RS Aggarwal Class 10 Solutions Chapter 3 - Linear Equations In Two Variables - Ex 3E(3.5)

RS Aggarwal Class 10 Chapter 3 - Linear Equations In Two Variables - Ex 3E(3.5) Solutions Free PDF

Students of Class 10 CBSE Board students should solve the exercise questions provided in the RS Aggarwal Class 10 Maths book. The solutions provided here can be referred by students if they get stuck while solving the questions. These solutions are comprehensive and designed in a way that the students can easily understand the steps and clear all their doubts.

These RS Aggarwal Solutions for class 10 are important to refer to as they can help students understand practice problems by adhering to the methods of solving math problems. The solutions are based on the latest CBSE syllabus following the CCE guidelines. With the help of different exercise questions, one can gain a better understanding of the different concepts in the syllabus.

Download PDF of RS Aggarwal Class 10 Solutions Chapter 3 – Linear Equations In Two Variables – Ex 3E (3.5)

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

Key Features of RS Aggarwal Class 10 Solutions Chapter 3 –  Linear Equations In Two Variables – Ex 3E(3.5)

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