ALGEBRA - National Defence Institute

ALGEBRA

National Defence Institute| SAINIK | Algebra

ALGEBRA

Like Arithmetic, Algebra is a science of numbers. In it,

besides the figures of arithmetic, the letters of the alphabet are extensively used to

represent number.

ALGEBRAIC EXPRESSION:

If we join two or more algebraic quantities by + or – , we

get algebraic expression. Thus 3x + 4y; 7a + 8b – 4c; 6a + 4b – 7c etc., are

algebraic expressions.

ADDITION OF ALGEBRAIC EXPRESSIONS:

We write 3x + 5x = 8x, 6a + 4a = 10a; 7b + 8b = 15b;

3y + 7y = 10y.

Same type quantities are added as simple addition. If we add

3x + 4y and 2x + 3y, we add the same type of quantities separately.

3x + 4y

2x + 3y

5x + 7y

SUBTRACTION OF ALGEBRAIC EXPRESSIONS:

Operation of subtraction is done in two steps. First, we

change the sign of the quantity which is to be subtracted and then we add it.

For example, Subtract 5x + 5y and 2x – 3y.


5x + 5y

2x – 3y

(–) (+)

7x + 8y

FORMULA:

➢ x2 = x × x

➢ x3 = x × x × x

➢ 2 x2 = 2 × x × x

➢ (a + b)2 = a2 + 2ab + b2

➢ (a – b)2 = a2 – 2ab + b2

➢ a2 – b2 = (a – b) (a + b)

➢ (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca

➢ (a + b)3 = a3 + b3 + 3ab (a + b) (or)

(a + b)3 = a3 + b3 + 3a2 b + 3ab2

➢ (a – b)3 = a3 – b3 – 3ab (a – b) (or)

(a – b)3 = a3 – b3 – 3a2 b + 3ab2


➢ a3 + b3 = (a + b) (a2 – ab + b2) (or)

a3 + b3 = (a + b)3 – 3ab (a + b)

➢ a3 – b3 = (a – b) (a2 + ab + b2) (or)

a3 – b3 = (a – b)3 + 3ab (a – b)

When We Add:

➢ Two like signs become a positive sign

➢ Two unlike signs become a negative sign


PRACTICE QUESTIONS:

1) Find the value of 3x + 4y if x = 5, y = 3.

(a) 27 (b) 37

(c) 21 (d) 17

Answer : (a) 27

Given that, x = 5

and y = 3

3x + 4y = 3 × x + 4 × y

= 3 × 5 + 4 × 3

= 15 + 12

⇒ 3x + 4y = 27.

2) If x = 5, y = 4, z = 3, find the value of 2x + 3y – 4z.

(a) 13 (b) 10

(c) 17 (d) 9

Answer : (b) 10

Given that, x = 5,

y = 4,

and z = 3.

2x + 3y – 4z = 2 × x + 3 × y – 4 × z

= 2 × 5 + 3 × 4 – 4 × 3

= 10 + 12 – 12


= 22 – 12

⇒ 2x + 3y – 4z = 10.

3) If x = 5, y = 3, z = 2, find the value of 𝟑𝒙 − 𝟒𝐲 + 𝐳

𝟐𝒙 + 𝐲 + 𝐳 .

(a) 2

5 (b)

1

2

(c) 1

3 (d)

1

5

Answer : (c) 1

3

Given that, x = 5,

y = 3,

and z = 2.

3𝑥 − 4y + z

2𝑥 + y + z =

3 × 𝑥 − 4 × y + z

2 × 𝑥 + y + z

= 3 × 5 − 4 × 3 + 2

2 × 5 + 3 + 2

= 15 − 12 + 2

10 + 3 + 2

= 17 − 12

15

= 5

15

⇒ 3𝑥 − 4y + z

2𝑥 + y + z =

1

3

4) If a = 3, b = 4, c = 2, find the value of 2a2 – 4ab + 15c2.

(a) 30 (b) 20

(c) 40 (d) 50


Answer : (a) 30

Given that, a = 3,

b = 4,

and c = 2.

2a2 – 4ab + 15c2 = 2 × a × a – 4 × a × b + 15 × c × c

= 2 × 3 × 3 – 4 × 3 × 4 + 15 × 2 × 2

= 18 – 48 + 60

= 78 – 48

⇒ 2a2 – 4ab + 15c2 = 30.

5) Find the value of x – 2y + 𝒙

𝐲 + 1, when x = 2, y = – 1.

(a) 1 (b) 2

(c) 3 (d) 4

Answer : (c) 3

Given that, x = 2,

y = – 1

x – 2y + 𝑥

y + 1 = 2 – 2 × (– 1) +

2

(– 1) + 1

= 2 + 2 – 2 + 1

= 5 – 2

⇒ x – 2y + 𝑥

y + 1 = 3.

6) Add the following :

2x + 3y, 5x + 7y, 3x + 4y


(a) 12x + 13y (b) 10x + 20y

(c) 15x + 7y (d) 10x + 14y

Answer : (d) 10x + 14y

2x + 3y

5x + 7y

3x + 4y

10x + 14y


8a – 7b + 6c ; 3a + 9b – 2c

(a) 3a + 7b – 4c (b) 7a – 5b + 3c

(c) 5a + 3b + 2c (d) 11a + 2b + 4c

Answer : (d) 11a + 2b + 4c

8a – 7b + 6c

3a + 9b – 2c

11a + 2b + 4c

8) Add (2a + 5b) and (a + 7b).

(a) 3a + 12b (b) 10a + 20b

(c) 15a + 7b (d) 5a + 14b


Answer : (a) 3a + 12b

2a + 5b

a + 7b

3a + 12b


7x – 4y; – 3x + 9y.

(a) 2x + 3y (b) 10x – 2y

(c) 4x + 5y (d) –10x + 4y

Answer : (c) 4x + 5y

7x – 4y

– 3x + 9y

4x + 5y

10) If a = 3 𝟔

𝟖 , b = 5

𝟓

𝟔 , c = 2

𝟏

𝟑 , then find the value of (𝐚 +

𝐛 + 𝐛

𝐜 + 𝟒) .

(a) 13 6

8 (b) 7

3

7

(c) 11 1

11 (d) 16

1

12

Answer : (d) 16 1

12


Given that, a = 3 6

8 ⇒ a =

30

8 =

15

4

b = 5 5

6 ⇒ b =

35

6

and c = 2 1

3 ⇒ c =

7

3

a + b + 𝑏

c + 4 =

15

4 +

35

6 +

35

6 ÷

7

3 + 4

= 15

4 +

35

6 +

35

6 ×

3

7 + 4

= 15

4 +

35

6 +

5

2 +

4

1

= 45

12 +

70

12 +

30

12 +

48

12

= 45 + 70 + 30 + 48

12

= 193

12

⇒ a + b + 𝑏

c + 4 = 16

1

12


x + 3y, 2x + 𝟏

𝟐 y,

𝟏

𝟐 x + 3y

(a) x + 11

2 y (b)

7

2 x +

13

2 y

(c) 15x + 7y (d) 10

2 x +

5

2 y

Answer : (b) 7

2 x +

13

2 y

x + 3y + 2x + 1

2 y +

1

2 x + 3y = x + 2x +

1

2 x + 3y +

1

2 y + 3y


= x ( 1 + 2 + 1

2 ) + y ( 3 +

1

2 + 3 )

= x ( 3 + 1

2 ) + y ( 6 +

1

2 )

= 7

2 x +

13

2 y

12) Subtract 2a + b – c from a – b + c

(a) 3a + 7b – 4c (b) 7a – 5b + 3c

(c) – a – 2b + 2c (d) 11a + 2b + 4c

Answer : (c) – a – 2b + 2c

a – b + c

2a + b – c

(–) (–) (+)

– a – 2b + 2c

13) Subtract 3a – 5b from 7a + b.

(a) 4a + 6b (b) 7a – 5b

(c) – a – 2b (d) a + 2b

Answer : (a) 4a + 6b

7a + b

3a – 5b

(–) (+)

4a + 6b


14) Subtract 3x + 4y from 5x + 9y.

(a) 2x + 5y (b) 10x – 2y

(c) 4x + 5y (d) –10x + 4y

Answer : (a) 2x + 5y

5x + 9y

3x + 4y

(–) (–)

2x + 5y

15) Subtract – 2a + 3b from 2a + 3b.

(a) 4a (b) – 6b

(c) – 4a (d) a + 2b

Answer : (a) 4a

2a + 3b

– 2a + 3b

(+) (–)

4a + 0 ⇒ 4a

16) Multiply (a + b) by c.

(a) abc (b) ab + bc

(c) (abc)2 (d) ac + bc


Answer : (d) ac + bc

Multiply (a + b) by c = (a + b) × c

= a × c + b × c

= a c + b c

17) Multiply 6xy and 3x2 y.

(a) 12 x2 y2 (b) 6 x3 y2

(c) 3 x y2 (d) 18 x3 y2

Answer : (d) 18 x3 y2

Multiply 6xy and 3x2 y = 6xy × 3x2 y

= 6 × x × y × 3 × x2 × y

= 6 × 3 × x × x2 × y × y

= 6 × 3 × x1 × x2 × y1 × y1

= 18 × x 1 + 2 × y 1 + 1

= 18 x3 y2

18) Multiply 2a + 3b + 4c by a – b + c

(a) 2a2 + ab + 6ac – 3b2 – bc + 4c2

(b) a2 + ab + 2ac – b2 + 3bc + c2

(c) a2 + ab + 5ac

(d) 2b2 – 4bc + 5c2

Answer : (a) 2a2 + ab + 6ac – 3b2 – bc + 4c2


2a + 3b + 4c × a – b + c

2a2 + 3ab + 4ac

– 2ab – 3b2 – 4bc

+ 2ac + 3bc + 4c2

2a2 + ab + 6ac – 3b2 – bc + 4c2

19) Multiply (3x – 4y) by (3x + 4y).

(a) 5 x2 + 7y2 (b) 12 x2 y2

(c) 9x2 – 16y2 (d) 2 x + y2

Answer : (c) 9x2 – 16y2

3x – 4y × 3x + 4y

9x2 – 12 xy

+ 12 xy – 16y2

9x2 – 16y2

20) Divide 3x + 42 by 3.

(a) x + 5 (b) x + 11

(c) x + 9 (d) x + 14

Answer : (d) x + 14


x + 14

3 ) 3x + 42 (

3x

42

42

0

21) The LCM of 12a3b, 6a2b3 and 4ab4 will be:

(a) 6a3b4 (b) 12a4b3

(c) 8a2b2 (d) 12a3b4

Answer : (d) 12a3b4

12a3b = 22 × 3 × a3 × b

6a2b3 = 2 × 3 × a2 × b3

and 4ab4 = 22 × a × b4

Required LCM = 22 × 3 × a3 × b4

= 12a3b4

22) Find the HCF of x3y2, x2y3 and x4y4.

(a) x3y4 (b) xy

(c) x2y2 (d) x4y4

Answer : (c) x2y2

x3y2 = x3 × y2

x2y3 = x2 × y3

and x4y4 = x4 × y4


Required HCF = x2 × y2 = x2y2

23) Factors of x2 – 25 are:

(a) (x – 1) (x – 25) (b) (x + 25) (x – 1)

(c) (x + 5) (x – 5) (d) (x – 5) (x – 5)

Answer : (c) (x + 5) (x – 5)

Formula: a2 – b2 = (a + b) (a – b)

x2 – 25 = (x)2 – (5)2

= (x + 5) (x – 5)

24) Factors of 36 – 9x2 will be:

(a) (6 + 3x) (6 – 3x) (b) (3x – 6) (6 – 3x)

(c) (3x + 6) (3x – 6) (d) (12x – 3x) (3 + 3x)

Answer : (a) (6 + 3x) (6 – 3x)

Formula: a2 – b2 = (a + b) (a – b)

36 – 9x2 = (6)2 – (3x)2

= (6 + 3x) (6 – 3x)

25) Factors of 8x3 + y3 are:

(a) (2x + y) (4x2 – 2xy + y2)

(b) (2x + y) (4x2 + 2xy + y2)

(c) (2x – y) (4x2 – 2xy + y2)

(d) (2x + y) (4x2 – 2xy – y2)


Answer : (a) (2x + y) (4x2 – 2xy + y2)

Formula: a3 + b3 = (a + b) (a2 – ab + b2)

8x3 + y3 = (2x)3 + (y)3

= (2x + y) [ (2x)2 – (2x).(y) + (y)2)

= (2x + y) (4x2 – 2xy + y2)

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