RS Aggarwal Solutions Class 10 Ex 3E

Q.1: 6(ax + by) = 3a + 2b, 6(bx – ay) = 3b – 2a

Solution:

6ax + 6by = 3a + 2b ……..(1)

6(bx – ay) = 3b – 2a

6bx – 6ay = 3b – 2a ……..(2)

Multiplying (1) by ‘a’ and (2) by ‘b’, we get

6a2x + 6aby = 3a2 + 2ab ……..(3)

6b2x – 6aby = 3b – 2ab ……..(4)

Adding (3) and  (4), we get

6a2x + 6b2 = 3a2 + 3b2

6(a2 + b2)x = 3(a2 + b2)

\( x = \frac{3(a^{2} + b^{2})}{6(a^{2} + b^{2})} = \frac{3}{6} = \frac{1}{2} \)

Substituting in (1), we get

(6a)(1/2) + 6by = 3a + 2b

3a + 6by = 3a + 2b

6by = 3a + 2b – 3a

6by = 2b

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

Hence, the solution is

x = (½) , y = (⅓)

Q.2: 2ax – 2by = -(a + 4b), 2bx + 2ay = -(b – 4a)

Solution:

2(ax – by) + (a + 4b) = 0

2ax – 2by = -(a + 4b) ……..(1)

2bx + 2ay = -(b – 4a) ……..(2)

Multiplying (1) by ‘a’ and (2) by ‘b’, we get

2a2x – 2aby = -a(a + 4ab) ……..(3)

2b2x + 2aby = -b(b – 4a) ……..(4)

Adding (3) and  (4), we get:

2a2x + 2b2x = – a(a + 4ab) – b(b – 4a)

2(a2 + b2)x = – a2 – 4ab – b2+ 4ab

2(a2+ b2) = – (a2 + b2)

\( x = – \frac{(a^{2} + b^{2})}{2(a^{2} + b^{2})} = – \frac{3}{6} = – \frac{1}{2} \)

Substituting in (1), we get

(2a)(-½) – 2by = -(a + 4b)

-a – 2by = -a – 4b

-2by = -a – 4b + a

-2by = -4b

\( y = \frac{-4b}{-2b} = 2 \)

Hence, the solution is

x = (-½), y = 2

Q.3: 71x + 37y = 253, 37x + 71y = 287

Solution:

The given equations are

71x + 37y = 253 ……..(1)

37x + 71y = 287 ……..(2)

Adding (1) and (2)

108x + 108y = 540

108(x + y) = 540

∴ x + y = (540/108) = 5 ……..(3)

Subtracting (2) from (1)

34x – 34y = 253 – 287 = -34

34(x – y) = -34

∴ x – y = -1 ……..(4)

Adding (3) and (4)

2x = 5 – 1 = 4

\( \Rightarrow \) x = 2

Subtracting (4) from (3)

2y = 5 + 1 = 6

\( \Rightarrow \) y = 3

Hence, solution is x = 2, y = 3

Q.4: 37x + 43y = 123, 43x + 37y = 117

Solution:

37x + 43y = 123 ……..(1)

43x + 37y = 117 ……..(2)

Adding (1) and (2)

80x + 80y = 240

80(x + y) = 240

x + y = \( \frac{240}{80} \) = 3 ……..(3)

Subtracting (1) from (2),

6x – 6y = -6

6(x – y) = -6

x – y = \( \frac{-6}{6} \) = -1 ……..(4)

Adding (3) and (4)

2x = 3 – 1 = 2

\( \Rightarrow \) x = 1

Subtracting (4) from (3)

2y = 3 + 1 = 4

\( \Rightarrow \) y = 2

Hence, solution is x = 1, y = 2

Q.5: 217x + 131y = 913, 131x + 217y = 827

Solution:

217x + 131y = 913 ……..(1)

131x + 217y = 827 ……..(2)

Adding (1) and (2)

348x + 348y = 1740

348(x + y) = 1740

x + y = 5 ……..(3)

Subtracting (2) from (1),

86x – 86y = 86

86(x – y) = 86

x – y = 1 ……..(4)

Adding (3) and (4)

2x = 6

\( \Rightarrow \) x = 3

Putting x = 3 in (3), we get

3 + y = 5

\( \Rightarrow \) y = 2

Hence, solution is x = 3, y = 2

Q.6: 41x – 17y = 99, 17x + 41y = 75

Solution:

41x – 17y = 99 ……..(1)

17x + 41y = 75 ……..(2)

Adding (1) and (2)

58x – 58y = 174

58(x – y) = 174

x – y = 3 ……..(3)

Subtracting (2) from (1),

24x + 24y = 24

24(x + y) = 24

x + y = 1 ……..(4)

Adding (3) and (4)

4x = 2

\( \Rightarrow \) x = 2

Putting x = 2 in (3), we get

2 – y = 3

\( \Rightarrow \) y = -1

Hence, solution is x = 2, y = -1

Solve each of the following systems of equations by using the method of cross multiplication:

Q.7: x + 2y + 1 = 0, 2x – 3y – 12 = 0

Solution:

x + 2y + 1 = 0 ……..(1)

2x – 3y – 12 = 0 ……..(2)

By cross multiplication, we have

\( \frac{x}{[2x(-12) – 1 \times (-3) ]} = \frac{y}{[1 \times 2 – 1 \times (-12) ]} = \frac{1}{[1 \times (-3) – 2 \times 2] } \)

\( \Rightarrow \frac{x}{(-24 + 3)} = \frac{y}{(2 + 12)} = \frac{1}{(-3 – 4)} \)

\( \Rightarrow \frac{x}{-21} = \frac{1}{-7} , \frac{y}{14} = \frac{1}{-7} \)

\( \Rightarrow x = \frac{-21}{-7} = 3 , y = \frac{14}{-7} = -2 \)

Hence, x = 3 and y = -2 is the solution.

Q.8: 2x + 5y – 1 = 0, 2x + 3y – 3 = 0

Solution:

2x + 5y – 1 = 0 ……..(1)

2x + 3y – 3 = 0 ……..(2)

By cross multiplication, we have

\( \frac{x}{[5x(-3) \times (-1) ]} = \frac{y}{[-1 \times 2 – (-3) \times (2) ]} = \frac{1}{[2 \times (3) – 2 \times 5] } \)

\( \Rightarrow \frac{x}{(-15 + 3)} = \frac{y}{(-2 + 6)} = \frac{1}{(6 – 10)} \)

\( \Rightarrow \frac{x}{-12} = \frac{1}{-4}, \frac{y}{4} = \frac{1}{-4} \)

\( \Rightarrow x = \frac{-12}{-4} = 3 , y = \frac{4}{-4} = -1 \)

Hence, x = 3 and y = -1 is the solution.

Q.9: 3x – 2y + 3 = 0, 4x + 3y – 47 = 0

Solution:

3x – 2y + 3 = 0 ……..(1)

4x + 3y – 47 = 0 ……..(2)

By cross multiplication, we have

\( \frac{x}{[(-2) \times (-47) – (3 \times 3) ]} = \frac{y}{[(3 \times 4) – (-47) \times (3) ]} = \frac{1}{[3 \times (3) – (-2) \times 4] } \)

\( \Rightarrow \frac{x}{(94 – 9)} = \frac{y}{(12 + 141)} = \frac{1}{(9 + 8)} \)

\( \Rightarrow \frac{x}{85} = \frac{y}{153} = \frac{1}{17} \)

17x = 85, 17y = 153

\( \Rightarrow x = \frac{85}{17}, y = \frac{153}{17} \)

Hence, x = 5 and y = 9 is the solution.

Q.10: 6x – 5y – 16 = 0, 7x – 13y + 10 = 0

Solution:

6x – 5y – 16 = 0 ……..(1)

7x – 13y + 10 = 0 ……..(2)

By cross multiplication, we have

\( \frac{x}{[-5 \times 10 – (-16) \times (-13) ]} = \frac{y}{[(-16 \times 7) – 10 \times 6 ]} = \frac{1}{[6 \times (-13) – (-5) \times 7] } \)

\( \Rightarrow \frac{x}{(-50 – 208)} = \frac{y}{(-112 – 60)} = \frac{1}{(-78 + 35)} \)

\( \Rightarrow \frac{x}{-258} = \frac{1}{-43} , \frac{y}{-172} = \frac{1}{-43} \)

\( \Rightarrow x = \frac{-258}{-43} = 6 , y = \frac{-172}{-43} = 4 \)

Hence, x = 6 and y = 4 is the solution.

Q.11: 3x + 2y + 25 = 0, 2x + y + 10 = 0

Solution:

3x + 2y + 25 = 0 ……..(1)

2x + y + 10 = 0 ……..(2)

By cross multiplication, we have

\( \frac{x}{[2 \times 10  – 25 \times 1 ]} = \frac{y}{[25 \times 2 – 10 \times 3 ]} = \frac{1}{[3 \times 1 – 2 \times 2] } \)

\( \Rightarrow \frac{x}{(20 – 25)} = \frac{y}{(50 – 30)} = \frac{1}{(3 – 4)} \)

\( \Rightarrow \frac{x}{-5} = \frac{1}{-1} , \frac{y}{20} = \frac{1}{-1} \)

Hence, x = 5 and y = -20 is the solution.

Q.12: 2x + y – 35 = 0, 3x – 4y – 65 = 0

Solution:

2x + y – 35 = 0 ……..(1)

3x – 4y – 65 = 0 ……..(2)

By cross multiplication, we have

\( \frac{x}{[1 \times (-65) – 4 \times (-35) ]} = \frac{y}{[(-35) \times 3 – (-65) \times 2 ]} = \frac{1}{[2 \times 4 – 3 \times 1] } \)

\( \Rightarrow \frac{x}{(-65 + 140)} = \frac{y}{(-105 + 130)} = \frac{1}{(8 – 3)} \)

\( \Rightarrow \frac{x}{75} = \frac{1}{5} , \frac{y}{25} = \frac{1}{5} \)

\( \Rightarrow x = \frac{75}{5}, y = \frac{25}{5} \)

Hence, x = 15 and y = 5 is the solution.

Q.13: 7x – 2y – 3 = 0, 22x – 3y – 16 = 0

Solution:

By cross multiplication, we have

\( \Rightarrow \frac{x}{(-16 – \frac{9}{2})} = \frac{y}{(-33 + 56)} = \frac{1}{\frac{-21}{2} +22} \)

\( \Rightarrow \frac{x}{\frac{23}{2}} = \frac{y}{23} = \frac{1}{\frac{23}{2}} \)

\( \Rightarrow \frac{x}{\frac{23}{2}} = \frac{1}{\frac{23}{2}}, \frac{y}{23} = \frac{1}{\frac{23}{2}} \)

Hence, x = 1 and y = 2 is the solution.

Q.14: \( \frac{x}{6} + \frac{y}{15} – 4 = 0 \), \( \frac{x}{3} – \frac{y}{12} – \frac{19}{4} = 0 \)

Solution:

\( \frac{x}{6} + \frac{y}{15} – 4 = 0 \) ……..(1)

\( \frac{x}{3} – \frac{y}{12} – \frac{19}{4} = 0 \) ……..(2)

By cross multiplication, we have

\( \frac{x}{[ \frac{1}{15} \times (- \frac{19}{4} ) – (- \frac{1}{12} ) (-4)]} = \frac{y}{(-4) \frac{1}{3} – \frac{1}{6} (- \frac{19}{4})} = \frac{1}{\frac{1}{6} \times (- \frac{1}{12} ) – \frac{1}{3} \times \frac{1}{15}} \)

Or \( \frac{x}{- \frac{19}{60} – \frac{1}{3} } = \frac{y}{- \frac{4}{3} + \frac{19}{24}} = \frac{1}{- \frac{1}{72} – \frac{1}{45}} \)

Or \( \frac{x}{- \frac{39}{60}} = \frac{y}{- \frac{13}{24}} = \frac{1}{- \frac{13}{360}} \)

\( x = – \frac{39}{60} \times (- \frac{360}{13}), y = -\frac{13}{24} \times (-\frac{360}{13}) \)

Hence, x = 18 and y = 15 is the solution.

Q.15: ax + by – (a – b) = 0, bx – ay – (a + b) = 0

Solution:

ax + by – (a – b) = 0

bx – ay – (a + b) = 0

By cross multiplication, we have:

\( \frac{x}{-ba – b^{2} – a^{2} + ab} = \frac{y}{-ba + b^{2} + a^{2} + ab} = \frac{1}{-a^{2} – b^{2}} \)

\( \Rightarrow \frac{x}{-b^{2} – a^{2}} = \frac{y}{b^{2} + a^{2}} = \frac{1}{-a^{2} – b^{2}} \)

\( \Rightarrow \frac{x}{-b^{2} – a^{2}} = \frac{1}{-(a^{2} + b^{2})} , \, \frac{y}{b^{2} + a^{2}} = \frac{1}{-(a^{2} + b{2})} \)

x = \( \frac{-(b^{2} + a^{2})}{-(a^{2} + b^{2})} \)  y = \( \frac{b^{2} + a^{2}}{-(a^{2} + b^{2})} \)

Hence, the solution is x = 1, y = -1

Q.16: 2ax + 3by = (a + 2b), 3ax + 2by = (2a + b)

Solution:

2ax + 3by – (a + 2b) = 0

3ax + 2by – (2a + b) = 0

By cross multiplication, we have

\( \frac{x}{[3b \times (-(2a + b)) – 2b \times (-(a + 2b)) ]} = \frac{y}{-(a + 2b) \times 3a – 2a \times (-(2a + b))} = \frac{1}{2a \times 2b – 3a \times 3b} \)

\( \frac{x}{[-6ab – 3b^{2} + 2ab + 4b^{2}]} = \frac{y}{-3a^{2} – 6ab + 4a^{2} + 2ab } = \frac{1}{4ab – 9ab} \)

\( \Rightarrow \frac{x}{b^{2} – 4ab} = \frac{y}{a^{2} – 4ab} = \frac{1}{-5ab} \)

\( \Rightarrow \frac{x}{-b(4a – b) } = \frac{y}{-a(4b – a)} = \frac{1}{-5ab} \)

\( \Rightarrow \frac{x}{-b(4a – b)} = \frac{1}{-5ab}, \, \frac{y}{-a(4b – 1)} = \frac{1}{-5ab} \)

\( x = \frac{-b(4a – b)}{-5ab}, \, y = \frac{-a(4b – a)}{-5ab} \)

Hence, \( x = \frac{4a – b}{5a} , y = \frac{4b – a}{5b} \) is the solution.

Q.17: \( \frac{x}{a} + \frac{y}{b} -2 = 0 \), ax – by – (a2 – b2) = 0

Solution:

\( \frac{x}{a} + \frac{y}{b} -2 = 0 \)

ax – by – (a2 – b2) = 0

By cross multiplication, we have

\( \frac{x}{[ \frac{1}{b} {-(a^{2} – b^{2})} – (-2)(-b)]} = \frac{y}{[(-2) – \frac{1}{a} \times {-(a^{2} – b^{2})}]} \frac{1}{- \frac{b}{a} – \frac{a}{b}} \)

\( \Rightarrow \frac{x}{[\frac{1}{b} { – (a^{2} – b^{2})} – (-2)(-b)]} = \frac{y}{[(-2a) – \frac{1}{a} \times {- (a^{2} – b^{2})}]} = \frac{1}{- \frac{b}{a} – \frac{a}{b}} \)

\( \Rightarrow \frac{x}{\frac{-a^{2} – b^{2}}{b}} = \frac{1}{\frac{-b^{2} – a^{2}}{ab}} , \, \frac{y}{\frac{-a^{2} – b^{2}}{a}} = \frac{1}{\frac{-b^{2} – a^{2}}{ab}} \)

\( x = \frac{-(a^{2} + b^{2})}{b} \times \frac{ab}{-(b^{2} – a^{2})} = a \)

\( y = \frac{-(a^{2} + b^{2})}{a} \times \frac{ab}{-(b^{2} + a^{2})} = b \)

∴ The solution is x = a, y = b.

Q.18: \( \frac{x}{a} + \frac{y}{b} – (a + b) = 0 \), \( \frac{x}{a^{2}} + \frac{y}{b^{2}} – 2 = 0 \)

Solution:

\( \frac{x}{a} + \frac{y}{b} – (a + b) = 0 \)

\( \frac{x}{a^{2}} + \frac{y}{b^{2}} – 2 = 0 \)

By cross multiplication, we have,

\( \Rightarrow \frac{x}{\frac{-2}{b} + \frac{a}{b^{2}} + \frac{b}{b^{2}}} = \frac{y}{\frac{-1}{a} – \frac{b}{a^{2}} + \frac{2}{a}} = \frac{1}{\frac{a – b}{a^{2} b^{2}}} \)

\( \Rightarrow \frac{x}{\frac{a – b}{b^{2}}} = \frac{y}{\frac{a – b}{a^{2}}} = \frac{1}{\frac{a – b}{a^{2} b^{2}}} \)

\( x = \frac{a – b}{b^{2}} \times \frac{a^{2} b^{2}}{a – b} = a^{2} \)

\( y = \frac{a – b}{a^{2}} \times \frac{a^{2} b^{2}}{a – b} = b^{2} \)

The solution is x = a2, y = b2


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The electron affinities of N, O, S and Cl are :