# NCERT Solutions for Class 7 Maths Chapter 10 Algebraic Expressions

NCERT Solutions for Class 7 Maths Chapter 10 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 10.

**Exercise 10.1 **

**1. Get the algebraic expressions in the following cases using variables, constants and arithmetic operations.**

**(i) Subtraction of z from y.**

**Solution:-**

= Y – z

**(ii) One-half of the sum of numbers x and y.**

**Solution:-**

= ½ (x + y)

= (x + y)/2

**(iii) The number z multiplied by itself.**

**Solution:-**

= z × z

= z^{2}

**(iv) One-fourth of the product of numbers p and q.**

**Solution:-**

= ¼ (p × q)

= pq/4

**(v) Numbers x and y, both squared and added.**

**Solution:-**

= x^{2 }+ y^{2}

**(vi) Number 5 is 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 – (y × z)

= 10 – yz

**(viii) Sum of numbers a and b subtracted from their product.**

**Solution:-**

= (a × b) – (a + b)

= ab – (a + b)

**2. (i) Identify the terms and their factors in the following expressions.**

**Show the terms and factors by tree diagrams.**

**(a) x – 3**

**Solution:-**

Expression: x – 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 – y ^{3}**

**Solution:-**

Expression: y – 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) – ab + 2b ^{2} – 3a^{2}**

**Solution:-**

Expression: -ab + 2b^{2} – 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) – 4x + 5 (b) – 4x + 5y (c) 5y + 3y ^{2} (d) xy + 2x^{2}y^{2}**

**(e) pq + q (f) 1.2 ab – 2.4 b + 3.6 a (g) ¾ x + ¼**

**(h) 0.1 p ^{2} + 0.2 q^{2}**

**Solution:-**

Expressions are defined as numbers, symbols and operators (such as +. –, × and ÷) grouped that show the value of something.

In algebra, a term is a single number or variable or numbers and variables multiplied together. Terms are separated by + or – signs or sometimes by division.

Factors are 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 + 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 – 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 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 – 3t ^{2} (ii) 1 + t + t^{2} + t^{3} (iii) x + 2xy + 3y (iv) 100m + 1000n (v) – 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 are defined as numbers, symbols and operators (such as +. –, × and ÷) grouped that show the value of something.

In algebra, a term is a single number or variable or numbers and variables multiplied together. Terms are separated by + or – signs or sometimes by division.

**NCERT Solutions for Class 7 Maths Chapter 9 Perimeter and Area**

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) have a coefficient of 1 (x is 1x).

Sl.No. | Expression | Terms | Coefficients |

(i) | 5 – 3t^{2} | – 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) | – p^{2}q^{2} + 7pq | -p^{2}q^{2}7pq | -1 7 |

(vi) | 1.2 a + 0.8 b | 1.2a0.8b | 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} – 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} – 8yx | – 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 – xy ^{2} (ii) 5y^{2} + 7x (iii) 2x^{2}y – 15xy^{2} + 7y^{2}**

**Solution:-**

Sl.No. | Expression | Terms | Coefficient of y^{2} |

(i) | 8 – xy^{2} | – xy^{2} | – x |

(ii) | 5y^{2} + 7x | 5y^{2} | 5 |

(iii) | 2x^{2}y – 15xy^{2} + 7y^{2} | – 15xy^{2}7y ^{2} | – 15x 7 |

**5. Classify into monomials, binomials and trinomials.**

**(i) 4y – 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 – 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 – a – b**

**Solution:-**

Trinomial.

An expression which contains three terms is called a trinomial.

**(vi) 5 – 3t**

**Solution:-**

Binomial.

An expression which contains two unlike terms is called a binomial.

**(vii) 4p ^{2}q – 4pq^{2}**

**Solution:-**

Binomial.

An expression which contains two unlike terms is called a binomial.

**(viii) 7mn**

**Solution:-**

Monomial.

An expression with only one term is called a monomial.

**(ix) z ^{2} – 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 terms have the same algebraic factors, they are like terms.

**(ii) –7x, (5/2)x**

**Solution:-**

Like term.

When terms have the same algebraic factors, they are like terms.

**(iii) – 29x, – 29y**

**Solution:-**

Unlike terms.

The terms have different algebraic factors, they are unlike terms.

**(iv) 14xy, 42yx**

**Solution:-**

Like term.

When terms have the same algebraic factors, they are like terms.

**(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) – xy ^{2}, – 4yx^{2}, 8x^{2}, 2xy^{2}, 7y, – 11x^{2}, – 100x, – 11yx, 20x^{2}y, – 6x^{2}, y, 2xy, 3x**

**Solution:-**

When terms have the same algebraic factors, they are like terms.

They are,

– xy^{2}, 2xy^{2}

– 4yx^{2}, 20x^{2}y

8x^{2}, – 11x^{2}, – 6x^{2}

7y, y

– 100x, 3x

– 11yx, 2xy

**(b) 10pq, 7p, 8q, – p ^{2}q^{2}, – 7qp, – 100q, – 23, 12q^{2}p^{2}, – 5p^{2}, 41, 2405p, 78qp,**

**13p ^{2}q, qp^{2}, 701p^{2}**

**Solution:-**

When terms have the same algebraic factors, they are like terms.

They are,

10pq, – 7qp, 78qp

7p, 2405p

8q, – 100q

– p^{2}q^{2}, 12q^{2}p^{2}

– 23, 41

– 5p^{2}, 701p^{2}

13p^{2}q, qp^{2}

**Exercise 10.2 **

**1. Simplify combining like terms.**

**(i) 21b – 32 + 7b – 20b**

**Solution:-**

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) – z ^{2} + 13z^{2} – 5z + 7z^{3} – 15z**

**Solution:-**

When terms have the same algebraic factors, they are like terms.

Then,

= 7z^{3} + (-z^{2} + 13z^{2}) + (-5z – 15z)

= 7z^{3 }+ z^{2 }(-1 + 13) + z (-5 – 15)

= 7z^{3 }+ z^{2} (12) + z (-20)

= 7z^{3 }+ 12z^{2} – 20z

**(iii) p – (p – q) – q – (q – p)**

**Solution:-**

When terms have the same algebraic factors, they are like terms.

Then,

= p – p + q – q – q + p

= p – q

**(iv) 3a – 2b – ab – (a – b + ab) + 3ab + b – a**

**Solution:-**

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) 5x ^{2}y – 5x^{2} + 3yx^{2} – 3y^{2} + x^{2} – y^{2} + 8xy^{2} – 3y^{2}**

**Solution:-**

When terms have the same algebraic factors, they are like terms.

Then,

= 5x^{2}y + 3yx^{2} – 5x^{2} + x^{2} – 3y^{2} – y^{2} – 3y^{2}

= x^{2}y (5 + 3) + x^{2} (- 5 + 1) + y^{2} (-3 – 1 -3) + 8xy^{2}

= x^{2}y (8) + x^{2} (-4) + y^{2} (-7) + 8xy^{2}

= 8x^{2}y – 4x^{2} – 7y^{2} + 8xy^{2}

**(vi) (3y ^{2} + 5y – 4) – (8y – y^{2} – 4)**

**Solution:-**

When terms have the same algebraic factors, they are like terms.

Then,

= 3y^{2} + 5y – 4 – 8y + y^{2} + 4

= 3y^{2} + y^{2} + 5y – 8y – 4 + 4

= y^{2} (3 + 1) + y (5 – 8) + (-4 + 4)

= y^{2} (4) + y (-3) + (0)

= 4y^{2} – 3y

**2. Add:**

**(i) 3mn, – 5mn, 8mn, – 4mn**

**Solution:-**

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) t – 8tz, 3tz – z, z – t**

**Solution:-**

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) – 7mn + 5, 12mn + 2, 9mn – 8, – 2mn – 3**

**Solution:-**

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) a + b – 3, b – a + 3, a – b + 3**

**Solution:-**

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

**THE NATURE CONSERVANCY: NURTURING A SUSTAINABLE FUTURE**

**(v) 14x + 10y – 12xy – 13, 18 – 7x – 10y + 8xy, 4xy**

**Solution:-**

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) 5m – 7n, 3n – 4m + 2, 2m – 3mn – 5**

**Solution:-**

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) 4x ^{2}y, – 3xy^{2}, –5xy^{2}, 5x^{2}y**

**Solution:-**

When terms 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 – 3xy^{2} – 5xy^{2}

= x^{2}y (4 + 5) + xy^{2 }(-3 – 5)

= x^{2}y (9) + xy^{2} (- 8)

= 9x^{2}y – 8xy^{2}

**(viii) 3p ^{2}q^{2} – 4pq + 5, – 10 p^{2}q^{2}, 15 + 9pq + 7p^{2}q^{2}**

**Solution:-**

When terms have the same algebraic factors, they are like terms.

Then, we have to add the like terms.

= 3p^{2}q^{2} – 4pq + 5 + (- 10p^{2}q^{2}) + 15 + 9pq + 7p^{2}q^{2}

= 3p^{2}q^{2} – 10p^{2}q^{2} + 7p^{2}q^{2} – 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 – 4a, 4b – ab, 4a – 4b**

**Solution:-**

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) x ^{2} – y^{2} – 1, y^{2} – 1 – x^{2}, 1 – x^{2} – y^{2}**

**Solution:-**

When terms have the same algebraic factors, they are like terms.

Then, we have to add the like terms.

= x^{2} – y^{2} – 1 + (y^{2} – 1 – x^{2}) + (1 – x^{2} – y^{2})

= x^{2} – y^{2} – 1 + y^{2} – 1 – x^{2} + 1 – x^{2} – y^{2}

= x^{2} – x^{2} – x^{2} – y^{2} + y^{2} – y^{2 }– 1 – 1 + 1

= x^{2} (1 – 1- 1) + y^{2} (-1 + 1 – 1) + (-1 -1 + 1)

= x^{2} (1 – 2) + y^{2} (-2 +1) + (-2 + 1)

= x^{2} (-1) + y^{2} (-1) + (-1)

= -x^{2} – y^{2} -1

**3. Subtract:**

**(i) –5y ^{2} from y^{2}**

**Solution:-**

When terms have the same algebraic factors, they are like terms.

Then, we have to subtract the like terms.

= y^{2} – (-5y^{2})

= y^{2} + 5y^{2}

= 6y^{2}

**(ii) 6xy from –12xy**

**Solution:-**

When terms have the same algebraic factors, they are like terms.

Then, we have to subtract the like terms.

= -12xy – 6xy

= – 18xy

**(iii) (a – b) from (a + b)**

**Solution:-**

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) a (b – 5) from b (5 – a)**

**Solution:-**

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) –m ^{2} + 5mn from 4m^{2} – 3mn + 8**

**Solution:-**

When terms have the same algebraic factors, they are like terms.

Then, we have to subtract the like terms.

= 4m^{2} – 3mn + 8 – (- m^{2} + 5mn)

= 4m^{2} – 3mn + 8 + m^{2} – 5mn

= 4m^{2} + m^{2} – 3mn – 5mn + 8

= 5m^{2 }– 8mn + 8

**(vi) – x ^{2} + 10x – 5 from 5x – 10**

**Solution:-**

When terms have the same algebraic factors, they are like terms.

Then, we have to subtract the like terms.

= 5x – 10 – (-x^{2} + 10x – 5)

= 5x – 10 + x^{2} – 10x + 5

= x^{2} + 5x – 10x – 10 + 5

= x^{2} – 5x – 5

**(vii) 5a ^{2} – 7ab + 5b^{2} from 3ab – 2a^{2} – 2b^{2}**

**Solution:-**

When terms have the same algebraic factors, they are like terms.

Then, we have to subtract the like terms.

= 3ab – 2a^{2} – 2b^{2} – (5a^{2} – 7ab + 5b^{2})

= 3ab – 2a^{2} – 2b^{2} – 5a^{2 }+ 7ab – 5b^{2}

= 3ab + 7ab – 2a^{2} – 5a^{2} – 2b^{2} – 5b^{2}

= 10ab – 7a^{2} – 7b^{2}

**(viii) 4pq – 5q ^{2} – 3p^{2} from 5p^{2} + 3q^{2} – pq**

**Solution:-**

When terms have the same algebraic factors, they are like terms.

Then, we have to subtract the like terms.

= 5p^{2} + 3q^{2} – pq – (4pq – 5q^{2} – 3p^{2})

= 5p^{2} + 3q^{2} – pq – 4pq + 5q^{2} + 3p^{2}

= 5p^{2} + 3p^{2} + 3q^{2} + 5q^{2} – pq – 4pq

= 8p^{2} + 8q^{2} – 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} + 3xy) – (x^{2} + xy + y^{2})

p = 2x^{2} + 3xy – x^{2} – xy – y^{2}

p = 2x^{2} – x^{2} + 3xy – xy – y^{2}

p = x^{2} + 2xy – y^{2}

131

**(b) What should be subtracted from 2a + 8b + 10 to get – 3a + 7b + 16?**

**Solution:-**

Let us assume x is 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

**5. What should be taken away from 3x ^{2} – 4y^{2} + 5xy + 20 to obtain – x^{2} – y^{2} + 6xy + 20?**

**Solution:-**

Let us assume a be the required term.

Then,

3x^{2} – 4y^{2} + 5xy + 20 – a = -x^{2} – y^{2} + 6xy + 20

a = 3x^{2} – 4y^{2} + 5xy + 20 – (-x^{2} – y^{2} + 6xy + 20)

a = 3x^{2} – 4y^{2} + 5xy + 20 + x^{2} + y^{2} – 6xy – 20

a = 3x^{2} + x^{2} – 4y^{2} + y^{2} + 5xy – 6xy + 20 – 20

a = 4x^{2} – 3y^{2} – xy

65

**6. (a) From the sum of 3x – y + 11 and – y – 11, subtract 3x – y – 11.**

**Solution:-**

First, we must find 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

60

**(b) From the sum of 4 + 3x and 5 – 4x + 2x ^{2}, subtract the sum of 3x^{2} – 5x and**

**–x ^{2} + 2x + 5.**

**Solution:-**

First, we have to find out the sum of 4 + 3x and 5 – 4x + 2x^{2}

= 4 + 3x + (5 – 4x + 2x^{2})

= 4 + 3x + 5 – 4x + 2x^{2}

= 4 + 5 + 3x – 4x + 2x^{2}

= 9 – x + 2x^{2}

= 2x^{2} – x + 9 … [equation 1]

Then, we have to find out the sum of 3x^{2} – 5x and – x^{2} + 2x + 5

= 3x^{2} – 5x + (-x^{2} + 2x + 5)

= 3x^{2} – 5x – x^{2} + 2x + 5

= 3x^{2} – x^{2} – 5x + 2x + 5

= 2x^{2} – 3x + 5 … [equation 2]

Now, we have to subtract equation (2) from equation (1)

= 2x^{2 }– x + 9 – (2x^{2} – 3x + 5)

= 2x^{2} – x + 9 – 2x^{2} + 3x – 5

= 2x^{2} – 2x^{2} – x + 3x + 9 – 5

= 2x + 4

Exercise 12.3 Page: 242

**1. If m = 2, find the value of:**

**(i) m – 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 – 5**

**Solution:-**

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) 9 – 5m**

**Solution:-**

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) 3m ^{2} – 2m – 7**

**Solution:-**

From the question, it is given that m = 2

Then, substitute the value of m in the question.

= (3 × 2^{2}) – (2 × 2) – 7

= (3 × 4) – (4) – 7

= 12 – 4 -7

= 12 – 11

= 1

**(v) (5m/2) – 4**

**Solution:-**

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

**2. If p = – 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 × (-2)) + 7

= -8 + 7

= -1

**(ii) – 3p ^{2} + 4p + 7**

**Solution:-**

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) – 2p ^{3} – 3p^{2} + 4p + 7**

**Solution:-**

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

**3. Find the value of the following expressions when x = –1:**

**(i) 2x – 7**

**Solution:-**

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) – x + 2**

**Solution:-**

From the question, it is given that x = -1

Then, substitute the value of x in the question.

= – (-1) + 2

= 1 + 2

= 3

**(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 × -1) + 1

= 1 – 2 + 1

= 2 – 2

= 0

**(iv) 2x ^{2} – x – 2**

**Solution:-**

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

**4. If a = 2, b = – 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

**(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 × -2) + (-2)^{2}

= 4 + (-4) + (4)

= 4 – 4 + 4

= 4

**(iii) 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)

= 4 – 4

= 0

**5. When a = 0, b = – 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 × 0) + (2 × -1)

= 0 – 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 × 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 × 0^{2} × -1) + (2 × 0 × (-1)^{2}) + (0 × -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 × (-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 – 5)**

**Solution:-**

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) 3 (x + 2) + 5x – 7**

**Solution:-**

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) 6x + 5 (x – 2)**

**Solution:-**

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) 4(2x – 1) + 3x + 11**

**Solution:-**

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

**7. Simplify these expressions and find their values if x = 3, a = – 1, b = – 2.**

**(i) 3x – 5 – x + 9**

**Solution:-**

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) 2 – 8x + 4x + 4**

**Solution:-**

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) 3a + 5 – 8a + 1**

**Solution:-**

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) 10 – 3b – 4 – 5b**

**Solution:-**

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) 2a – 2b – 4 – 5 + a**

**Solution:-**

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

**8. (i) If z = 10, find the value of z ^{3} – 3(z – 10).**

**Solution:-**

From the question, it is given that z = 10

We have,

= z^{3} – 3z + 30

Then, substitute the value of z in the equation.

= (10)^{3} – (3 × 10) + 30

= 1000 – 30 + 30

= 1000

**(ii) If p = – 10, find the value of p ^{2} – 2p – 100**

**Solution:-**

From the question, it is given that p = -10

We have,

= p^{2} – 2p – 100

Then, substitute the value of p in the equation.

= (-10)^{2} – (2 × (-10)) – 100

= 100 + 20 – 100

= 20

**9. What should be the value of a** if the value of 2×2 + x – a equals **5, when x = 0?**

**Solution:-**

From the question, it is given that x = 0

We have,

2x^{2} + x – a = 5

a = 2x^{2} + x – 5

Then, substitute the value of x in the equation.

a = (2 × 0^{2}) + 0 – 5

a = 0 + 0 – 5

a = -5

36

**10. Simplify the expression and find its value when a = 5 and b = – 3.**

**2(a ^{2} + ab) + 3 – ab**

**Solution:-**

From the question, it is given that a = 5 and b = -3

We have,

= 2a^{2} + 2ab + 3 – ab

= 2a^{2} + ab + 3

Then, substitute the values of a and b in the equation.

= (2 × 5^{2}) + (5 × (-3)) + 3

= (2 × 25) + (-15) + 3

= 50 – 15 + 3

= 53 – 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 number of segments required to form n digits of the kind

is (5n + 1)

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 number of segments required to form n digits of the kind

is (3n + 1)

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 number of segments required to form n digits of the kind

is (5n + 2)

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

**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} | … | 10^{th} | … | 100^{th} | … | ||

(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) | n^{2} + 1 | 2 | 5 | 10 | 17 | – | – | – | – | 10001 | – |

**Solution:-**

**(i)** From the table (2n – 1)

Then, 100^{th }term =?

Where n = 100

= (2 × 100) – 1

= 200 – 1

= 199

**(ii)** From the table (3n + 2)

5^{th }term =?

Where n = 5

= (3 × 5) + 2

= 15 + 2

= 17

Then, 10^{th }term =?

Where n = 10

= (3 × 10) + 2

= 30 + 2

= 32

Then, 100^{th }term =?

Where n = 100

= (3 × 100) + 2

= 300 + 2

= 302

**(iii)** From the table (4n + 1)

5^{th }term =?

Where n = 5

= (4 × 5) + 1

= 20 + 1

= 21

Then, 10^{th }term =?

Where n = 10

= (4 × 10) + 1

= 40 + 1

= 41

Then, 100^{th }term =?

Where n = 100

= (4 × 100) + 1

= 400 + 1

= 401

**(iv)** From the table (7n + 20)

5^{th }term =?

Where n = 5

= (7 × 5) + 20

= 35 + 20

= 55

Then, 10^{th }term =?

Where n = 10

= (7 × 10) + 20

= 70 + 20

= 90

Then, 100^{th }term =?

Where n = 100

= (7 × 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} | … | 10^{th} | … | 100^{th} | … | ||

(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) | n^{2} + 1 | 2 | 5 | 10 | 17 | 26 | – | 101 | – | 10001 | – |