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 × 5y × 7z. First find 4x × 5y and multiply it by 7z; or fIrst find 5y × 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 × 5y × 7z = (4x × 5y) × 7z

= 20xy × 7z = 140 xyz

Also 4x × 5y × 7z = 4x × (5y × 7z)

= 4x × 35yz = 140 xyz

We observe that

(4x × 5y) × 7x = 4 × (5y × 7z)

∴ 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 × 3x + 2x × 5xy

= (2 × 3) × x × x + (2 × 5) × x × xy

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

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

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

Try These (Page 145)

Question 1.

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

Solution:

(4p^{2} + 5p + 7) × 3p

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

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

= 12 × p^{3} + 15 × p^{2} + 21 × 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 × x] + [2x × (-1)] + 5

= [(2 × 1) × x × x] + [2 × (-1) × x] + 5

= [2 × x^{2}] + [-2 × x] + 5

= 2x^{2} – 2x + 5

For x = -1, 2x^{2} – 2x + 5

= 2(-1)^{2} – 2(-1) + 5

= (2 × 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 × (q + r) = (4p × q) + (4p × r)

= (4 × 1) × p × q + (4 × 1) × p × r

= 4 × pq + 4 × pr = 4pq + 4pr

2. ab × (a – b) = ab × a + ab × (-b)

= (1 × 1) × ab × a + [1 × (-1)] × ab × b

= 1 × a^{2} × b + (-1) × a × b^{2}

= a^{2}b – ab^{2}

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

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

= 7 × a^{3}b^{2} + 7 × a^{2}b^{3} = 7a^{3}b^{2} + 7a^{2}b^{3}

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

= (1 × 4) × a^{2} × a + (-9 × 4) × a

= 4 × a^{3} + (-36) × a

= 4a^{3} – 36a

5. (pq + qr + rp) × 0

= pq × 0 + qr × 0 + rp × 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) × (4x – 3)

= 2x(4x – 3) + 5(4x – 3)

= (2x × 4x) – (2x × 3) + (5 × 4x) – (5 × 3)

= (8x^{2}) – 6x + 20x – 15

= 8x^{2} + (-6 + 20)x – 15

= 8x^{2} + 14x – 15

2. (y – 8) × (3y – 4) = y(3y – 4) – 8(3y – 4)

= (3y × y) – (4 × y) – 8 × 3y – 8 × (-4)

= 3y^{2} – 4y – 24y + 32

= 3y^{2} – 28y + 32

3. (2.5l – 0.5m) × (2.5l + 0.5m)

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

= (2.5l × 2.5l) + (2.5l × 0.5m) – (2.5l × 0.5m) – (0.5m × 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) × (x + 5)

= a × (x × 5) + 3b × (x + 5)

= (a × x) + (a × 5) + (3b × x) + (3b × 5)

= ax + 5a + 3bx + 15b

5. (2pq + 3q^{2}) × (3pq – 2q^{2})

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

= (2pq × 3pq) – (2pq × 2q^{2}) + (3q^{2} × 3pq) – (3q^{2} × 3pq) – (3q^{2} × 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}) × 4(a^{2} – \(\frac{2}{3}\)b^{2})

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

= (3a^{2} × a^{2}) – (3a^{2} × \(\frac{2}{3}\)b^{2}) + (12b^{2} × a^{2}) – (12b^{2} × \(\frac{2}{3}\)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) × (x + 3) = 5(x + 3) – 2x(x + 3)

= 5x + 15 – (2x × x) – (2x × 3)

= 5x + 15 – 2x^{2} – 6x

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

= -2x^{2} – x + 15 = 15 – x – 2x^{2}

2. (x + 7y) × (7x – y) = x(7x – y) + 7y(7x – y)

= (x × 7x) – (x × y) + (7y × 7x) – (7y × 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} × a) + (a^{2} × b^{2}) + (b × a) + (b × b^{2})

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

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

= (p^{2} × 2p) + (p^{2} × q) – (q^{2} × 2p) – (q^{2} × 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} × x) + (x^{2} × 5) – (5 × x) – (5 × 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} × b^{3}) + (a^{2} × 3) + (5 × b^{3}) + (5 × 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}) × (t^{2} – s) = t(t^{2} – s) + s^{2}(t^{2} – s)

= (t × t^{2}) – (t × s) + (s^{2} × t^{2}) + [s^{2} × (-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 × 2x) + (x × y) + (y × 2x) + (y × y) + (x × x) – (x × y) + (2y × x) – (2y × 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 × x^{2}) – (x × xy) + (x × y^{2}) + (y × 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 × x^{2}y) + (0 × 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 × a) + (a × b) – (a × c) + (b × a) + (b × b) – (b × c) + (c × a) + (c × b) – (c × 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 × 8 = 56

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

= (5)^{2} + (2 + 3) × 5 + (2 × 3)

= 25 + (5) × 5 + 6

= 25 + 25 + 6 = 56

∴ LHS = RHS

∴ 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 × 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 ‘a’ by (-c) and ‘b’ by (-c), we have (x – c)(x – c)

= x^{2} + [(-c) + (-c)] x + [(-c) × (-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 ‘b’ by (-a), we have:

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

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

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

which is same as the Identity III.