NCERT Solutions Class 7 Mathematics Chapter 12 Algebraic Expressions

Exercise 12.1

Q.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 m and n.
(vii) Product of numbers y and z subtracted from 10.
(viii) Sum of numbers a and b subtracted from their product.
Sol.1) i) 𝑦 − 𝑧      ii) 𝑥+𝑦/2
iii) 𝑧²                      iv) 𝑝𝑞/4
v) 𝑥² + 𝑦²             vi) 3𝑚𝑛 + 5         
vii) 10 − 𝑦𝑧        viii) 𝑎𝑏 − (𝑎 + 𝑏)

Q.2) (i) Identify the terms and their factors in the following expressions, show the terms and factors by tree diagram:
(a) 𝑥 -3                  (b) 1 + 𝑥 + 𝑥²             (c) 𝑦 − 𝑦³
(d) 5𝑥𝑦² + 7𝑥²𝑦    (e) −𝑎𝑏 + 2𝑏² − 3𝑎²
(ii) Identify the 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. 𝑎𝑏 − 2.4𝑏 + 3.6𝑎   (g) 3/4 𝑥 + 1/4   (h) 0.1𝑝² + 0.2𝑞² 
Sol.2)

ii) (a) -4𝑥 + 5                    (b) −4𝑥 + 5𝑦
Terms −4𝑥, 5                        Terms: −4𝑥, 5𝑦
Factors: −4, 𝑥;                      5 factors: −4, 𝑥; 5, 𝑦

(c) 5𝑦 + 3𝑦²                     (d) 𝑥𝑦 + 2𝑥²𝑦²
Terms: 5𝑦, 3𝑦²                     Terms: 𝑥𝑦, 2𝑥²𝑦²
Factors: 5, 𝑦; 3, 𝑦, 𝑦             Factors: 𝑥, 𝑦 ; 2𝑥, 𝑥, 𝑦, 𝑦

(e) 𝑝𝑞 + 𝑞                         (f) 1.2𝑎𝑏 − 2.4𝑏 + 3.6𝑎
Terms: 𝑝𝑞, 𝑞                        Terms: 1.2𝑎𝑏, −2.4𝑏, 3.6𝑎
Factors: 𝑝, 𝑞 ; 𝑞                    Factors: 1.2, 𝑎 , 𝑏 ; −2.4, 𝑏 ; 3.6, 𝑎

(g) 3/4 𝑥 + 1/4                  (h) 0.1𝑝2 + 0.2𝑞²
                                           Terms: 3/4 𝑥, 1/4
                                           Terms: 0.1𝑝², 0.2𝑞²
                                           Factors: 3/4, 𝑥 ; 1/4
                                           Factors: 0.1, 𝑝, 𝑝; 0.2, 𝑞, 𝑞
 

Q.3) 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) 1.2𝑎 + 0.8𝑏
(vii) 3.14𝑟²              (viii) 2(𝑙 + 𝑏)            (ix) 0.1𝑦 + 0.01𝑦²
Sol.3)

Q.4) (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) 2𝑥² − 15𝑥𝑦² + 7𝑦²
Sol.4)

Q.5) Classify into monomials, binomials and trinomials:
(i) 4𝑦 − 7𝑥 (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
Sol.5) Type of Polynomial
(i) 4𝑦 − 7𝑥                          Binomial
(ii) 𝑦²                                 Monomial
(iii) 𝑥 + 𝑦 − 𝑥𝑦                    Trinomial
(iv) 100                             Monomial
(v) 𝑎𝑏 − 𝑎 − 𝑏                    Trinomial
(vi) 5 − 3𝑡                          Binomial
(vii) 4𝑝²𝑞 − 4𝑝𝑞²               Binomial
(viii) 7𝑚𝑛                          Monomial
(ix) 𝑧² − 3𝑧 + 8                 Trinomial
(x) 𝑎² + 𝑏²                        Binomial
(xi) 𝑧² + 𝑧                         Binomial
(xii) 1 + 𝑥 + 𝑥²                  Trinomial

Q.6) 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
Sol.6)

The terms which have the same algebraic factors are called like terms and when the terms have different algebraic factors, they are called unlike terms.

Q.7) Identify like terms in the following:
a) −𝑥²𝑦, −4𝑦𝑥², 8𝑥², 2𝑥𝑦², 7𝑦, −11𝑥² − 100𝑥, −11𝑦𝑥, 20𝑥²𝑦, −6𝑥², 𝑦 , 2𝑥𝑦 , 3𝑥
b) 10𝑝𝑞, 7𝑝, 8𝑝, −𝑝²𝑞², −7𝑝𝑞 , −100𝑞 , −23, 12𝑝²𝑞², −5𝑝², 41, 2405𝑝,
78𝑞𝑝, 13𝑝²𝑞, 𝑞𝑝²701𝑝²
Sol.7) a) Like terms
i) −𝑥𝑦², 2𝑥𝑦² ii) −4𝑦𝑥², 20𝑥²𝑦 iii) 8𝑥², −11𝑥², −6𝑥²
iv) 7𝑦, 𝑦 v) −100𝑥, 3𝑥 vi) −11𝑦𝑥, 2𝑥𝑦
b) Like terms are:
i) 10𝑝𝑞, −7𝑝𝑞 , 78𝑝𝑞 ii) 7𝑝, 2405𝑝 iii) 8𝑞, −100𝑞
iv) −𝑝²𝑞², 12𝑝²𝑞² v) −12, 41 vi) −5𝑝², 701𝑝²
vii) 13𝑝²𝑞, 𝑞𝑝²

Exercise 12.2

Q.1) Simplify combining like terms:
i) 21𝑏 − 32 + 7𝑏 − 20𝑏 ii) −𝑧² + 13𝑧² − 5𝑥 + 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)
Sol.1) i) 21𝑏 − 32 + 7𝑏 − 20𝑏
= 21𝑏 + 7𝑏 − 20𝑏 − 32
= 28𝑏 − 20𝑏 − 32 = 8𝑏 − 32

ii) −z² + 13z² − 5𝑥 + 7z³ − 15𝑧
= 7z³ + (−z² + 13z²) − (5𝑧 + 15𝑧)
= 7z³ + 12z² − 20𝑧

iii) 𝑝 − (𝑝 − 𝑞) − 𝑞 − (𝑞 − 𝑝)
= 𝑝 − 𝑝 + 𝑞 − 𝑞 − 𝑞 + 𝑝
= 𝑝 − 𝑝 + 𝑝 + 𝑞 − 𝑞 − 𝑞 = 𝑝 − 𝑞

iv) 3𝑎 − 2𝑏 − 𝑎𝑏 − (𝑎 − 𝑏 + 𝑎𝑏) + 3𝑎𝑏 + 𝑏 − 𝑎
= 3𝑎 − 2𝑏 − 𝑎𝑏 − 𝑎 + 𝑏 − 𝑎𝑏 + 3𝑎𝑏 + 𝑏 − 𝑎
= 3𝑎 − 𝑎 − 𝑎 − 2𝑏 + 𝑏 + 𝑏 − 𝑎𝑏 − 𝑎𝑏 + 3𝑎𝑏
= (3𝑎 − 𝑎 − 𝑎) − (2𝑏 − 𝑏 − 𝑏) − (𝑎𝑏 + 𝑎𝑏 − 3𝑎𝑏)
= 𝑎 − 0 − (𝑎 − 𝑏)
= 𝑎 + 𝑎𝑏

v) 5𝑥²𝑦 − 5𝑥² + 3𝑦𝑥² − 3𝑦² + 𝑥² − 𝑦² + 8𝑥𝑦² − 3𝑦²
= 5𝑥²𝑦 + 3𝑦𝑥² + 8𝑥𝑦² − 5𝑥² + 𝑥² − 3𝑦² − 𝑦² − 3𝑦²
= (5𝑥²𝑦 + 3𝑥²𝑦) + 8𝑥𝑦² − (5𝑥² − 𝑥²) − (3𝑦² + 𝑦² + 3𝑦²)
= 8𝑥²𝑦 + 8𝑥𝑦² − 4𝑥² − 7𝑦²

vi) (3𝑦² + 5𝑦 − 4) − (8𝑦 − 𝑦² − 4)
= 3𝑦² + 5𝑦 − 4 − 8𝑦 + 𝑦2 + 4
= (3𝑦² + 𝑦²) + (5𝑦 − 8𝑦) − (4 − 4)
= 4𝑦² − 3𝑦 − 0 = 4𝑦² − 3𝑦

Q.2) 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, −10𝑝²𝑞², 15 + 9𝑝𝑞 + 7𝑝²𝑞²
ix) 𝑎𝑏 − 4𝑎, 4𝑏 − 𝑎𝑏, 4𝑎 − 4𝑏
x) 𝑥² − 𝑦² − 1, 𝑦² − 1 − 𝑥² , 1 − 𝑥² − 𝑦²
Sol.2) i) 3𝑚𝑛, −5𝑚𝑛, 8𝑚𝑛 − 4𝑚𝑛
= 3𝑚𝑛 + (−5𝑚𝑛) + 8𝑚𝑛 + (−4𝑚𝑛)
= (3 − 5 + 8 − 4)𝑚𝑛 = 2𝑚𝑛

ii) 𝑡 − 8𝑡𝑧, 3𝑡𝑧 − 𝑧, 𝑧 − 𝑡
= 𝑡 − 8𝑡𝑧 + 3𝑡𝑧 − 𝑧 + 𝑧 − 𝑡
= 𝑡 − 𝑡 − 8𝑡𝑧 + 3𝑡𝑧 − 𝑧 + 𝑧
= (1 − 1)𝑡 + (−8 + 3)𝑡𝑧 + (−1 + 1)𝑧
= 0 − 5𝑡𝑧 + 0 = −5𝑡𝑧

iii) −7𝑚𝑛 + 5, 12𝑚𝑛 + 2, 9𝑚𝑛 − 8, −2𝑚𝑛 − 3
= −7𝑚𝑛 + 5 + 12𝑚𝑛 + 2 + 9𝑚𝑛 − 8 + (−2𝑚𝑛) − 3
= −7𝑚𝑛 + 12𝑚𝑛 + 9𝑚𝑛 − 2𝑚𝑛 + 5 + 2 − 8 − 3
= (−7 + 12 + 9 − 2)𝑚𝑛 + 7 − 11
= 12𝑚𝑛 − 4

iv) 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 + 𝑏 + 3

v) 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
= 7𝑥 + 0𝑦 + 0𝑥𝑦 + 5 = 7𝑥 + 5

vi) 5𝑚 − 7𝑛, 3𝑛 − 4𝑚 + 2, 2𝑚 − 3𝑚𝑛 − 5
= 5𝑚 − 7𝑛 + 3𝑛 − 4𝑚 + 2 + 2𝑚 − 3𝑚𝑛 − 5
= 5𝑚 − 4𝑚 + 2𝑚 − 7𝑛 + 3𝑛 − 3𝑚𝑛 + 2 − 5
= (5 − 4 + 2)𝑚 + (−7 + 3)𝑛 − 3𝑚𝑛 − 3
= 3𝑚 − 4𝑛 + 3𝑚𝑛 − 3

vii) 4𝑥²𝑦, −3𝑥𝑦², −5𝑥𝑦², 5𝑥²𝑦
= 4𝑥²𝑦 + (−3𝑥𝑦²) + (−5𝑥𝑦²) + 5𝑥²𝑦
= 4𝑥²𝑦 + 5𝑥²𝑦 − 3𝑥𝑦² − 5𝑥𝑦²
= 9𝑥²𝑦 − 8𝑥𝑦²

viii) 3𝑝²𝑞² − 4𝑝𝑞 + 5, −10𝑝²𝑞², 15 + 9𝑝𝑞 + 7𝑝²𝑞²
= 3𝑝²𝑞² − 4𝑝𝑞 + 5 + (−10𝑝²𝑞²) + 15 + 9𝑝𝑞 + 7𝑝²𝑞²
= 3𝑝²𝑞² − 10𝑝²𝑞² + 7𝑝²𝑞² + 4𝑝𝑞 + 9𝑝𝑞 + 5 + 15
= (3 − 10 + 7)𝑝²𝑞² + (−4 + 9)𝑝𝑞 + 20
= 0𝑝²𝑞² + 5𝑝𝑞 + 20 = 5𝑝𝑞 + 20

ix) 𝑎𝑏 − 4𝑎, 4𝑏 − 𝑎𝑏, 4𝑎 − 4𝑏
= 𝑎𝑏 − 4𝑎 + 4𝑏 − 𝑎𝑏 + 4𝑎 − 4𝑏
= −4𝑎 + 4𝑎 + 4𝑏 − 4𝑏 + 𝑎𝑏 − 𝑎𝑏
= 0 + 0 + 0 = 0

x) 𝑥² − 𝑦² − 1, 𝑦² − 1 − 𝑥² , 1 − 𝑥² − 𝑦²
= 𝑥² − 𝑦² − 1 + 𝑦² − 1 − 𝑥² + 1 − 𝑥² − 𝑦^2
= 𝑥² − 𝑥² − 𝑥² − 𝑦² + 𝑦² − 𝑦² − 1 − 1 + 1
= (1 − 1 − 1)𝑥² + (−1 + 1 − 1)𝑦² − 1 − 1 + 1
= −𝑥² − 𝑦² − 1

Q.3) 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
Sol.3) i) 𝑦² − (−5𝑦²) = 𝑦² + 5𝑦²
= 6𝑦²

ii) −12𝑥𝑦 − (6𝑥𝑦) = −12𝑥𝑦 − 6𝑥𝑦
= −18𝑥𝑦

iii) (𝑎 + 𝑏) − (𝑎 − 𝑏) = 𝑎 + 𝑏 − 𝑎 + 𝑏
= 𝑎 − 𝑎 + 𝑏 + 𝑏 = 2𝑏

iv) 𝑏(5 − 𝑎) − 𝑎(𝑏 − 5) = 5𝑏 − 𝑎𝑏 − 𝑎𝑏 + 5𝑎
= 5𝑏 − 2𝑎𝑏 + 5𝑎
= 5𝑎 + 5𝑏 − 2𝑎𝑏

v) 4𝑚² − 3𝑚𝑛 + 8 − (−𝑚² + 5𝑚𝑛)
= 4𝑚² − 3𝑚𝑛 + 8 + 𝑚² − 5𝑚𝑛
= 4𝑚² + 𝑚² − 3𝑚𝑛 − 5𝑚𝑛 + 8
= 5𝑚² − 8𝑚𝑛 + 8

vi) 5𝑥 − 10 − (−𝑥² + 10𝑥 − 5)
= 5𝑥 − 10 + 𝑥² − 10𝑥 + 5
= 𝑥² + 5𝑥 − 10𝑥 − 10 + 5
= 𝑥² − 5𝑥 − 5

vii) 3𝑎𝑏 − 2𝑎² − 2𝑏² − (5𝑎² − 7𝑎𝑏 + 5𝑏²)
= 3𝑎𝑏 − 2𝑎² − 2𝑏² − 5𝑎² − 7𝑎𝑏 − 5𝑏²
= 3𝑎𝑏 + 7𝑎𝑏 − 2𝑎² − 5𝑎² − 2𝑏² − 5𝑏²
= 10𝑎𝑏 − 7𝑎² − 7𝑏²
= −7𝑎² − 7𝑏² + 10𝑎𝑏

viii) 5𝑝² + 3𝑞² − 𝑝𝑞 − (4𝑞 − 5𝑞² − 3𝑝²)
= 5𝑝² + 3𝑞² − 𝑝𝑞 − 4𝑝𝑞 + 5𝑞² + 3𝑝²
= 5𝑝² + 3𝑝² + 3𝑞² + 5𝑞² − 𝑝𝑞 − 4𝑝𝑞
= 8𝑝² + 8𝑞² − 5𝑝𝑞

Q.4) (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 ?
Sol.4) A) Let 𝑝 should be added
Then according to the question,
𝑥² + 𝑥𝑦 + 𝑦² + 𝑝 = 2𝑥² + 3𝑥𝑦
⇒ 𝑝 = 2𝑥² + 3𝑥𝑦 − (𝑥² + 𝑥𝑦 + 𝑦²)
⇒ 𝑝 = 2𝑥² + 3𝑥𝑦 − 𝑥² − 𝑥𝑦 − 𝑦²
⇒ 𝑝 = 2𝑥² − 𝑥² − 𝑦² + 3𝑥𝑦 − 𝑥𝑦
⇒ 𝑝 = 𝑥² − 𝑦² + 2𝑥𝑦
Hence, 𝑥² − 𝑦² + 2𝑥𝑦 should be added

B) let 𝑞 should be subtracted
Then according to question,
2𝑎 + 8𝑏 + 10 − 𝑞 = −3𝑎 + 7𝑏 + 16
⇒ −𝑞 = −3𝑎 + 7𝑏 + 16 − (2𝑎 + 8𝑏 + 10)
⇒ −𝑞 = −3𝑎 + 7𝑏 + 16 − 2𝑎 − 8𝑏 − 10
⇒ −𝑞 = −3𝑎 − 2𝑎 + 7𝑏 − 8𝑏 + 16 − 10
⇒ −𝑞 = −5𝑎 − 𝑏 + 6
⇒ 𝑞 = −(−5𝑎 − 𝑏 + 6)
⇒ 𝑞 = 5𝑎 + 𝑏 − 6

Q.5) What should be taken away from 3𝑥² − 4𝑦² + 5𝑥𝑦 + 20 to obtain −𝑥² − 𝑦² + 6𝑥𝑦 + 20 ?
Sol.5) Let 𝑞 should be subtracted
Then according to question,
3𝑥² − 4𝑦² + 5𝑥𝑦 + 20 − 𝑞 = −𝑥² − 𝑦² + 6𝑥𝑦 + 20
⇒ 𝑞 = 3𝑥² − 4𝑦² + 5𝑥𝑦 + 20 − (−𝑥² − 𝑦² + 6𝑥𝑦 + 20)
⇒ 𝑞 = 3𝑥² − 4𝑦² + 5𝑥𝑦 + 20 + 𝑥² + 𝑦² − 6𝑥𝑦 − 20
⇒ 𝑞 = 3𝑥² + 𝑥² − 4𝑦² + 𝑦² + 5𝑥𝑦 − 6𝑥𝑦 + 20 − 20
⇒ 𝑞 = 4𝑥² − 3𝑦² − 𝑥𝑦 + 0
Hence, 4𝑥² − 3𝑦² − 𝑥𝑦 should be subtracted.

Q.6) (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.
Sol.6) a) According to question,
(3𝑥 − 𝑦 + 11) + (−𝑦 − 11) − (3𝑥 − 𝑦 − 11) = 3𝑥 − 𝑦 + 11 − 𝑦 − 11 − 3𝑥 + 𝑦 + 11
= 3𝑥 − 3𝑥 − 𝑦 + 𝑦 + 11 − 11 + 11
= (3 − 3)𝑥 − (1 + 1 − 1)𝑦 + 11 + 11 − 11
= 0𝑥 − 𝑦 + 11 = −𝑦 + 11

b) According to the question,
[(4 + 3𝑥) + (5 − 4𝑥 + 2𝑥²)] − [(3𝑥² − 5𝑥) + (−𝑥² + 2𝑥 + 5)]
= [4𝑥 + 3𝑥 + 5 − 4𝑥 + 2𝑥²] − [3𝑥² − 5𝑥 − 𝑥² + 2𝑥 + 5]
= [2𝑥² + 3𝑥 − 4𝑥 + 5 + 4] − [3𝑥² − 𝑥² + 2𝑥 − 5𝑥 + 5]
= [2𝑥² − 𝑥 + 9] − [2𝑥² − 3𝑥 + 5]
= 2𝑥² − 𝑥 + 9 − 2𝑥² + 3𝑥 − 5
= 2𝑥² − 2𝑥² − 𝑥 + 3𝑥 + 9 − 5
= 2𝑥 + 4

Exercise 12.3

Q.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
Sol.1) i) 𝑚 − 2
= 2 − 2 = 0                          [putting 𝑚 = 2]
ii) 3𝑚 − 5
= 3 × 2 − 5                          [putting 𝑚 = 2]
= 6 − 5 = 1
iii) 9 − 5𝑚
= 9 − 5 × 2                          [putting 𝑚 = 2]
= 9 − 10 = −1

iv) 3𝑚2 − 2𝑚 − 7
= 3(2)² − 2(2) − 7               [putting 𝑚 = 2]
= 3 × 4 − 2 × 2 − 7
= 12 − 4 − 7
= 12 − 11 = 1

v) (5𝑚/2) − 4
= (5×2/2) − 4                       [putting 𝑚 = 2]
= (10/2) − 4
= 5 − 4 = 1

Q.2) If 𝑥 = −2, find
(i) 4𝑝 + 7
(ii) −3𝑝² + 4𝑝 + 7
(iii) −2𝑝³ − 3𝑝² + 4𝑝 + 7
Sol.2) (i) 4𝑝 + 7
= 4(−2) + 7                                         [putting 𝑝 = −2]
= −8 + 7 = −1

(ii) −3𝑝² + 4𝑝 + 7
= −3(−2)² + 4(−2) + 7                        [putting 𝑝 = −2]
= −3 × 4 − 8 + 7
= −12 − 8 + 7
= −20 + 7 = −13

(iii) −2𝑝³ − 3𝑝² + 4𝑝 + 7
= −2(−2)³ − 3(−2)² + 4(−2) + 7         [putting 𝑝 = −2]
= −2 × (−8) − 3 × 4 − 8 + 7
= 16 − 12 − 8 + 7
= −20 + 23 = 3

Q.3) Find the value of the following expressions, when 𝑥 =- 1:
(i) 2𝑥 − 7 ii) −𝑥 + 2 iii) 𝑥² + 2𝑥 + 1 iv) 2𝑥² − 𝑥 − 2
Sol.3) i) 2𝑥 − 7
= 2(−1) − 7                             [putting 𝑚 = 2]
= −2 − 7 = −9

ii) −𝑥 + 2
= −(−1) + 2                             [putting 𝑚 = 2]
= 1 + 2 = 3

iii) 𝑥² + 2𝑥 + 1
= (−1)² + 2(−1) + 1                 [putting 𝑚 = 2]
= 1 − 2 + 1
= 2 − 2 = 0

iv) 2𝑥² − 𝑥 − 2
= 2(−1)² − (−1) − 2                 [putting 𝑚 = 2]
= 2 × 1 + 1 − 2
= 2 + 1 − 2
= 3 − 2 = 1

Q.4) If 𝑎 = 2, 𝑏 = −2, find the value of:
i) 𝑎² + 𝑏² ii) 𝑎² + 𝑎𝑏 + 𝑏² iii) 𝑎² − 𝑏²
Sol.4) i) 𝑎² + 𝑏²
= (2)² + (2)²
= 4 + 4 = 8

ii) 𝑎² + 𝑎𝑏 + 𝑏²
= (2)² + (2)(−2) + (−2)²
= 4 − 4 + 4 = 4

iii) 𝑎² − 𝑏²
= (2)² − (−2)²
= 4 − 4 = 0

Q.5) 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
Sol.5) i) 2𝑎 + 2𝑏
= 2(0) + 2(−1)
= 0 − 2 = −2

ii) 2𝑎² + 𝑏² + 1
= 2(0)² + (−1)² + 1
= 2 × 0 + 1 + 1 = 0 + 2 = 0            [putting 𝑎 = 0, 𝑏 = −1]

iii) 2𝑎²𝑏 + 2𝑎𝑏² + 𝑎𝑏
= 2(0)²(−1) + 2(0)(−1)² + (0)(−1)
= 0 + 0 + 0 = 0

iv) 𝑎² + 𝑎𝑏 + 2
= (0)² + (0)(−1) + 2
= 0 + 0 + 2 = 2

Q.6) Simplify the following expressions and find the value at 𝑥 = 2:
(i) 𝑥 + 7 + 4(𝑥 − 5)
(ii) 3(𝑥 + 2) + 5𝑥 − 7
(iii) 10𝑥 + 4(𝑥 − 2)
(iv) 5(3𝑥 − 2) + 4𝑥 + 8
Sol.6) (i) 𝑥 + 7 + 4(𝑥 − 5)
= 𝑥 + 7 + 4𝑥 − 20
= 4𝑥 + 𝑥 + 7 − 20 = 5𝑥 − 13
= 5(2) − 13 = 10 − 13                    [𝑃𝑢𝑡𝑡𝑖𝑛𝑔 𝑥 = 2 ]
= −3

(ii) 3(𝑥 + 2) + 5𝑥 − 7
= 3𝑥 + 6 + 5𝑥 − 7
= 3𝑥 + 5𝑥 + 6 − 7 = 8𝑥 − 1
= 8( 2) – 1                                    [𝑃𝑢𝑡𝑡𝑖𝑛𝑔 𝑥 = 2 ]
= 16 – 1 = 15

(iii) 10𝑥 + 4(𝑥 − 2)
= 10𝑥 + 4𝑥 − 8
= 14𝑥 − 8
= 14( 2) – 8                                   [𝑃𝑢𝑡𝑡𝑖𝑛𝑔 𝑥 = 2 ]
28 – 8 = 20

(iv) 5(3𝑥 − 2) + 4𝑥 + 8
= 15𝑎 − 10 + 4𝑥 + 8
= 15𝑎 + 4𝑎 − 10 + 8 = 19𝑎 − 2 [𝑃𝑢𝑡𝑡𝑖𝑛𝑔 = 2 ]
= 19(2) − 2 = 38 − 2
= 36

Q.7) 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
Sol.7) Putting values 𝑥 = 3, 𝑎 =– 1, 𝑏 =– 2.

i) 3𝑥 – 5 – 𝑥 + 9
= 3𝑥 − 𝑥 − 5 + 9 = 2𝑥 + 4
= 2 × 3 + 4
= 6 + 4 = 10

ii) 2– 8𝑥 + 4𝑥 + 4
= −8𝑥 + 4𝑥 + 2 + 4 = −4𝑥 − 6
= −4 × 3 + 6
= −12 + 6 = −12

iii) 3𝑎 + 5 – 8𝑎 + 1
= 3𝑎 − 8𝑎 + 5 + 1 = −5𝑎 + 6
= −5(−1) + 6
= 5 + 6 = 11

iv) 10– 3𝑏– 4– 5𝑏
= −3𝑏 − 5𝑏 + 10 − 4 = −8𝑏 + 6
= −8(−2) + 6
= 16 + 6 = 22

v) 2𝑎– 2𝑏– 4– 5 + 𝑎
= 2𝑎 + 𝑎 − 2𝑏 − 4 − 5
= 3𝑎 − 2𝑏 − 9 = 3(−1) − 2(−2) − 9
= −3 + 4 − 9 = −8

Q.8) (i) If 𝑧 = 10, find the value of 𝑧³ − 3(𝑧 − 10) .
(ii) 𝐼𝑓 𝑝 = -10, find the value of 𝑝² − 2𝑝 − 100.
Sol.8) i) 𝑧³ − 3(𝑧 − 10)
= 𝑧³ − 3𝑧 + 30                                        [putting 𝑧 = 10]
= (10 × 10 × 10) − (3 × 10) + 30
= 1000 − 30 + 30 = 1000
ii) 𝑝² − 2𝑝 − 100
= (−10) × (−10) − 2(−10) − 100              [putting 𝑝 = −10]
= 100 + 20 − 100 = 20

Q.9) What should be the value of 𝑎 if the value of 2𝑥² + 𝑥 – 𝑎 equals to 5, when 𝑥 = 0?
Sol.9) 2𝑥² + 𝑥 – 𝑎 = 5 , when 𝑥 = 0
(2 × 0) + 0 − 𝑎 = 5
= 0 − 𝑎 = 5
= 𝑎 = −5

Q.10) Simplify the expression and find its value when 𝑎 = 5 and 𝑏 =– 3.
2(𝑎² + 𝑎𝑏) + 3– 𝑎𝑏
Sol.10) 2(𝑎² + 𝑎𝑏) + 3– 𝑎𝑏 , where 𝑎 = 5 & 𝑏 = −3
= 2𝑎² + 2𝑎𝑏 + 3 − 𝑎𝑏
= 2𝑎² + 2𝑎𝑏 − 𝑎𝑏 + 3
= 2𝑎² + 𝑎𝑏 + 3
Now, putting the values of 𝑎 𝑎𝑛𝑑 𝑏
= 2 × (5 × 5) + 5 × (−3) + 3
= 50 − 15 + 3 = 38

Exercise 12.4

Q.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.

Sol.1)

(i) 5𝑛 + 1
Putting 𝑛 = 5,               5 × 5 + 1 = 25 + 1 = 26
Putting 𝑛 = 10,             5 × 10 + 1 = 50 + 1 = 51
Putting 𝑛 =100,            5 × 100 + 1 = 500 + 1 = 501

(ii) 3𝑛 + 1
Putting 𝑛 = 5,               3 × 5 + 1 = 15 + 1 = 16
Putting 𝑛 = 10,             3 × 10 + 1 = 30 + 1 = 31
Putting 𝑛 =100,            3 × 100 + 1 = 300 + 1 = 301

(iii) 5𝑛 + 2
Putting 𝑛 = 5,               5 × 5 + 2 = 25 + 2 = 27
Putting 𝑛 = 10,             5 × 10 + 2 = 50 + 2 = 52
Putting 𝑛 =100,            5 × 100 + 2 = 500 + 2 = 502

Q.2) Use the given algebraic expression to complete the table of number patterns.

Sol.2) (i) 2𝑛 − 1
Putting 𝑛 =100,              2 × 100 – 1 = 200 – 1 = 199

(ii) 3𝑛 + 2
Putting 𝑛 = 5,                 3 × 5 + 2 = 15 + 2 = 17
Putting 𝑛 = 10,               3 × 10 + 2 = 30 + 2 = 32
Putting 𝑛 =100,              3 × 100 + 2 = 300 + 2 = 302

(iii) 4𝑛 + 1
Putting 𝑛 = 5,                 4 × 5 + 1 = 20 + 1 = 21
Putting 𝑛 = 10,               4 × 10 + 1 = 40 + 1 = 41
Putting 𝑛 =100,              4 × 100 + 1 = 400 + 1 = 401

(iv) 7𝑛 + 20
Putting 𝑛 = 5,                 7 × 5 + 20 = 25 + 20 = 55
Putting 𝑛 = 10,               7 × 10 + 20 = 70 + 20 = 90
Putting 𝑛 =100,              7 × 100 + 20 = 700 + 20 = 720

(v) 𝑛² + 1
Putting 𝑛 = 5,                 5 × 5 + 1 = 25 + 1 = 26
Putting 𝑛 = 10,               10 × 10 + 1 = 100 + 1 = 101
Putting 𝑛 =100,              100 × 100 + 1 = 10000 + 1 = 10001