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**Karnataka State Syllabus Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.5**

Question 1.

Which of the following pairs of linear equations has unique solution, no solution, or infinitely many solutions. In case there is a unique solution, find it by using cross multiplication method.

(i) x – 3y – 3 = 0

3x – 9y – 2 = 0

(ii) 2x + y = 5

3x + 2y = 8

(iii) 3x – 5y = 20

6x – 10y = 40

(iv) x – 3y – 7 = 0

3x – 3y – 15 = 0

Solution:

(i) x – 3y – 3 = 0 …………. (i)

3x – 9y – 2 = 0 ……………… (ii)

Here, a_{1} = 1, b_{1} = -3, c_{1} = -3

a_{2} = 3, b_{2} = -9, c_{2} = -2

a1a2=13b1b2=−3−9=13

c1c2=−3−2=32

Here, a1a2=b1b2≠c2c2

∴ No solution for these.

(ii) 2x + y = 5

3x + 2y = 8

2x + y – 5 = 0 → (1)

3x + 2y – 8 = 0 → (2)

Here, a_{1} = 2, b_{1} = 1, c_{1}= – 5

a_{2} = 3, b_{2} = 2, c_{2} = – 8

Here,

∴ Many solutions are there for these

x1x−8−(2x−5)=y−5×3−(−8×2)=12×2−3×1

x−8+10=y−15+16=14−3

x+2=y1=11

x+2=11x=2

y1=11y=1

∴ Unique solution: x = 2, y = 1

(iii) 3x – 5y = 20

6x – 10y = 40

3x – 5y – 20 = 0 → (1)

6x – 10y – 40 = 0 → (2)

Here, a_{1} = 3 b_{1} = – 5, c_{1} = – 20

a_{2} = 6, b_{2 }= – 10, c_{2} = – 40

Here,

Infinite solutions are there for these.

(iv) x – 3y – 7 = 0 …………… (i)

3x – 3y – 15 = 0 …………….. (ii)

a_{1} = 1, b_{1} = -3, c_{1} = -7

a_{2} = 3, b_{2} = -3, c_{2} = -15

a1a2=13b1b2=−3−3=11

c1c2=−7−15=715

Here, a1a2≠b1b2

∴ Many solutions are there

x−3x−15−(−3x−7)=y−7×3−(−15×1)=11x−3−(3x−3)

x45−21=y−21+15=1−3+9

x24=y−6=16

If x24=16 then

6x = 24

x=246=4

If y−6=16 then

6y = -6

y=−66

∴ y = -1

∴ Unique solutions are there for these x = 4, y = -1.

Question 2.

(i) For which values of a and b does the following pair of linear equations have an infinite number of solutions ?

2x + 3y = 7

(a – b) x + (a + b) y = 3a + b – 2

(ii) For which value of k will the following pair of linear equations have no solution ?

3x + y = 1

(2k – 1)x + (k – 1)y = 2k + 1

Solution:

(i) 2x + 3y = 7

2x + 3y – 7 = 0 ………… (i)

(a – b)x + (a + b)y = (3a + b – 2) ………… (ii)

(a – b)x + (a + b)y – (3a + b – 2) ……………. (ii)

Here, a_{1} =2, b_{1} = 3, c_{1} = -7

a_{2} = (a – b), b_{2} = (a + b), c_{2} = -(3a + b – 2)

If it has infinite solutions, then we have

a1a2=b1b2=c1c2

2(a−b)=3(a+b)=−7−(3a+b−2)

If 2(a−b)=−7(−3a−b+2) then

2(-3a – b + 2) = -7(a – b)

-6a – 2b + 4 = -7a + 7b

-6a + 7a – 2b – 7b = -4

a – 9b = 4 ………… (i)

If 3(a+b)=−7(−3a−b+2) then

3(-3a – b + 2) = -7(a + b)

-9a- 3b + 6 = -7a – 7b

-9a + 7a – 3b + 7b = -6

-2a + 4b = -6

-a + 2b = -3 …………. (ii)

From eqn. (i) + eqn. (ii),

7b = 7

b=77=1

Substituting the value of ‘b’ in eqn. (i),

a – 9b = -4

a – 9(1) = -4

a – 9 = -4

a = -4 + 9

a = 5

∴ a = 5, b= 1.

(ii) If the linear pair do not have solutions, then we have

3x + y – 1 = 0 → (1)

(2k – 1)x + (k – 1)y – (2k + 1) = 0 → (2)

Here a_{1} = 3, b_{1} = 1, c_{1}= 1

a_{2} = (2k – 1), b_{2} = (k – 1), c_{2} = -(2k + 1)

3(k – 1) = 1(2k – 1)

3k – 3 = 2k – 2

3k – 2k = – 2 + 3

∴ k = 1.

Question 3.

Solve the following pair of linear equations by the substitution and cross-multiplication methods:

8x + 5y = 9

3x + 2y = 4

Solution:

8x + 5y = 9 ……………. (i)

3x + 2y = 4 …………….. (ii)

(1) Substitution Method:

8x + 5y = 9

8x = 9 – 5y

x=9−5y8

Substituting the value of ‘x’ in eqn (i)

3x + 2y = 4

3(9−5y8)+2y=4

27−15y8+2y1=4

27−15y+1648=4

27 + y = 8 × 4

27 + y = 32

y = 32 – 27

∴ y = 5

Substituting the value of ‘y’ in

x=9−5y8

=9−5×58

=9−258

=−168

∴ x = -2

∴ x = -2, y = 5

(2) Cross-Multiplication Method:

8x + 5y – 9 = 0 …………. (i)

3x + 2y – 4 = 0 …………. (ii)

Here, a_{1} = 8 b_{1} = 5 c_{1}=-9

a_{2} = 3, b_{2} = 2, c_{2} = -4

8x + 5y = 9 3x + 2y = 4

x5x−4−(2x−9)=y−9×3−(−4×8)=18×2−3×5

x−20+18=y−27+32=116−15

x−2=y5=11

x−2=1x=−2

y5=1y=5

Question 4.

Form the pair of linear equations in the following problems and find their solutions (if they exist) by any algebraic method :

(i) A part of monthly hostel charges is fixed and the remaining depends on the number of days one has taken food in the mess. When a student A takes food for 20 days she has to pay Rs. 1000 as hostel charges whereas a student B, who takes food for 26 days, pays Rs. 1180 as hostel charges. Find the fixed charges and the cost of food per day.

Solution:

Fixed hostel charges are Rs. x.

Amount to be paid to be Rs. y.

x + 20y = 1000 …………. (i)

x + 26y = 1180 …………. (ii)

By elimination method, we can find solution.

Subtracting eqn. (ii) from eqn. (i),

– 6y =- 180

y=1806

∴ y = Rs. 30

Substituting the value of ‘y’ in eqn. (i),

x + 20y = 1000

x + 20(30) = 1000

x + 600 = 1000

∴ x = 1000 – 600

∴ x = Rs. 400

Fixed charge, x = Rs. 400

Amount to be paid, y = Rs. 30.

(ii) A fraction becomes 13 when 1 is subtracted from the numerator and it becomes 14 when 8 is added to its 4 denominator. Find the fraction.

Solution:

Let the fraction be

3(x – 1) = y

3x – 3 = y

3x – y – 3 = 0 → (1)

4x = y + 8

4x – y – 8 = 0 → (2)

By Substitution method we can find out solution:

From eqn. (i),

3x – y – 3 = 0

y = 3x – 3

Substituting the value of y’ in eqn. (2),

4x – y – 8 = 0

4x – (3x – 3) – 8 = 0

4x – 3x + 3 – 8 = 0

x – 5 = 0

∴ x = 5.

Substituting the value of ‘x’ in

y = 3x – 3,

= 3(5) – 3

= 15 – 3

∴ y = 12

∴ x = 5, y = 12

∴ Fraction is

(iii) Yash scored 40 marks in a test, getting 3 marks for each right answer and losing 1 mark for each wrong answer. Had 4 marks been awarded for each correct answer and 2 marks been deducted for each incorrect answer, then Yash would have scored 50 marks. How many questions were there in the test?

Solution:

Let the number of right answers be ‘x’, and the number of wrong answers be y’.

3x – y = 40 ……….. (i)

4x – 2y = 50 ………. (ii)

By Elimination method we can find our solution :

3x – y = 40 ……… (i)

2x – y = 25 ……….. (ii)

Subtracting eqn. (ii) from eqn. (i),

Substituting the value of ‘x’ in eqn. (i),

3x – y = 40

3(15) – y = 40

45 – y = 40

-y = 40 – 45

-y = -5

∴ y = 5

Questions of right answers, x = 15

Questions of wrong answers, y = 5

∴ Total number of questions = x + y = 15 + 5 = 20

(iv) Places A and B are 100 km apart on a highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of the two cars ?

Solution:

Distance between places A and B is 100 km.

Speed of A car us ‘u’ km/hr.

Speed of B car is ‘v’ km/hr.

If the cars travel in the same direction at different speeds they meet in 5 hours about 20 km.

∴ u – v = 20 ………. (i)

If the cars travel in the same direction, total distance covered =100 km.

∴ u + v = 100 …….. (ii)

We can solve by Elimination method:

From eqn. (i) + eqn. (ii),

u=1202

∴ u = 60 km/hr.

Substituting the value of ‘u’ in Eqn. (ii),

u + v = 100

60 + v = 100

v = 100 – 60

v = 40 km/hr.

∴ u = 60 km/hr.

v = 40 km/hr.

(v) The area of a rectangle gets reduced by 9 square units, if its length is reduced by 5 units and breadth is increased by 3 units. If we increase the length by 3 units and the breadth by 2 units, the area increases by 67 square units. Find the dimensions of the rectangle.

Solution:

Length of rectangle is ‘x’ unit.

Breadth of rectangle is ‘y’ unit.

Area of rectangle = length × breadth = xy

(x – 5) (y + 3) = xy – 9

xy + 3x – 5y – 15 = xy – 9

3x – 5y = -9 + 15

3x – 5y = 6 …………. (i)

(x + 3) (y + 2) = xy + 67

xy + 2x + 3y + 6 = xy + 67

2x + 3y = 67 – 6

2x + 3y = 61 ……….. (ii)

Solving the equation by Elimination method :

Multiplying eqn. (i) by and eqn. (ii) by 5

9x – 15y = 18 ………….. (iii)

10x + 15y = 305 ………… (iv)

From eqn. (iii) + eqn. (iv)

x=32319

∴ x = 17 unit.

Substituting the value of ‘x’ in eqn. (i),

3x – 5y = 6

3(17) – 5y = 6

51 – 5y = 6

-5y = 6 – 51

– 5y = – 45

5y = 45

y=455

∴ y = 9 units.

∴ Length of that rectangle, x = 17 unit

breadth, y = 9 unit.

**All Chapter KSEEB Solutions For Class 10 Maths**

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