**Exercise 3.5 Page: 62**

**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 and 3x – 9y – 2 = 0 (ii) 2x + y = 5 and 3x + 2y = 8**

**(iii) 3x – 5y = 20 and 6x – 10y = 40 (iv) x – 3y – 7 = 0 and 3x – 3y – 15 = 0**

**Solutions:**

(i) Given, x – 3y – 3 =0 and 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})

Since, the given set of lines are parallel to each other they will not intersect each other and therefore there is no solution for these equations.

(ii) Given, 2x + y = 5 and 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})

Since they intersect at a unique point these equations will have a unique solution by cross multiplication method:

x/(b_{1}c_{2}-c_{1}b_{2}) = y/(c_{1}a_{2} – c_{2}a=) = 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 and y =1

(iii) Given, 3x – 5y = 20 and 6x – 10y = 40

(a_{1}/a_{2}) = 3/6 = 1/2

(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}

Since the given sets of lines are overlapping each other there will be infinite number of solutions for this pair of equation.

(iv) Given, x – 3y – 7 = 0 and 3x – 3y – 15 = 0

(a_{1}/a_{2}) = 1/3

(b_{1}/b_{2}) = -3/-3 = 1

(c_{1}/c_{2}) = -7/-15

a_{1}/a_{2} ≠ b_{1}/b_{2}

Since this pair of lines are intersecting each other at a unique point, there will be a unique solution.

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.

**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) 3y + 2x -7 =0

(a + b)y + (a-b)y – (3a + 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)

For infinitely many solutions,

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

Thus 2/(a-b) = 7/(3a+b– 2)

6a + 2b – 4 = 7a – 7b

a – 9b = -4 ……………………………….(i)

2/(a-b) = 3/(a+b)

2a + 2b = 3a – 3b

a – 5b = 0 ……………………………….….(ii)

Subtracting (i) from (ii), we get

4b = 4

b =1

Substituting this eq. in (ii), we get

a -5 x 1= 0

a = 5

Thus at a = 5 and b = 1 the given equations will have infinite 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 solutions

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

Therefore, for k = 2 the given pair of linear equations will have no solution.

**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 …………………..(1)

3x + 2y = 4 ……………….….(2)

From equation (2) we get

x = (4 – 2y )/ 3 ……………………. (3)

Using this value in equation 1, we get

8(4-2y)/3 + 5y = 9

32 – 16y +15y = 27

-y = -5

y = 5 ……………………………….(4)

Using this value in equation (2), we get

3x + 10 = 4

x = -2

Thus, x = -2 and y = 5.

Now, Using Cross Multiplication method:

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 and y =5.

Pair of Linear Equations in Two Variables

**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.**

**(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.**

**Solutions:**

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

According to the question,

x + 20y = 1000……………….. (i)

x + 26y = 1180………………..(ii)

Subtracting (i) from (ii) we get

6y = 180

y = Rs.30

Using this value in equation (ii) we get

x = 1180 -26 x 30

x= Rs.400.

Therefore, fixed charges is Rs.400 and charge per day is Rs.30.

**(ii) **L*et* the fraction be x/y.

So, as per the question given,

(x -1)/y = 1/3 => 3x – y = 3…………………(1)

x/(y + 8) = 1/4 => 4x –y =8 ………………..(2)

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

x = 5 ………………………………………….(3)

Using this value in equation (2), we get,

(4×5)– y = 8

y= 12

Therefore, the fraction is 5/12.

(iii) Let the number of right answers is x and number of wrong answers be y

According to the given question;

3x−y=40……..(1)

4x−2y=50

⇒2x−y=25…….(2)

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

x = 15 ….….(3)

Putting this in equation (2), we obtain;

30 – y = 25

Or y = 5

Therefore, number of right answers = 15 and number of wrong answers = 5

Hence, total number of questions = 20

**(iv)** Let x be the number of correct numbers and y be the number of incorrect answers.

According to the question given,

3x – y = 40 …………………………………(i)

4x – 2y = 50

And 2x-y = 25………………………………(ii)

Subtracting equation (ii) from (i), we get

x = 15………………………………………(iii)

Using this in equation (i), we get,

3(15) – 40 = y

y = 5

Therefore, the number of correct answers = 15

And the number of incorrect answers = 5

The total number of questions = 20

(v) Let,

The length of rectangle = x unit

And breadth of the rectangle = y unit

Now, as per the question given,

(x – 5) (y + 3) = xy -9

3x – 5y – 6 = 0……………………………(1)

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

2x + 3y – 61 = 0…………………………..(2)

Using cross multiplication method, we get,

x/(305 +18) = y/(-12+183) = 1/(9+10)

x/323 = y/171 = 1/19

Therefore, x = 17 and y = 9.

Hence, the length of rectangle = 17 units

And breadth of the rectangle = 9 units

Pair of Linear Equations in Two Variables