 32
= b (8) â€“ 32
= 8b â€“ 32
(ii) â€“ z^{2} + 13z^{2} â€“ 5z + 7z^{3} â€“ 15z
Solution:-
When term have the same algebraic factors, they are like terms.
Then,
= 7z^{3} + (-z^{2} + 13z^{2}) + (-5z â€“ 15z)
= 7z^{3 }+ z^{2 }(-1 + 13) + z (-5 â€“ 15)
= 7z^{3 }+ z^{2} (12) + z (-20)
= 7z^{3 }+ 12z^{2} â€“ 20z
(iii) p â€“ (p â€“ q) â€“ q â€“ (q â€“ p)
Solution:-
When term have the same algebraic factors, they are like terms.
Then,
= p â€“ p + q â€“ q â€“ q + p
= p â€“ q
(iv) 3a â€“ 2b â€“ ab â€“ (a â€“ b + ab) + 3ab + b â€“ a
Solution:-
When term have the same algebraic factors, they are like terms.
Then,
= 3a â€“ 2b â€“ ab â€“ a + b â€“ ab + 3ab + b â€“ a
= 3a â€“ a â€“ a â€“ 2b + b + b â€“ ab â€“ ab + 3ab
= a (1 â€“ 1- 1) + b (-2 + 1 + 1) + ab (-1 -1 + 3)
= a (1 â€“ 2) + b (-2 + 2) + ab (-2 + 3)
= a (1) + b (0) + ab (1)
= a + ab
(v) 5x^{2}y â€“ 5x^{2} + 3yx^{2} â€“ 3y^{2} + x^{2} â€“ y^{2} + 8xy^{2} â€“ 3y^{2}
Solution:-
When term have the same algebraic factors, they are like terms.
Then,
= 5x^{2}y + 3yx^{2} â€“ 5x^{2} + x^{2} â€“ 3y^{2} â€“ y^{2} â€“ 3y^{2}
= x^{2}y (5 + 3) + x^{2} (- 5 + 1) + y^{2} (-3 â€“ 1 -3) + 8xy^{2}
= x^{2}y (8) + x^{2} (-4) + y^{2} (-7) + 8xy^{2}
= 8x^{2}y â€“ 4x^{2} â€“ 7y^{2} + 8xy^{2}
(vi) (3y^{2} + 5y â€“ 4) â€“ (8y â€“ y^{2} â€“ 4)
Solution:-
When term have the same algebraic factors, they are like terms.
Then,
= 3y^{2} + 5y â€“ 4 â€“ 8y + y^{2} + 4
= 3y^{2} + y^{2} + 5y â€“ 8y â€“ 4 + 4
= y^{2} (3 + 1) + y (5 â€“ 8) + (-4 + 4)
= y^{2} (4) + y (-3) + (0)
= 4y^{2} â€“ 3y
2. Add:
(i) 3mn, â€“ 5mn, 8mn, â€“ 4mn
Solution:-
When term have the same algebraic factors, they are like terms.
Then, we have to add the like terms
= 3mn + (-5mn) + 8mn + (- 4mn)
= 3mn â€“ 5mn + 8mn â€“ 4mn
= mn (3 â€“ 5 + 8 â€“ 4)
= mn (11 â€“ 9)
= mn (2)
= 2mn
(ii) t â€“ 8tz, 3tz â€“ z, z â€“ t
Solution:-
When term have the same algebraic factors, they are like terms.
Then, we have to add the like terms
= t â€“ 8tz + (3tz â€“ z) + (z â€“ t)
= t â€“ 8tz + 3tz â€“ z + z â€“ t
= t â€“ t â€“ 8tz + 3tz â€“ z + z
= t (1 â€“ 1) + tz (- 8 + 3) + z (-1 + 1)
= t (0) + tz (- 5) + z (0)
= â€“ 5tz
(iii) â€“ 7mn + 5, 12mn + 2, 9mn â€“ 8, â€“ 2mn â€“ 3
Solution:-
When term have the same algebraic factors, they are like terms.
Then, we have to add the like terms
= â€“ 7mn + 5 + 12mn + 2 + (9mn â€“ 8) + (- 2mn â€“ 3)
= â€“ 7mn + 5 + 12mn + 2 + 9mn â€“ 8 â€“ 2mn â€“ 3
= â€“ 7mn + 12mn + 9mn â€“ 2mn + 5 + 2 â€“ 8 â€“ 3
= mn (-7 + 12 + 9 â€“ 2) + (5 + 2 â€“ 8 â€“ 3)
= mn (- 9 + 21) + (7 â€“ 11)
= mn (12) â€“ 4
= 12mn â€“ 4
(iv) a + b â€“ 3, b â€“ a + 3, a â€“ b + 3
Solution:-
When term have the same algebraic factors, they are like terms.
Then, we have to add the like terms
= a + b â€“ 3 + (b â€“ a + 3) + (a â€“ b + 3)
= a + b â€“ 3 + b â€“ a + 3 + a â€“ b + 3
= a â€“ a + a + b + b â€“ b â€“ 3 + 3 + 3
= a (1 â€“ 1 + 1) + b (1 + 1 â€“ 1) + (-3 + 3 + 3)
= a (2 -1) + b (2 -1) + (-3 + 6)
= a (1) + b (1) + (3)
= a + b + 3
(v) 14x + 10y â€“ 12xy â€“ 13, 18 â€“ 7x â€“ 10y + 8xy, 4xy
Solution:-
When term have the same algebraic factors, they are like terms.
Then, we have to add the like terms
= 14x + 10y â€“ 12xy â€“ 13 + (18 â€“ 7x â€“ 10y + 8xy) + 4xy
= 14x + 10y â€“ 12xy â€“ 13 + 18 â€“ 7x â€“ 10y + 8xy + 4xy
= 14x â€“ 7x + 10yâ€“ 10y â€“ 12xy + 8xy + 4xy â€“ 13 + 18
= x (14 â€“ 7) + y (10 â€“ 10) + xy(-12 + 8 + 4) + (-13 + 18)
= x (7) + y (0) + xy(0) + (5)
= 7x + 5
(vi) 5m â€“ 7n, 3n â€“ 4m + 2, 2m â€“ 3mn â€“ 5
Solution:-
When term have the same algebraic factors, they are like terms.
Then, we have to add the like terms
= 5m â€“ 7n + (3n â€“ 4m + 2) + (2m â€“ 3mn â€“ 5)
= 5m â€“ 7n + 3n â€“ 4m + 2 + 2m â€“ 3mn â€“ 5
= 5m â€“ 4m + 2m â€“ 7n + 3n â€“ 3mn + 2 â€“ 5
= m (5 â€“ 4 + 2) + n (-7 + 3) â€“ 3mn + (2 â€“ 5)
= m (3) + n (-4) â€“ 3mn + (-3)
= 3m â€“ 4n â€“ 3mn â€“ 3
(vii) 4x^{2}y, â€“ 3xy^{2}, â€“5xy^{2}, 5x^{2}y
Solution:-
When term have the same algebraic factors, they are like terms.
Then, we have to add the like terms
= 4x^{2}y + (-3xy^{2}) + (-5xy^{2}) + 5x^{2}y
= 4x^{2}y + 5x^{2}y â€“ 3xy^{2} â€“ 5xy^{2}
= x^{2}y (4 + 5) + xy^{2 }(-3 â€“ 5)
= x^{2}y (9) + xy^{2} (- 8)
= 9x^{2}y â€“ 8xy^{2}
(viii) 3p^{2}q^{2} â€“ 4pq + 5, â€“ 10 p^{2}q^{2}, 15 + 9pq + 7p^{2}q^{2}
Solution:-
When term have the same algebraic factors, they are like terms.
Then, we have to add the like terms
= 3p^{2}q^{2} â€“ 4pq + 5 + (- 10p^{2}q^{2}) + 15 + 9pq + 7p^{2}q^{2}
= 3p^{2}q^{2} â€“ 10p^{2}q^{2} + 7p^{2}q^{2} â€“ 4pq + 9pq + 5 + 15
= p^{2}q^{2} (3 -10 + 7) + pq (-4 + 9) + (5 + 15)
= p^{2}q^{2 }(0) + pq (5) + 20
= 5pq + 20
(ix) ab â€“ 4a, 4b â€“ ab, 4a â€“ 4b
Solution:-
When term have the same algebraic factors, they are like terms.
Then, we have to add the like terms
= ab â€“ 4a + (4b â€“ ab) + (4a â€“ 4b)
= ab â€“ 4a + 4b â€“ ab + 4a â€“ 4b
= ab â€“ ab â€“ 4a + 4a + 4b â€“ 4b
= ab (1 -1) + a (4 â€“ 4) + b (4 â€“ 4)
= ab (0) + a (0) + b (0)
= 0
Â
(x) x^{2} â€“ y^{2} â€“ 1, y^{2} â€“ 1 â€“ x^{2}, 1 â€“ x^{2} â€“ y^{2}
Solution:-
When term have the same algebraic factors, they are like terms.
Then, we have to add the like terms
= x^{2} â€“ y^{2} â€“ 1 + (y^{2} â€“ 1 â€“ x^{2}) + (1 â€“ x^{2} â€“ y^{2})
= x^{2} â€“ y^{2} â€“ 1 + y^{2} â€“ 1 â€“ x^{2} + 1 â€“ x^{2} â€“ y^{2}
= x^{2} â€“ x^{2} â€“ x^{2} â€“ y^{2} + y^{2} â€“ y^{2 }â€“ 1 â€“ 1 + 1
= x^{2} (1 â€“ 1- 1) + y^{2} (-1 + 1 â€“ 1) + (-1 -1 + 1)
= x^{2} (1 â€“ 2) + y^{2}Â (-2 +1) + (-2 + 1)
= x^{2} (-1) + y^{2}Â (-1) + (-1)
= -x^{2} â€“ y^{2} -1
Â
3. Subtract:
(i) â€“5y^{2} from y^{2}
Solution:-
When term have the same algebraic factors, they are like terms.
Then, we have to subtract the like terms
= y^{2} â€“ (-5y^{2})
= y^{2}Â + 5y^{2}
= 6y^{2}
(ii) 6xy from â€“12xy
Solution:-
When term have the same algebraic factors, they are like terms.
Then, we have to subtract the like terms
= -12xy â€“ 6xy
= â€“ 18xy
(iii) (a â€“ b) from (a + b)
Solution:-
When term have the same algebraic factors, they are like terms.
Then, we have to subtract the like terms
= (a + b) â€“ (a â€“ b)
= a + b â€“ a + b
= a â€“ a + b + b
= a (1 â€“ 1) + b (1 + 1)
= a (0) + b (2)
= 2b
(iv) a (b â€“ 5) from b (5 â€“ a)
Solution:-
When term have the same algebraic factors, they are like terms.
Then, we have to subtract the like terms
= b (5 -a) â€“ a (b â€“ 5)
= 5b â€“ ab â€“ ab + 5a
= 5b + ab (-1 -1) + 5a
= 5a + 5b â€“ 2ab
(v) â€“m^{2} + 5mn from 4m^{2} â€“ 3mn + 8
Solution:-
When term have the same algebraic factors, they are like terms.
Then, we have to subtract the like terms
= 4m^{2} â€“ 3mn + 8 â€“ (- m^{2} + 5mn)
= 4m^{2} â€“ 3mn + 8 + m^{2} â€“ 5mn
= 4m^{2} + m^{2} â€“ 3mn â€“ 5mn + 8
= 5m^{2 }â€“ 8mn + 8
(vi) â€“ x^{2} + 10x â€“ 5 from 5x â€“ 10
Solution:-
When term have the same algebraic factors, they are like terms.
Then, we have to subtract the like terms
= 5x â€“ 10 â€“ (-x^{2} + 10x â€“ 5)
= 5x â€“ 10 + x^{2} â€“ 10x + 5
= x^{2} + 5x â€“ 10x â€“ 10 + 5
= x^{2} â€“ 5x â€“ 5
(vii) 5a^{2} â€“ 7ab + 5b^{2} from 3ab â€“ 2a^{2} â€“ 2b^{2}
Solution:-
When term have the same algebraic factors, they are like terms.
Then, we have to subtract the like terms
= 3ab â€“ 2a^{2} â€“ 2b^{2} â€“ (5a^{2} â€“ 7ab + 5b^{2})
= 3ab â€“ 2a^{2} â€“ 2b^{2} â€“ 5a^{2 }+ 7ab â€“ 5b^{2}
= 3ab + 7ab â€“ 2a^{2} â€“ 5a^{2} â€“ 2b^{2} â€“ 5b^{2}
= 10ab â€“ 7a^{2} â€“ 7b^{2}
(viii) 4pq â€“ 5q^{2} â€“ 3p^{2} from 5p^{2} + 3q^{2} â€“ pq
Solution:-
When term have the same algebraic factors, they are like terms.
Then, we have to subtract the like terms
= 5p^{2} + 3q^{2} â€“ pq â€“ (4pq â€“ 5q^{2} â€“ 3p^{2})
= 5p^{2} + 3q^{2} â€“ pq â€“ 4pq + 5q^{2} + 3p^{2}
= 5p^{2} + 3p^{2} + 3q^{2} + 5q^{2} â€“ pq â€“ 4pq
= 8p^{2} + 8q^{2} â€“ 5pq
4. (a) What should be added to x^{2} + xy + y^{2} to obtain 2x^{2} + 3xy?
Solution:-
Let us assume p be the required term
Then,
p + (x^{2} + xy + y^{2}) = 2x^{2} + 3xy
p = (2x^{2}Â + 3xy) â€“ (x^{2} + xy + y^{2})
p = 2x^{2} + 3xy â€“ x^{2} â€“ xy â€“ y^{2}
p = 2x^{2} â€“ x^{2} + 3xy â€“ xy â€“ y^{2}
p = x^{2}Â + 2xy â€“ y^{2}
(b) What should be subtracted from 2a + 8b + 10 to get â€“ 3a + 7b + 16?
Solution:-
Let us assume x be the required term
Then,
2a + 8b + 10 â€“ x = -3a + 7b + 16
x = (2a + 8b + 10) â€“ (-3a + 7b + 16)
x = 2a + 8b + 10 + 3a â€“ 7b â€“ 16
x = 2a + 3a + 8b â€“ 7b + 10 â€“ 16
x = 5a + b â€“ 6
5. What should be taken away from 3x^{2} â€“ 4y^{2} + 5xy + 20 to obtain â€“ x^{2} â€“ y^{2} + 6xy + 20?
Solution:-
Let us assume a be the required term
Then,
3x^{2} â€“ 4y^{2} + 5xy + 20 â€“ a = -x^{2} â€“ y^{2} + 6xy + 20
a = 3x^{2} â€“ 4y^{2} + 5xy + 20 â€“ (-x^{2} â€“ y^{2} + 6xy + 20)
a = 3x^{2} â€“ 4y^{2} + 5xy + 20 + x^{2} + y^{2} â€“ 6xy â€“ 20
a = 3x^{2} + x^{2} â€“ 4y^{2} + y^{2} + 5xy â€“ 6xy + 20 â€“ 20
a = 4x^{2} â€“ 3y^{2} â€“ xy
6. (a) From the sum of 3x â€“ y + 11 and â€“ y â€“ 11, subtract 3x â€“ y â€“ 11.
Solution:-
First we have to find out the sum of 3x â€“ y + 11 and â€“ y â€“ 11
= 3x â€“ y + 11 + (-y â€“ 11)
= 3x â€“ y + 11 â€“ y â€“ 11
= 3x â€“ y â€“ y + 11 â€“ 11
= 3x â€“ 2y
Now, subtract 3x â€“ y â€“ 11 from 3x â€“ 2y
= 3x â€“ 2y â€“ (3x â€“ y â€“ 11)
= 3x â€“ 2y â€“ 3x + y + 11
= 3x â€“ 3x â€“ 2y + y + 11
= -y + 11
(b) From the sum of 4 + 3x and 5 â€“ 4x + 2x^{2}, subtract the sum of 3x^{2} â€“ 5x and
â€“x^{2} + 2x + 5.
Solution:-
First we have to find out the sum of 4 + 3x and 5 â€“ 4x + 2x^{2}
= 4 + 3x + (5 â€“ 4x + 2x^{2})
= 4 + 3x + 5 â€“ 4x + 2x^{2}
= 4 + 5 + 3x â€“ 4x + 2x^{2}
= 9 â€“ x + 2x^{2}
= 2x^{2} â€“ x + 9 â€¦ [equation 1]
Then, we have to find out the sum of 3x^{2} â€“ 5x and â€“ x^{2} + 2x + 5
= 3x^{2} â€“ 5x + (-x^{2} + 2x + 5)
= 3x^{2} â€“ 5x â€“ x^{2} + 2x + 5
= 3x^{2} â€“ x^{2} â€“ 5x + 2x + 5
= 2x^{2} â€“ 3x + 5 â€¦ [equation 2]
Now, we have to subtract equation (2) from equation (1)
= 2x^{2 }â€“ x + 9 â€“ (2x^{2} â€“ 3x + 5)
= 2x^{2} â€“ x + 9 â€“ 2x^{2}Â + 3x â€“ 5
= 2x^{2} â€“ 2x^{2} â€“ x + 3x + 9 â€“ 5
= 2x + 4
Exercise 12.3 Page: 242
1. If m = 2, find the value of:
(i) m â€“ 2
Solution:-
From the question it is given that m = 2
Then, substitute the value of m in the question
= 2 -2
= 0
(ii) 3m â€“ 5
Solution:-
From the question it is given that m = 2
Then, substitute the value of m in the question
= (3 Ã— 2) â€“ 5
= 6 â€“ 5
= 1
(iii) 9 â€“ 5m
Solution:-
From the question it is given that m = 2
Then, substitute the value of m in the question
= 9 â€“ (5 Ã— 2)
= 9 â€“ 10
= â€“ 1
(iv) 3m^{2} â€“ 2m â€“ 7
Solution:-
From the question it is given that m = 2
Then, substitute the value of m in the question
= (3 Ã— 2^{2}) â€“ (2 Ã— 2) â€“ 7
= (3 Ã— 4) â€“ (4) â€“ 7
= 12 â€“ 4 -7
= 12 â€“ 11
= 1
(v) (5m/2) â€“ 4
Solution:-
From the question it is given that m = 2
Then, substitute the value of m in the question
= ((5 Ã— 2)/2) â€“ 4
= (10/2) â€“ 4
= 5 â€“ 4
= 1
2. If p = â€“ 2, find the value of:
(i) 4p + 7
Solution:-
From the question it is given that p = -2
Then, substitute the value of p in the question
= (4 Ã— (-2)) + 7
= -8 + 7
= -1
(ii) â€“ 3p^{2} + 4p + 7
Solution:-
From the question it is given that p = -2
Then, substitute the value of p in the question
= (-3 Ã— (-2)^{2}) + (4 Ã— (-2)) + 7
= (-3 Ã— 4) + (-8) + 7
= -12 â€“ 8 + 7
= -20 + 7
= -13
Â
(iii) â€“ 2p^{3} â€“ 3p^{2} + 4p + 7
Solution:-
From the question it is given that p = -2
Then, substitute the value of p in the question
= (-2 Ã— (-2)^{3}) â€“ (3 Ã— (-2)^{2}) + (4 Ã— (-2)) + 7
= (-2 Ã— -8) â€“ (3 Ã— 4) + (-8) + 7
= 16 â€“ 12 â€“ 8 + 7
= 23 â€“ 20
= 3
3. Find the value of the following expressions, when x = â€“1:
(i) 2x â€“ 7
Solution:-
From the question it is given that x = -1
Then, substitute the value of x in the question
= (2 Ã— -1) â€“ 7
= â€“ 2 â€“ 7
= â€“ 9
(ii) â€“ x + 2
Solution:-
From the question it is given that x = -1
Then, substitute the value of x in the question
= â€“ (-1) + 2
= 1 + 2
= 3
Â
(iii) x^{2} + 2x + 1
Solution:-
From the question it is given that x = -1
Then, substitute the value of x in the question
= (-1)^{2} + (2 Ã— -1) + 1
= 1 â€“ 2 + 1
= 2 â€“ 2
= 0
(iv) 2x^{2} â€“ x â€“ 2
Solution:-
From the question it is given that x = -1
Then, substitute the value of x in the question
= (2 Ã— (-1)^{2}) â€“ (-1) â€“ 2
= (2 Ã— 1) + 1 â€“ 2
= 2 + 1 â€“ 2
= 3 â€“ 2
= 1
4. If a = 2, b = â€“ 2, find the value of:
(i) a^{2} + b^{2}
Solution:-
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
Â
(ii) a^{2} + ab + b^{2}
Solution:-
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 Ã— -2) + (-2)^{2}
= 4 + (-4) + (4)
= 4 â€“ 4 + 4
= 4
(iii) a^{2} â€“ b^{2}
Solution:-
From the question it is given that a = 2, b = -2
Then, substitute the value of a and b in the question
= 2^{2} â€“ (-2)^{2}
= 4 â€“ (4)
= 4 â€“ 4
= 0
5. When a = 0, b = â€“ 1, find the value of the given expressions:
(i) 2a + 2b
Solution:-
From the question it is given that a = 0, b = -1
Then, substitute the value of a and b in the question
= (2 Ã— 0) + (2 Ã— -1)
= 0 â€“ 2
= -2
(ii) 2a^{2} + b^{2} + 1
Solution:-
From the question it is given that a = 0, b = -1
Then, substitute the value of a and b in the question
= (2 Ã— 0^{2}) + (-1)^{2} + 1
= 0 + 1 + 1
= 2
(iii) 2a^{2}b + 2ab^{2} + ab
Solution:-
From the question it is given that a = 0, b = -1
Then, substitute the value of a and b in the question
= (2 Ã— 0^{2} Ã— -1) + (2 Ã— 0 Ã— (-1)^{2}) + (0 Ã— -1)
= 0 + 0 +0
= 0
(iv) a^{2} + ab + 2
Solution:-
From the question it is given that a = 0, b = -1
Then, substitute the value of a and b in the question
= (0^{2}) + (0 Ã— (-1)) + 2
= 0 + 0 + 2
= 2
6. Simplify the expressions and find the value if x is equal to 2
(i) x + 7 + 4 (x â€“ 5)
Solution:-
From the question it is given that x = 2
We have,
= x + 7 + 4x â€“ 20
= 5x + 7 â€“ 20
Then, substitute the value of x in the equation
= (5 Ã— 2) + 7 â€“ 20
= 10 + 7 â€“ 20
= 17 â€“ 20
= â€“ 3
(ii) 3 (x + 2) + 5x â€“ 7
Solution:-
From the question it is given that x = 2
We have,
= 3x + 6 + 5x â€“ 7
= 8x â€“ 1
Then, substitute the value of x in the equation
= (8 Ã— 2) â€“ 1
= 16 â€“ 1
= 15
(iii) 6x + 5 (x â€“ 2)
Solution:-
From the question it is given that x = 2
We have,
= 6x + 5x â€“ 10
= 11x â€“ 10
Then, substitute the value of x in the equation
= (11 Ã— 2) â€“ 10
= 22 â€“ 10
= 12
(iv) 4(2x â€“ 1) + 3x + 11
Solution:-
From the question it is given that x = 2
We have,
= 8x â€“ 4 + 3x + 11
= 11x + 7
Then, substitute the value of x in the equation
= (11 Ã— 2) + 7
= 22 + 7
= 29
7. Simplify these expressions and find their values if x = 3, a = â€“ 1, b = â€“ 2.
(i) 3x â€“ 5 â€“ x + 9
Solution:-
From the question it is given that x = 3
We have,
= 3x â€“ x â€“ 5 + 9
= 2x + 4
Then, substitute the value of x in the equation
= (2 Ã— 3) + 4
= 6 + 4
= 10
Â
(ii) 2 â€“ 8x + 4x + 4
Solution:-
From the question it is given that x = 3
We have,
= 2 + 4 â€“ 8x + 4x
= 6 â€“ 4x
Then, substitute the value of x in the equation
= 6 â€“ (4 Ã— 3)
= 6 â€“ 12
= â€“ 6
Â
(iii) 3a + 5 â€“ 8a + 1
Solution:-
From the question it is given that a = -1
We have,
= 3a â€“ 8a + 5 + 1
= â€“ 5a + 6
Then, substitute the value of a in the equation
= â€“ (5 Ã— (-1)) + 6
= â€“ (-5) + 6
= 5 + 6
= 11
Â
(iv) 10 â€“ 3b â€“ 4 â€“ 5b
Solution:-
From the question it is given that b = -2
We have,
= 10 â€“ 4 â€“ 3b â€“ 5b
= 6 â€“ 8b
Then, substitute the value of b in the equation
= 6 â€“ (8 Ã— (-2))
= 6 â€“ (-16)
= 6 + 16
= 22
Â
(v) 2a â€“ 2b â€“ 4 â€“ 5 + a
Solution:-
From the question it is given that a = -1, b = -2
We have,
= 2a + a â€“ 2b â€“ 4 â€“ 5
= 3a â€“ 2b â€“ 9
Then, substitute the value of a and b in the equation
= (3 Ã— (-1)) â€“ (2 Ã— (-2)) â€“ 9
= -3 â€“ (-4) â€“ 9
= â€“ 3 + 4 â€“ 9
= -12 + 4
= -8
Â
8. (i) If z = 10, find the value of z^{3} â€“ 3(z â€“ 10).
Solution:-
From the question it is given that z = 10
We have,
= z^{3} â€“ 3z + 30
Then, substitute the value of z in the equation
= (10)^{3} â€“ (3 Ã— 10) + 30
= 1000 â€“ 30 + 30
= 1000
(ii) If p = â€“ 10, find the value of p^{2} â€“ 2p â€“ 100
Solution:-
From the question it is given that p = -10
We have,
= p^{2} â€“ 2p â€“ 100
Then, substitute the value of p in the equation
= (-10)^{2} â€“ (2 Ã— (-10)) â€“ 100
= 100 + 20 â€“ 100
= 20
9. What should be the value of a if the value of 2x^{2} + x â€“ a equals to 5, when x = 0?
Solution:-
From the question it is given that x = 0
We have,
2x^{2} + x â€“ a = 5
a = 2x^{2} + x â€“ 5
Then, substitute the value of x in the equation
a = (2 Ã— 0^{2}) + 0 â€“ 5
a = 0 + 0 â€“ 5
a = -5
10. Simplify the expression and find its value when a = 5 and b = â€“ 3.
2(a^{2} + ab) + 3 â€“ ab
Solution:-
From the question it is given that a = 5 and b = -3
We have,
= 2a^{2} + 2ab + 3 â€“ ab
= 2a^{2} + ab + 3
Then, substitute the value of a and b in the equation
= (2 Ã— 5^{2}) + (5 Ã— (-3)) + 3
= (2 Ã— 25) + (-15) + 3
= 50 â€“ 15 + 3
= 53 â€“ 15
= 38
Exercise 12.4 Page: 246
1. 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
Solution:-
(a) From the question it is given that the numbers of segments required to form n digits of the kind
is (5n + 1)
Then,
The number of segments required to form 5 digits = ((5 Ã— 5) + 1)
= (25 + 1)
= 26
The number of segments required to form 10 digits = ((5 Ã— 10) + 1)
= (50 + 1)
= 51
The number of segments required to form 100 digits = ((5 Ã— 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
is (3n + 1)
Then,
The number of segments required to form 5 digits = ((3 Ã— 5) + 1)
= (15 + 1)
= 16
The number of segments required to form 10 digits = ((3 Ã— 10) + 1)
= (30 + 1)
= 31
The number of segments required to form 100 digits = ((3 Ã— 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
is (5n + 2)
Then,
The number of segments required to form 5 digits = ((5 Ã— 5) + 2)
= (25 + 2)
= 27
The number of segments required to form 10 digits = ((5 Ã— 10) + 2)
= (50 + 2)
= 52
The number of segments required to form 100 digits = ((5 Ã— 100) + 1)
= (500 + 2)
= 502
2. Use the given algebraic expression to complete the table of number patterns.
S. No. | Expression | Terms | |||||||||
1^{st} | 2^{nd} | 3^{rd} | 4^{th} | 5^{th} | â€¦ | 10^{th} | â€¦ | 100^{th} | â€¦ | ||
(i) | 2n â€“ 1 | 1 | 3 | 5 | 7 | 9 | â€“ | 19 | â€“ | â€“ | â€“ |
(ii) | 3n + 2 | 5 | 8 | 11 | 14 | â€“ | â€“ | â€“ | â€“ | â€“ | â€“ |
(iii) | 4n + 1 | 5 | 9 | 13 | 17 | â€“ | â€“ | â€“ | â€“ | â€“ | â€“ |
(iv) | 7n + 20 | 27 | 34 | 41 | 48 | â€“ | â€“ | â€“ | â€“ | â€“ | â€“ |
(v) | n^{2} + 1 | 2 | 5 | 10 | 17 | â€“ | â€“ | â€“ | â€“ | 10001 | â€“ |
Solution:-
(i) From the table (2n â€“ 1)
Then, 100^{th }term =?
Where n = 100
= (2 Ã— 100) â€“ 1
= 200 â€“ 1
= 199
(ii) From the table (3n + 2)
5^{th }term =?
Where n = 5
= (3 Ã— 5) + 2
= 15 + 2
= 17
Then, 10^{th }term =?
Where n = 10
= (3 Ã— 10) + 2
= 30 + 2
= 32
Then, 100^{th }term =?
Where n = 100
= (3 Ã— 100) + 2
= 300 + 2
= 302
(iii) From the table (4n + 1)
5^{th }term =?
Where n = 5
= (4 Ã— 5) + 1
= 20 + 1
= 21
Then, 10^{th }term =?
Where n = 10
= (4 Ã— 10) + 1
= 40 + 1
= 41
Then, 100^{th }term =?
Where n = 100
= (4 Ã— 100) + 1
= 400 + 1
= 401
(iv) From the table (7n + 20)
5^{th }term =?
Where n = 5
= (7 Ã— 5) + 20
= 35 + 20
= 55
Then, 10^{th }term =?
Where n = 10
= (7 Ã— 10) + 20
= 70 + 20
= 90
Then, 100^{th }term =?
Where n = 100
= (7 Ã— 100) + 20
= 700 + 20
= 720
(v) From the table (n^{2} + 1)
5^{th }term =?
Where n = 5
= (5^{2}) + 1
= 25+ 1
= 26
Then, 10^{th }term =?
Where n = 10
= (10^{2}) + 1
= 100 + 1
= 101
So the table is completed below.
S. No. | Expression | Terms | |||||||||
1^{st} | 2^{nd} | 3^{rd} | 4^{th} | 5^{th} | â€¦ | 10^{th} | â€¦ | 100^{th} | â€¦ | ||
(i) | 2n â€“ 1 | 1 | 3 | 5 | 7 | 9 | â€“ | 19 | â€“ | 199 | â€“ |
(ii) | 3n + 2 | 5 | 8 | 11 | 14 | 17 | â€“ | 32 | â€“ | 302 | â€“ |
(iii) | 4n + 1 | 5 | 9 | 13 | 17 | 21 | â€“ | 41 | â€“ | 401 | â€“ |
(iv) | 7n + 20 | 27 | 34 | 41 | 48 | 55 | â€“ | 90 | â€“ | 720 | â€“ |
(v) | n^{2} + 1 | 2 | 5 | 10 | 17 | 26 | â€“ | 101 | â€“ | 10001 | â€“ |
Â