**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

3x – 9y – 2 = 0

a_{1}/a_{2} = ^{1}/_{3} ; b_{1}/b_{2} = ^{-3}/_{-9} = ^{1}/_{3} ; c_{1}/c_{2} = ^{-3}/_{-2} = ^{3}/_{2}

a_{1}/a_{2} = b_{1}/b_{2} ≠ c_{1}/c_{2}

Therefore, the given sets of lines are parallel to each other. Therefore, they will not intersect each other and thus, there will not be any solution for these equations.

(ii) 2x + y = 5

3x + 2y = 8

a_{1}/a_{2} = ^{2}/_{3} ; b_{1}/b_{2} = ^{1}/_{2} ; c_{1}/c_{2} = ^{-5}/_{-8}

a_{1}/a_{2} ≠ b_{1}/b_{2}

Therefore, they will intersect each other at a unique point and thus, there will be a unique solution for these equations. By cross-multiplication method,

^{x}/(b_{1}c_{2}-b_{2}c_{1}) = ^{y}/(c_{1}a_{2}-c_{2}a_{1}) = ^{1}/(a_{1}b_{2}-a_{2}b_{1})

^{x}/[-8-(-10)] = ^{y}/[-15+16] = ^{1}/(4-3)

^{x}/_{2} = ^{y}/_{1} = 1

^{x}/_{2} = 1 ; ^{y}/_{1} = 1

x = 2 ; y = 1

Therefore, x = 2, y = 1

(iii) 3x – 5y = 20

6x – 10y = 40

a_{1}/a_{2} = ^{3}/_{6} = ^{½} ; b_{1}/b_{2} = ^{-5}/_{-10} = ^{1}/_{2} ; c_{1}/c_{2} = ^{-20}/_{-40} = ^{1}/_{2}

a_{1}/a_{2} = b_{1}/b_{2} = c_{1}/c_{2}

Therefore, the given sets of lines will be overlapping each other i.e., the lines will be coincident to each other and thus, there are infinite solutions possible for these equations.

(iv) x – 3y – 7 = 0

3x – 3y – 15 = 0

a_{1}/a_{2} = ^{1}/_{3} _{ }; b_{1}/b_{2} = ^{-3}/_{-3} = 1 ; c_{1}/c_{2} = ^{-7}/_{-15} = ^{7}/_{15}

a_{1}/a_{2} ≠ b_{1}/b_{2}

Therefore, they will intersect each other at a unique point and thus, there will be a unique solution for these equations.

By cross-multiplication,

^{x}/_{(45-(21)} = ^{y}/_{(-21-(15))} = ^{1}/_{(-3-(-9)}

^{x}/_{24} = ^{y}/_{-6} = ^{1}/_{6}

^{x}/_{24} = ^{1}/_{6} and ^{y}/_{-6} = ^{1}/_{6}

x = 4 and y = -1

Thus, x = 4 and y = -1

**2. **

**(i) For which values of a and b will 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 = 0

(a-b)x + (a+b)y – (3x +b -2) = 0

a_{1}/a_{2} = ^{2}/_{(a-b)} ; b_{1}/b_{2} = ^{3}/_{(a+b)} ; c_{1}/c_{2} = ^{-7}/_{-(3a+b-2)} = ^{7}/_{(3a+b-2)} ;

For infinitely many solutions,

a_{1}/a_{2} = b_{1}/b_{2} = c_{1}/c_{2}

^{2}/_{(a-b)} = ^{7}/_{(3a+b-2)}

6a+2b-4=7a-7b

a-9b = -4 ————(1)

^{2}/_{(a-b)} = ^{3}/_{(a+b)}

2a + 2b = 3a – 3b

a – 5b = 0 ————(2)

Subtracting (1) from (2), we obtain

4b = 4

b = 1

Substituting this in equation(2), we obtain

a – 5×1 = 0

a = 5

Hence, *a *= 5 and *b *= 1 are the values for which the given equations give infinitely many solutions.

(ii) 3x + y – 1 = 0

(2k-1)x +(k-1)y – 2k -1 = 0

a_{1}/a_{2} = ^{3}/_{(2k-1)} ; b_{1}/b_{2} = ^{1}/_{(k-1)} ; c_{1}/c_{2} = ^{-1}/_{-2k-1} = ^{1}/_{2k+1} ;

For no solution;

a_{1}/a_{2} = b_{1}/b_{2} ≠ c_{1}/c_{2}

^{3}/_{(2k-1)} = ^{1}/_{(k-1)} ≠ ^{1}/_{(2k+1)}

^{3}/_{(2k-1)} = ^{1}/_{(k-1)}

3k – 3 = 2k – 1

k = 2

Hence, for *k *= 2, the given equation has no solution

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

**8x + 5y = 9**

**3x + 2y = 4**

Solution:

8x + 5y = 9 ———-(1)

3x + 2y = 4 ———–(2)

From equation (ii), we get

x = ^{(4-2y)}/_{3} ————(3)

Substituting this value in equation (1), we get

8(^{[4-2y)}/_{3}) + 5y = 9

32 – 16y + 15y = 27

-y = -5

y = 5

Substituting the value of y in equation (2)

3x + 10 = 4

x = -2

Hence x = -2 , y = 5

Again by cross multiplication method, we get

8x + 5y – 9 = 0

3x + 2y – 4 = 0

^{x}/_{(-20-(-18))} = ^{y}/_{[-27-(-32)]} = ^{1}/_{(16-15)}

^{x}/_{-2 }= ^{y}/_{5} = ^{1}/_{1}

^{x}/_{-2} = 1 and ^{y}/_{5} = 1

x = -2 and y = 5

**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.**

**(ii). A fraction becomes 1/3 when 1 is subtracted from the numerator and it becomes ¼ 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.**

Solution:

(i) Let *x *be the fixed charge of the food and *y *be the charge for food per day.

Given

x + 20y = 1000 ————-(1)

x + 26 y = 1180 ———–(2)

Subtracting equation (1) from equation (2), we get

6y = 180

y = 30

Substituting this value in equation (1),

x + 20 x 30 = 1000

x = 1000 – 600

x = 400

Hence, fixed charge = Rs 400

And charge per day = Rs 30

(ii)Let the fraction be x/y.

Given

^{x-1}/_{y} = ^{1}/_{3} i.e., 3x – y = 3 ———–(1)

^{x}/_{(y+8)} = ^{¼} i.e., 4x – y = 8 ————(2)

Subtracting equation (1) from equation (2),

x = 5 ———(3)

Putting this value in equation (1), we obtain

15 – y = 3

y = 12

Hence the fraction is ^{5}/_{12}.

(iii)Let the number of right answers and wrong answers be *x *and *y *respectively.

Given,

3x – y = 40 ————-(1)

4x – 2y = 50

i.e., 2x – y = 25 ————-(2)

Subtracting equation (2) from equation (1),

*x *= 15 ———–(3)

Substituting the value of x in equation (2),

30 – y = 25

y = 5

Therefore, number of right answers = 15

And number of wrong answers = 5

Total number of questions = 20

(iv)Let the speed of 1st car and 2nd car be *u *km/h and *v *km/h.

Respective speed of both cars while they are travelling in same direction = (u – v) km/h

Respective speed of both cars while they are travelling in opposite directions i.e., travelling towards each other = (u + v) km/h

Given

5 ( u – v) = 100

i.e., u – v = 20 ————-(1)

1(u+v) = 100 ————-(2)

Adding both the equations,

2u = 120

u = 60 km/h ———(3)

Substituting the value of u in equation (2),

*v *= 40 km/h

Hence, speed of one car = 60 km/h and speed of other car = 40 km/h

(v) Let length and breadth of rectangle be *x *unit and *y *unit respectively.

Area = *xy *

Given,

(x – 5)(y+3) = xy – 9

i.e., 3x – 5y – 6 = 0 ———-(1)

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

i.e., 2x + 3y – 61 = 0 ———-(2)

By cross-multiplication method, we obtain

^{x}/_{[305-(-18)]} = ^{y}/_{[-12-(-183)] }= ^{1}/_{[9-(10)]}

^{x}/_{323 }= ^{y}/_{171} = ^{1}/_{19}

x = 17, y = 9

Hence, the length and breadth of the rectangle are 17 units and 9 units respectively.

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