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**Ques 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****Sol:** Comparing the equations of each of the given systems with the general system of

We get:

(i) For x - 3y - 3 = 0

3x - 9y - 2 = 0

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

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

We find that

âˆ´ The given system has no solution.

(ii) For

a_{1} = 2, b_{1} = 1, c_{1} = - 5

a_{2} = 3, b_{2} = 2, c_{2} = - 8

We find that

âˆ´ The given system has a unique solution.

To solve the equations, we have:

â‡’

â‡’

â‡’

â‡’ x = 2 and y = 1

(iii) For

a_{1} = 3, b_{1} = âˆ’5, c_{1} = âˆ’ 20

a_{2} = 6, b_{2} = âˆ’ 10, c_{2} = âˆ’ 40

Since,

âˆ´ The given system of linear equations has infinitely many solutions.

(iv) For

a_{1} = 1, b_{1} = âˆ’ 3, c_{1} = âˆ’ 7

a_{2} = 3, b_{1} = âˆ’ 3, c_{2} = âˆ’ 15

âˆ´

Since

âˆ´ The given system has a unique solution.

Now, to use cross multiplication method, we have:

â‡’

â‡’

Thus, x = 4 and y = -1**Ques ****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****Sol:** (i) Comparing

2x + 3y = 7

(a - b) x + (a + b) y = (3a + b - 2)

with A_{1} x + B_{1} y = C_{1}

A_{2} x + B_{2} y = C2

we get:

A_{1} = 2, B_{1} = 3, C_{1} = 7

A_{2} = (a - b), B_{2} = (a + b), C_{2} = (3a + b - 2)

For an infinite number of solutions,

â‡’

From the first two equations, we get:

â‡’ 2 (a + b)=3 (a âˆ’ b)

â‡’ 2a + 2b =3a âˆ’ 3b

â‡’ 2a âˆ’ 3a + 2b + 3b = 0

â‡’ âˆ’ a + 5b = 0 â‡’ a âˆ’ 5b = 0 ...(1)

From the last two equations,

â‡’ 3 (3a + b âˆ’ 2) = 7 (a + b)

â‡’ 9a + 3b âˆ’ 6=7a + 7b

â‡’ 9a âˆ’ 7a + 3b âˆ’ 7b âˆ’ 6 = 0

â‡’ 2a âˆ’ 4b = 6

â‡’ a âˆ’ 2b = 3 ...(2)

Now, to solve by cross multiplication method:

A_{1} = 1, B_{1} = - 5, C_{1} = 0

A_{2} = 1, B_{2} = - 2, C_{2} = 0

â‡’

â‡’

â‡’

â‡’

Thus, a = 5 and b = 1

(ii) For no solution, we have

Let us consider,

â‡’ 3 (k âˆ’ 1) = 1 (2k âˆ’ 1)

â‡’ 3k âˆ’ 3 = 2k âˆ’ 1

â‡’ 3k âˆ’ 2k = âˆ’1 +

â‡’ k = 2

âˆ´ For k = 2, the given pair of linear equations will have no solution.**Ques ****3: Solve the following pair of linear equations by the substitution and cross-multiplication methods:****8x + 5y = 9****3x + 2y = 4****Sol: **Method - 1 [using Substitution Method]

8x + 5y = 9 ...(1)

3x + 2y = 4 ...(2)

From (2),

Substituting this value of y in (1), we have:

â‡’ 2 Ã— 8x + 5 Ã— x - 5 Ã— 3x = 9 Ã— 2

â‡’ 16x + 20 - 15x = 18

â‡’ x + 20 - 18 = 0

â‡’ x + 2 = 0 â‡’ x = - 2

Now, putting x = (- 2) in

â‡’

Thus, x = - 2 and y = 5**Method - II** [Cross Multiplication Method]

Comparing with general pair of equations,

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

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

âˆ´

â‡’

â‡’

â‡’

â‡’ x = 1 Ã— - 2 = - 2 and y = 1 Ã— 5 = 5

Thus x = - 2 and y = 5**Ques ****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 â‚¹ 1000 as hostel charges whereas a student B, who takes food for 26 days, pays â‚¹ 1180 as hostel charges. Find the fixed charges and the cost of food per day.****(ii) A fraction becomes 1/3 when 1 is subtracted from the numerator and it becomes 1/4 when 8 is added to its denominator. Find the fraction.****(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?****(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?****(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.****Sol:** (i) Let the fixed charges = â‚¹ x

And charges of food per day = â‚¹ y

For student A:

Number of days = 20

âˆ´ Cost of food for 20 days = â‚¹ 20y

According to the problem,

x + 20y = 1000 ...(1)

For student B:

Number of days = 26

âˆ´ Cost of food for 26 days = â‚¹ 26y

According to the problem:

x + 26y = 1180 ...(2)

Now, we have the system of linear equations:

We have: a_{1} = 1, b_{1} = 20, c_{1} = - 1000

a_{2} = 1, b_{2} = 26, c_{2} = - 1800

â‡’

â‡’

âˆ´

Thus, x = 400 and y = 30

âˆ´Fixed charges = â‚¹ 400 and cost of food per day = â‚¹ 30

(ii) Let the numerator = x

And the denominator = y

âˆ´ Fraction = x/y

Case-I

â‡’

â‡’ 3 (x âˆ’ 1) = y

â‡’ 3x âˆ’ 3 = y

â‡’ 3x âˆ’ y âˆ’ 3 = 0 ..(1)

Case-II

â‡’

â‡’ 4 x = y + 8

â‡’ 4x âˆ’ y âˆ’ 8 = 0 ...(2)

Comparing (1) and with the general pair of linear equations,

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

a_{2} = 4, b_{2} = - 1, c_{2} = - 8

â‡’

â‡’

âˆ´

Thus, x = 5 and y = 12

âˆ´ Fraction = 5/12

(iii) Let the number of correct answers = x

And the number of wrong answers = y

Case-I

Marks for all correct answer (3 Ã— x) = 3x

Marks for all wrong answer (1 Ã— y) = y

âˆ´ According to the conditions:

3x - y = 40 ...(1)

Case-II

Marks for all correct answers (4 Ã— x) = 4x

Marks for all wrong answers (2 Ã— y) = 2y

âˆ´ According to the condition:

4x - 2y = 50

â‡’ 2x - y = 25 ...(2)

From (1) and (2) we have:

a_{1} = 3, b_{1} = - 1, c_{1} = - 40

a_{2} = 2, b_{2} = - 1, c_{2} = - 25

âˆ´

â‡’

â‡’

â‡’

âˆ´ x = 15 and y = 5

Now, total questions = [Number of correct answers] + [Number of wrong answers]

= 15 + 5

= 20

Thus, required number of questions = 20.

(iv) Let the speed of car-I be x km/hr

And the speed of car-II be y km/hr

Case-I

Distance travelled by car-I = AC

â‡’ speed Ã— time = 5 Ã— x km

AC = 5x

Distance travelled by car-II = BC = 5y

Since AB = AC - BC

100 = 5x - 5y

â‡’ 5x - 5y - 100 = 0

â‡’ x - y - 20 = 0 ...(1)

Case-II

Distance travelled by car-I in 1 hour = AD

âˆ´ AD = 1 Ã— x = x

Distance travelled by car-II in 1 hour = BD

âˆ´ BD = 1 Ã— y = y

Now AB = AD + DB

â‡’ 100 = x + y

â‡’ x + y = 100 ...(2)

Using cross-multiplication,

x - y - 20 = 0

x + y - 100 = 0

a_{1} = 1, b_{1} = - 1, c_{1} = - 20

a_{2} = 1, b_{2} = 1, c_{2} = - 100

â‡’

â‡’

â‡’

â‡’

Thus, speed of car-I = 60 km/hr

speed of car-II = 40 km/hr.

(v) Let the length of the rectangle = x units

And the breadth of the rectangle = y units.

âˆ´ Area of the rectangle = x Ã— y = xy**Condition-I:**

(Length - 5) Ã— (Breadth + 3) = Area - 9

â‡’ (x - 5) (y + 3) = xy - 9

â‡’ xy + 3x - 5y - 15 = xy - 9

â‡’ xy + 3x - 5y - 15 - xy + 9 = 0

â‡’ 3x - 5y - 6 = 0 ...(1)**Condition-II:**

(Length + 3) Ã— (Breadth + 2) = Area + 67

â‡’ (x + 3) (y + 2) = xy + 67

â‡’ xy + 2x + 3y + 6 = xy + 67

â‡’ xy + 2x + 3y + 6 - xy - 67 = 0

â‡’ 2x + 3y - 61 = 0 ...(2)

Now, using cross multiplication method for (1) and (2), we have:

a_{1} = 3, b_{1} = - 5, c_{1} = - 6

a_{2} = 2, b_{2} = 3, c_{2} = - 61

âˆ´

â‡’

â‡’

âˆ´

Thus, length of the rectangle = 17 units

And breadth of the rectangle = 9 units.

Equations Reducible to Pair of Linear Equations in Two Variables

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