S.chand publication New Learning Composite mathematics solution of class 8 Chapter 5 Algebraic Expressions Exercise 5G

 Exercise 5G

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Question 

(1) x² – 3x – 28 by x – 7

(2) 6x² – 28x + 30 by 2x – 6

(3) 6x³ + x² – 19x + 6 by 2x – 3

(4) a³ – 5a² + 2a + 8 by a -2

(5) x³ – 8 by x – 2

(6) 3a² – 7ab + 4b² by a- 2b

(7) x⁴ + 6x³ + 13x² + 13x + 8 by x² + 3x + 2

(8) x³ – 7x²y + 13xy² + 5y³ by x – 2y

Sol :

S.chand publication New Learning Composite mathematics solution of class 8 Chapter 5 Algebraic Expressions Exercise 5F

 Exercise 5F

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Q1 | Ex-5F | Class 8 |Algebraic Expressions | S.chand  | New Learning Composite mathematics | Chapter 5  | myhelper

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Question 1

Divide:

(a) 7a+7b by 7

Sol :
$=\frac{7a+7b}{7}=\frac{7(a+b)}{7}$
=a+b

(b) ax+ay by a

Sol :

$=\frac{ax+ay}{a}=\frac{a(x+y)}{a}$
=x+y

(c) 6y–2 by 2

Sol :

$=\frac{6y-2}{2}=\frac{2(3y-1)}{2}$
=3y-1

(d) b–a by b

Sol :
$=\frac{b-a}{b}=\dfrac{b\left(1-\frac{a}{b}\right)}{b}$
$=1-\frac{a}{b}$

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Q2 | Ex-5F | Class 8 |Algebraic Expressions | S.chand  | New Learning Composite mathematics | Chapter 5  | myhelper

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Question 2

(a) 8y–8 by -8

Sol :
$=\frac{8y-8}{-8}=\frac{8(y-1)}{-8}$
=-y+1

(b) 6x⁴+3x³+3x² by 3x²

Sol :
$=\frac{6x^4+3x^3+3x^2}{3x^2}$

$=\frac{3x^2(2x^2+x+1)}{3x^2}$

=2x²+x+1

(c) 25x⁸–20x⁵ by 5x³y³

Sol :
$=\frac{25x^8-20x^5}{5x^3y^3}=\frac{5x^5(5x^3-4)}{5x^3y^3}$

$=\frac{x^2}{y^3}(5x^3-4)$

$=\frac{5x^4}{y^3}-\frac{4x^2}{y^3}$

(d) x²y²z²–xyz+1 by xyz

Sol :

$=\frac{x^2y^2z^2-xyz+1}{xyz}=\frac{x^2y^2z^2}{xyz}-\frac{xyz}{xyz}+\frac{1}{xyz}$

$xyz-1+\frac{1}{xyz}$

$=xyz+\frac{1}{xyz}-1$

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Q3 | Ex-5F | Class 8 |Algebraic Expressions | S.chand  | New Learning Composite mathematics | Chapter 5  | myhelper

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Question 3

Express each quotient as a sum.

(a) $\frac{35a^5+28a^4b^2-14a^3b^3}{-7a^2}$

Sol :

$=\frac{35a^5+28a^4b^2-14a^3b^3}{-7a^2}$

$=-5a^3-4a^2b^2+2ab^3$

(b) $\frac{16x^2y-48xy^2+8x^2y^2}{8x^2y^2}$

Sol :

$=\frac{16x^2y-48xy^2+8x^2y^2}{8x^2y^2}$

$=\frac{2}{y}-\frac{6}{x}+1$

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Q4 | Ex-5F | Class 8 |Algebraic Expressions | S.chand  | New Learning Composite mathematics | Chapter 5  | myhelper

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Question 4

Evaluate

$\frac{x}{y}$ when $=\frac{x+y}{y}=21$

Sol :

$=\frac{x+y}{y}=21$

x+y=21y

x=21y-y=20y

$\frac{x}{y}=20$

S.chand publication New Learning Composite mathematics solution of class 8 Chapter 5 Algebraic Expressions Exercise 5E

 Exercise 5E

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Q1 | Ex-5E | Class 8 |Algebraic Expressions | S.chand  | New Learning Composite mathematics | Chapter 5  | myhelper

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Question 1

Simplify:

(a) $\frac{9x^7}{3x^5}$
Sol :

=3x^7-5=3x²

(b) $\frac{-20y^{12}}{4y^8}$
Sol :

=-5y^12-8=-5y⁴

(c) $\frac{-15a^7}{-3a^{15}}$

Sol :

=5a^7-15=5a^-8

$=\frac{5}{a^8}$

(d) $\frac{18x^4y^2}{9x^3y}$
Sol :

=2xy

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Q2 | Ex-5E | Class 8 |Algebraic Expressions | S.chand  | New Learning Composite mathematics | Chapter 5  | myhelper

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Question 2

(a) $\frac{-35x^8y^9}{-7x^3y^2}$
Sol :

=5x⁵y⁷

(b) $\frac{48a^2b^2c^4}{-12abc^3}$

Sol :
=-4abc

(c) $\frac{10x^{20}y^7z^4}{0.1x^{16}y^3z^9}$
Sol :

=100x⁴y⁴z^-5

$=\frac{100x^4y^4}{z^5}$ 

(d) $\frac{0.6a^4b^3c^2}{0.3ab^5c^7}$

Sol :

=2a³b^-2c^-5

$=\frac{2a^3}{b^2c^5}$

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Question 3

(a) $\frac{(3a^3b)(5ab^3)}{(5ab)(6ab^7)}$
Sol :

$=\frac{3\times a^3 \times b\times 5\times a\times b^3}{5\times a\times b\times 6\times a\times b^7}$

$=\frac{a^2}{2b^4}$

(b) $\frac{4ab(-7a^3b^7)}{-14a^2b^2}$

Sol :

$=4\times a\times b \times (-7)\times a^3 \times b^7$
=2a²b⁶

(c) $\frac{(5a^2b)^2(-100b^3)}{(5^2b)^2}$

Sol :

$=\frac{5^2 \times a^4 \times b^2 \times (-100)\times b^3}{5^4 b^2}$

=-4a⁴b³

(d) $\frac{(-2x^2y)^3}{(6xy^2)^2}$

Sol :

$=\frac{-8x^6y^3}{36x^2y^4}=\frac{-2x^4}{9y}$

(e) $\frac{(3x^2)^3(3y)}{(3x)^3(3x)^2}$

Sol :

$=\frac{3\times 3\times 3\times 3\times x^6\times y}{3\times 3\times 3\times 3^2\times x^2\times x^3 }=\frac{xy}{3}$

(f) $\frac{(-2pq^2)^8}{(4p^2q)^4}$
Sol :

$=\frac{-2^8 \times p^8 \times q^{16}}{2^8 \times p^8 \times q^4}$

=q¹²

S.chand publication New Learning Composite mathematics solution of class 8 Chapter 5 Algebraic Expressions Exercise 5D

 Exercise 5D

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Question 1

(a) (y – 6) (2y + 7)

Sol :

=y(2y+7)-6(2y+7)

=2y²+7y-12y-42

=2y²-5y-42

(b) (3x -7) (5x – 8)

Sol :

=3x(5x-8)-7(5x-8)

=15x²-24x-35x+56

=15x²-59x+56

(c) (x – 2) (6x + 9)

Sol :

=x(6x+9)-2(6x+9)

=6x²+9x-12x-18

=6x²-3x-18

(d) (3m² – n) (m² – 2n)

Sol :

=3m²(m²-2n)-n(m²-2n)

=3m⁴-6m²n-m²n+2n²

=3m⁴-7m²n+2n²

(e) (3a – b²) (2a + 7b²)

Sol :

=3a(2a+7b²)-b²(2a+7b²)

=6a²+21ab²-2ab²-7b⁴

=6a²+19ab²-7b⁴

(f) (x² – 2xy) (3xy + y²)

Sol :

=x²(3xy+y²)-2xy(3xy+y²)

=3x³y+x²y²-6x²y²-2xy³

=3x³y-5x²y²-2xy³

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Question 2

(a) (x – 2) (x² – 3x + 2)

Sol :

=x(x²-3x+2)-2(x²-3x+2)

=x³-3x²+2x-2x²+6x-4

=x³-5x²+8x-4

(b) (3x + 4) (x² – 5x + 2)

Sol :

=3x(x²-5x+2)+4(x²-5x+2)

=3x³-15x²+6x+4x²-20x+8

=3x³-11x²-14x+8

(c) (3 – y) (5y² – 2 – 3y)

Sol :

=3(5y²-2-3y)-4(5y²-2-3y)

=15y²-6-9y-5y³+2y+3y²

=-5y³+18y²-7y-6

(d) (4a – 3) (3a² + 2 – 5a)

Sol :

=4a(3a²+2–5a)-3(3a²+2–5a)

=12a³+8a-20a²-9a²-6+15a

=12a³+23a-29a²-6

=12a³-29a²+23a-6

(e) (5x – 1) (-2x³ + 4x – 3)

Sol :

=5x(-2x³+4x–3)-1(-2x³+4x–3)

=-10x⁴+20x²-15x+2x³-4x+3

=-10x⁴+20x²+2x³-19x+3

=-10x⁴+2x³+20x²-19x+3

(f) (2p² – 3) (4p³– p²+7)

Sol :

=2p²(4p³– p²+7)-3(4p³– p²+7)

=8p⁵-2p⁴+14p²-12p³+3p²-21

=8p⁵-2p⁴-12p³+17p²-21

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Question 3

(a) (x + 1) (x²–x+1)

Sol :

=x(x²–x+1)+1(x²–x+1)

=x³-x²+x+x²-x+1

=x³-1

(b) (x – y) (x²+xy+y²)

Sol :

=x(x²+xy+y²)-y(x²+xy+y²)

=x³+x²y+xy²-x²y-xy²-y³

=x³-y³

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Question 4

(a) (x³–5x²+2) (x²–2x)

Sol :

=x²(x³–5x²+2)-2x(x³–5x²+2)

=x⁵-5x⁴+2x²-2x⁴+10x³-4x

=x⁵-7x⁴+10x³+2x²-4x

(b) (2x²–3x+4) (x²–5x–3)

Sol :

=2x²(x²–5x–3)-3x(x²–5x–3)+4(x²–5x–3)

=2x⁴-10x³-6x²-3x³+15x²+9x+4x²-20x-12

=2x⁴-13x³+13x²-11x-12

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Question 5

Simplify:

(x+2) (x²–6x+8)–(2x–3)(2x²–x–1)

Sol :

=[x(x²–6x+8)+2(x²–6x+8)]-[2x(2x²–x–1)-3(2x²–x–1)]

=(x³–6x²+8x+2x²-12x+16)-[(4x³–2x²–2x-6x²+3x+3)]

=(x³–6x²+8x+2x²–12x+16)-4x³+2x²+2x+6x²–3x-3

=x³–6x²+8x+2x²–12x+16-4x³+2x²+2x+6x²–3x-3

=-3x³+4x²-5x+13

S.chand publication New Learning Composite mathematics solution of class 8 Chapter 5 Algebraic Expressions Exercise 5C

 Exercise 5C

Q1 | Ex-5C | Class 8 |Algebraic Expressions | S.chand  | New Learning Composite mathematics | Chapter 5  | myhelper

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Question 1

Find the Product:

(a) -4(2y + 7)

Sol : -8y-28

(b) -1(3x – y)

Sol : -3x+y

(c) x²(x – 4)

Sol : x³-4x²

(d) 3y²(5y³ + 1)

Sol : 15y⁵+3y²

(e) -2p(p² – p -7)

Sol : -2p³+2p²+14p

(f) (x² – x – 10) (-3x)

Sol : -3x³+3x²+30x

Q2 | Ex-5C | Class 8 |Algebraic Expressions | S.chand  | New Learning Composite mathematics | Chapter 5  | myhelper

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Question 2

(a) (6 – 5m – m⁷) (-2m²)

Sol : 

=-12m²+10m³+2m⁹

(b) y⁴ (y⁷ -3y² + 6)

Sol : 

=y¹¹-3y⁶+6y⁴

(c) (3 – q + 4q³) (-q⁴)

Sol : 

=-3q⁴+q⁵-4q⁷

(d) -2x² (3x² – 4xy + y²)

Sol : 

=-6x⁴+8x³y-2x²y²

(e) 5a³b²(7 –  2ab⁴ + 3a²b⁵)

Sol : 

=35a³b²-10a⁴b⁶+15a⁵b⁷