Gujarat Board GSEB Textbook Solutions Class 8 Maths Chapter 9 Algebraic Expressions and Identities InText Questions and Answers.

## Gujarat Board Textbook Solutions Class 8 Maths Chapter 9 Algebraic Expressions and Identities InText Questions

Try These (Page 138)

Question 1.

Give five examples of expressions containing one variable and five examples of expressions containing two variables?

Solution:

(a) Five examples containing one variable are:

- x – 4
- 5 – y
- 2x + 5
- 10x + 4
- 11 – z

(b) Five examples containing two variables are:

- 7x – 2y
- 5x + 2y – 10
- 6x + z – 2
- 15y – z
- 9x – 10z

Question 2.

Show on the number line x, x – 4, 2x + 1, 3x – 2.

Solution:

(i)

The point X represent the variable ‘x’.

(ii)

(iii)

(iv)

Try These (Page 138)

Question 1.

Identify the coefficient of each term in the expression

x^{2}y^{2} – 10x^{2}y + 5xy^{2} – 20?

Solution:

Coefficient of x^{2}y^{2} is 1.

Coefficient of x^{2}y is -10.

Coefficient of xy^{2} is 5.

Try These (Page 138)

Question 1.

Classify the following polynomials as monomials, binomials, trinomials -z + 5, x + y + z, y + z + 100, ab – ac, 17

Solution:

Question 2.

Construct:

(a) 3 binomials with only x as a variable

(b) 3 binomials with x and y as variable

(c) 3 monomials with x and y as variables

(d) 2 polynomials with 4 or more terms

Solution:

(a)

- 5x – 4
- 3x + 2
- 10 + 2x

(b)

- 2x + 3y
- 5x – y
- x – 3y

(c)

- 2xy
- xy
- -7xy

(d)

- 2x
^{3}– x^{2}+ 6x – 8 - 10 + 7x – 2x
^{2}+ 4x^{3}

Try These (Page 139)

Question 1.

Write two terms which are like

- 7xy
- 4mn
^{2} - 2l

Solution:

- Two terms like 7xy are: -3xy and 8xy.
- Two terms like 4mn
^{2}are: 6mn^{2}and -2n^{2}m. - Two terms like 2l are: 5l and -7b.

Try These (Page 143)

Question 1.

Find 4x Ã— 5y Ã— 7z. First find 4x Ã— 5y and multiply it by 7z; or fIrst find 5y Ã— 7z and multiply it by 4x. Is the result the same? What do you observe? Does the order in which you carry out the multiplication marier?

Solution:

We have

4x Ã— 5y Ã— 7z = (4x Ã— 5y) Ã— 7z

= 20xy Ã— 7z = 140 xyz

Also 4x Ã— 5y Ã— 7z = 4x Ã— (5y Ã— 7z)

= 4x Ã— 35yz = 140 xyz

We observe that

(4x Ã— 5y) Ã— 7x = 4 Ã— (5y Ã— 7z)

âˆ´ The product of monomials is associative, i.e. the order in which we multiply the monomials does not matter.

Try These (Page 144)

Question 1.

Find the product

- 2 x (3x + 5xy)
- a
^{2 }(2ab – 5c)

Solution:

1. 2x(3x + 5xy) = 2x Ã— 3x + 2x Ã— 5xy

= (2 Ã— 3) Ã— x Ã— x + (2 Ã— 5) Ã— x Ã— xy

2. a^{2}(2ab – 5c) = a^{2} Ã— 2ab + a^{2} Ã— (-5c)

= (1 Ã— 2) Ã— a^{2} Ã— ab + [1 Ã— (-5)] Ã— a^{2} Ã— c

= 2 Ã— a^{3}b + (-5) Ã— a^{2}c = 2a^{3}b – 5a^{2}c

Try These (Page 145)

Question 1.

Find the product: (4p^{2} + 5p + 7) Ã— 3p

Solution:

(4p^{2} + 5p + 7) Ã— 3p

= (4p^{2} Ã— 3p) + (5p Ã— 3p) + (7 Ã— 3p)

= [(4 Ã— 3) Ã— p^{2} Ã— p] + [(5 Ã— 3) Ã— p Ã— p] + (7 Ã— 3) Ã— p

= 12 Ã— p^{3} + 15 Ã— p^{2} + 21 Ã— p

= 12p^{3} + 15p^{2} + 21p

Question 2.

Simplify: 2x(x – 1) + 5 and find its value at x = -1.

Solution:

2x(x – 1) + 5 = [2x Ã— x] + [2x Ã— (-1)] + 5

= [(2 Ã— 1) Ã— x Ã— x] + [2 Ã— (-1) Ã— x] + 5

= [2 Ã— x^{2}] + [-2 Ã— x] + 5

= 2x^{2} – 2x + 5

For x = -1, 2x^{2} – 2x + 5

= 2(-1)^{2} – 2(-1) + 5

= (2 Ã— 1) – (-2) + 5

= 2 + 2 + 5 = 9

Question 3.

Carry out the multiplication of the expressions in each of the following pairs?

- 4p, q + r
- ab, a – b
- a + b, 7a
^{2}b^{2} - a
^{2}– 9, 4a - pq + qr + rp, 0

Solution:

1. 4p Ã— (q + r) = (4p Ã— q) + (4p Ã— r)

= (4 Ã— 1) Ã— p Ã— q + (4 Ã— 1) Ã— p Ã— r

= 4 Ã— pq + 4 Ã— pr = 4pq + 4pr

2. ab Ã— (a – b) = ab Ã— a + ab Ã— (-b)

= (1 Ã— 1) Ã— ab Ã— a + [1 Ã— (-1)] Ã— ab Ã— b

= 1 Ã— a^{2} Ã— b + (-1) Ã— a Ã— b^{2}

= a^{2}b – ab^{2}

3. (a + b) Ã— 7a^{2}b^{2} = a Ã— 7a^{2}b^{2} + b + 7a^{2}b^{2}

= (1 Ã— 7) Ã— a Ã— a^{2}b^{2} + (1 Ã— 7) Ã— b Ã— a^{2}b^{2}

= 7 Ã— a^{3}b^{2} + 7 Ã— a^{2}b^{3} = 7a^{3}b^{2} + 7a^{2}b^{3}

4. (a^{2} – 9) Ã— 4a = a^{2} Ã— 4a + (-9) Ã— 4a

= (1 Ã— 4) Ã— a^{2} Ã— a + (-9 Ã— 4) Ã— a

= 4 Ã— a^{3} + (-36) Ã— a

= 4a^{3} – 36a

5. (pq + qr + rp) Ã— 0

= pq Ã— 0 + qr Ã— 0 + rp Ã— 0

= 0 + 0 + 0 = 0

Try These (Page 148)

Question 1.

Multiply the binomials.

- (2x + 5) and (4x – 3)
- (y – 8) and (3y – 4)
- (2.51 – 0.5m) and (2.51 + 0.5m)
- (a + 3b) and (x + 5)
- (2pq + 3q
^{2}) and (3pq – 2q^{2}) - (\(\frac{3}{4}\) a
^{2}+ 3b^{2}) and 4 (a^{2}– \(\frac{2}{3}\)b^{2})

Solution:

1. (2x + 5) Ã— (4x – 3)

= 2x(4x – 3) + 5(4x – 3)

= (2x Ã— 4x) – (2x Ã— 3) + (5 Ã— 4x) – (5 Ã— 3)

= (8x^{2}) – 6x + 20x – 15

= 8x^{2} + (-6 + 20)x – 15

= 8x^{2} + 14x – 15

2. (y – 8) Ã— (3y – 4) = y(3y – 4) – 8(3y – 4)

= (3y Ã— y) – (4 Ã— y) – 8 Ã— 3y – 8 Ã— (-4)

= 3y^{2} – 4y – 24y + 32

= 3y^{2} – 28y + 32

3. (2.5l – 0.5m) Ã— (2.5l + 0.5m)

= 2.5l(2.5l + 0.5m) – 0.5m(2.5l + 0.5m)

= (2.5l Ã— 2.5l) + (2.5l Ã— 0.5m) – (2.5l Ã— 0.5m) – (0.5m Ã— 0.5m)

= 6.25l^{2} + 1.25lm – 1.25lm – 0.25m^{2}

= 6.25l^{2} + (0)lm – 0.25m^{2} = 6.25l^{2} – 0.25m^{2}

4. (a + 3b) Ã— (x + 5)

= a Ã— (x Ã— 5) + 3b Ã— (x + 5)

= (a Ã— x) + (a Ã— 5) + (3b Ã— x) + (3b Ã— 5)

= ax + 5a + 3bx + 15b

5. (2pq + 3q^{2}) Ã— (3pq – 2q^{2})

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

= (2pq Ã— 3pq) – (2pq Ã— 2q^{2}) + (3q^{2} Ã— 3pq) – (3q^{2} Ã— 3pq) – (3q^{2} Ã— 2q^{2})

= 6p^{2}q^{2} – 4pq^{3} + 9pq^{3} – 6q^{4}

= 6p^{2}q^{2} + (-4 + 9)pq^{3} – 6q^{4}

= 6p^{2}q^{2} + 5pq^{3} – 6q^{4}

6. (\(\frac{3}{4}\)a^{2} + 3b^{2}) Ã— 4(a^{2} – \(\frac{2}{3}\)b^{2})

= \(\frac{3}{4}\)a^{2} Ã— 4(a^{2} – \(\frac{2}{3}\)b^{2}) + 3b^{2} Ã— 4(a^{2} – \(\frac{2}{3}\)b^{2})

= (3a^{2} Ã— a^{2}) – (3a^{2} Ã— \(\frac{2}{3}\)b^{2}) + (12b^{2} Ã— a^{2}) – (12b^{2} Ã— \(\frac{2}{3}\)b^{2})

= 3a^{4} – 2a^{2}b^{2} + 12a^{2}b^{2} – 8b^{4}

= 3a^{4} + (-2 + 12)a^{2}b^{2} – 8b^{4}

= 3a^{4} + 10a^{2}b^{2} – 8b^{4}

Question 2.

Find the product:

- (5 – 2x)(3 + x)
- (x + 7y)(7x – y)
- (a
^{2}+ b)(a + b^{2}) - (p
^{2}– q^{2}) (2p + q)

Solution:

1. (5 – 2x) Ã— (x + 3) = 5(x + 3) – 2x(x + 3)

= 5x + 15 – (2x Ã— x) – (2x Ã— 3)

= 5x + 15 – 2x^{2} – 6x

= -2x^{2} + (-6 + 5)x + 15

= -2x^{2} – x + 15 = 15 – x – 2x^{2}

2. (x + 7y) Ã— (7x – y) = x(7x – y) + 7y(7x – y)

= (x Ã— 7x) – (x Ã— y) + (7y Ã— 7x) – (7y Ã— y)

= 7x^{2} – xy + 49xy – 7y^{2}

= 7x^{2} + 48xy – 7y^{2}

3. (a^{2} + b) (a + b^{2}) = a^{2}(a + b^{2}) + b(a + b^{2})

= (a^{2} Ã— a) + (a^{2} Ã— b^{2}) + (b Ã— a) + (b Ã— b^{2})

= a^{3} + a^{2}b^{2} + ab + b^{3}

4. (p^{2} – q^{2}) Ã— (2p + q) = p^{2}(2p + q) – q^{2}(2p + q)

= (p^{2} Ã— 2p) + (p^{2} Ã— q) – (q^{2} Ã— 2p) – (q^{2} Ã— q)

= 2p^{3} + p^{2}q – 2pq^{2} – q^{3}

Question 3.

Simplify:

- (x
^{2}– 5)(x + 5) + 25 - (a
^{2}+ 5) (b^{3}+ 3) + 5 - (t + s
^{2})(t^{2}– s) - (a + b)(c – d) + (a – b)(c + d) + 2(ac + bd)
- (x + y) (2x + y) + (x + 2y) (x – y)
- (x + y)(x
^{2}– xy + y^{2}) - (1.5x – 4y)(1.5x + 4y + 3) – 4.5x + 12y
- (a + b + c)(a + b – c)

Solution:

1. (x^{2} – 5)(x + 5) + 25

= x^{2}(x + 5) -5(x + 5) + 25

= (x^{2} Ã— x) + (x^{2} Ã— 5) – (5 Ã— x) – (5 Ã— 5) + 25

= x^{3} + 5x^{2} – 5x – 25 + 25

= x^{3} + 5x^{2} – 5x

2. (a^{2} + 5)(b^{3} + 3) + 5 = a^{2}(b^{3} + 3) + 5(b^{3} + 3) + 5

= (a^{2} Ã— b^{3}) + (a^{2} Ã— 3) + (5 Ã— b^{3}) + (5 Ã— 3) + 5

= a^{2}b^{3} + 3a^{2} + 5b^{3} + 15 + 5

= a^{3}b^{3} + 3a^{2} + 5b^{3} + 20

3. (t + s^{2}) Ã— (t^{2} – s) = t(t^{2} – s) + s^{2}(t^{2} – s)

= (t Ã— t^{2}) – (t Ã— s) + (s^{2} Ã— t^{2}) + [s^{2} Ã— (-s)]

= t^{3} – ts + s^{2}t^{2} – s^{3}

= t^{3} – st + s^{2}t^{2} – s^{3}

= t^{3} – st + s^{2}t^{2} – s^{3}

4. (a + b)(c – d) + (a – b)(c + d) + 2(ac + bd)

= a(c – d) + b(c – d) + a(c + d) – b(c + d) + 2(ac + bd)

= ac – ad + bc – bd + ac + ad – bc – bd + 2ac + 2bd

= ac + ac + 2ac = 4ac

5. (x + y)(2x + y) + (x + 2y)(x – y)

= x(2x + y) + y(2x + y) + x(x – y) + 2y(x – y)

= (x Ã— 2x) + (x Ã— y) + (y Ã— 2x) + (y Ã— y) + (x Ã— x) – (x Ã— y) + (2y Ã— x) – (2y Ã— y)

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

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

= 3x^{2} + 4xy – y^{2}

6. (x + y)(x^{2} – xy + y^{2})

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

= (x Ã— x^{2}) – (x Ã— xy) + (x Ã— y^{2}) + (y Ã— x^{2})

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

= x^{3} + (-x^{2}y + x^{2}y) + yx^{2} – xy^{2} + y^{3}

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

= x^{3} + (0 Ã— x^{2}y) + (0 Ã— xy^{2}) + y^{3}

= x^{3} + 0 + 0 + y^{3} = x^{3} + y^{3}

7. (1.5x – 4y)(1.5x + 4y + 3) – 4.5x + 12y

= 1.5x(1.5x + 4y + 3) – 4y(1.5x + 4y + 3) – 4.5x + 12y

= 2.25x^{2} + 6xy + 4.5x – 6xy – 16y^{2} – 12y – 4.5x + 12y

= 2.25x^{2} + (6 – 6)xy + (4.5 – 4.5)x – 16y^{2} + (12 – 12)y

= 2.25x^{2} + (0)xy + (0)x – 16y^{2} + (0)y

= 2.25x^{2} + 0 + 0 – 16y^{2} + 0

= 2.25x^{2} – 16y^{2}

8. (a + b + c)(a + b – c)

= a(a + b – c) + b(a + b – c) + c(a + b – c)

= (a Ã— a) + (a Ã— b) – (a Ã— c) + (b Ã— a) + (b Ã— b) – (b Ã— c) + (c Ã— a) + (c Ã— b) – (c Ã— c)

= a^{2} + ab – ac + ab + b^{2} – bc + ac + bc – c^{2}

= a^{2} + (ab + ab) + (-ac + ac) + b^{2} + (-bc + bc) – c^{2}

= a^{2} + (2ab) + (0) + b^{2} + 0 – c^{2}

= a^{2} + 2ab + b^{2} – c^{2}

= a^{2} + b^{2} – c^{2} + 2ab

Try These (Page 149)

Question 1.

Verify Identity (IV), for a = 2. b = 3, x = 5.

Solution:

We have

(x + a) (x + b) = x^{2} + (a + b)x + ab

Putting a = 2, b = 3 and x = 5 in the identity:

LHS = (x + a)(x + b)

= (5 + 2) (5 + 3)

= 7 Ã— 8 = 56

RHS = x^{2} + (a + b)x + ab

= (5)^{2} + (2 + 3) Ã— 5 + (2 Ã— 3)

= 25 + (5) Ã— 5 + 6

= 25 + 25 + 6 = 56

âˆ´ LHS = RHS

âˆ´ The given identity is true for the given values.

Question 2.

Consider the special case of Identity (IV) with a = b, what do you get? Is it related to Identity (I)?

Solution:

When a = b (each y)

(x + a) (x + b) = x^{2} + (a + b)x + ab becomes

(x + y)(x + y) = x^{2} + (y + y)x + (y Ã— y)

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

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

= Yes, it is the same as Identity I

Question 3.

Consider the special case of Identity (IV) with a = -c and b = -c. What do you get? Is it related to Identity (II)?

Solution:

Identity IV is given by

(x + a)(x + b) = x^{2} + (a + b)x + ab

Replacing â€˜aâ€™ by (-c) and â€˜bâ€™ by (-c), we have (x – c)(x – c)

= x^{2} + [(-c) + (-c)] x + [(-c) Ã— (-c)]

= x^{2} + [-2c]x + (c^{2}) = x^{2} – 2cx + c^{2}

Question 4.

Consider the special case of Identity (IV) with b = -a. What do you get? It is related to Identity (III).

Solution:

The Identity IV is given by

(x + a)(x + b) = x^{2} + (a + b)x + ab

Replacing â€˜bâ€™ by (-a), we have:

(x + a)(x – a) = x^{2} + [a + (-a)]x + [a Ã— (-a)]

= x^{2} + [0]x + [-a^{2}]

= x^{2} + 0 + (-a^{2}) = x^{2} – a^{2}

which is same as the Identity III.