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

Answer:

(i) x − 3y – 3 = 0

3x − 9y – 2 = 0

Comparing equation x − 3y – 3 = 0 with a1x +b1y + c1 = 0 and 3x − 9y – 2 = 0 with a2x +b2y + c2 = 0,

We get

Here 

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

Comparing equation 2x + y -5= 0 with a1x +b1y + c1 = 0 and 3x + 2y -8= 0 with a2x +b2y + c2 = 0,

We get

Herethis means that there is unique solution for the given equations.

By cross-multiplication method,

⇒

⇒ x = 2 and y = 1

(iii) 3x − 5y = 20

6x − 10y = 40

Comparing equation 3x − 5y = 20 with a1x +b1y + c1 = 0 and 6x − 10y = 40 with a2x +b2y + c2 = 0 ,

We get

Here 

It means lines coincide with each other.

Hence, there are infinite many solutions.

(iv) x − 3y – 7 = 0

3x − 3y – 15 = 0

Comparing equation x − 3y – 7 = 0 with a1x +b1y + c1 = 0 and 3x − 3y – 15 = 0 with a2x +b2y + c2 = 0,

We get

Here this means that we have unique solution for these equations.

By cross-multiplication,

⇒

⇒

⇒ 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

Answer:

(i) Comparing equation 2x + 3y – 7 = 0 with a1x +b1y + c1 = 0 and (a − b) x + (a + b)

y − 3a – b + 2 = 0 with a2x +b2y + c2 = 0

We get andand 

Linear equations have infinite many solutions if 

⇒

⇒

⇒ 2a + 2b = 3a − 3b and 6 − 3b − 9a = −7a − 7b

⇒ a = 5b… (1) and −2a = −4b – 6… (2)

Putting (1) in (2), we get

−2 (5b) = −4b – 6

⇒ −10b + 4b = −6

⇒ −6b = –6 ⇒ b = 1

Putting value of b in (1), we get

a = 5b = 5 (1) = 5

Therefore, a = 5 and b = 1

(ii) Comparing (3x + y – 1 = 0) with a1x +b1y + c1 = 0 and (2k − 1)x + (k − 1)y −2k – 1 = 0) with a2x +b2y + c2 = 0,

We get andand c2 = −2k − 1

Linear equations have no solution if 

⇒

⇒ 

⇒ 3 (k − 1) = 2k – 1

⇒ 3k – 3 = 2k − 1

⇒ k = 2

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

8x + 5y = 9

3x + 2y = 4

Answer.

Substitution Method:

8x + 5y = 9 … (i)

3x + 2y = 4 … (ii)

From equation (ii),we get

x = … (iii)

Substituting this value in equation (i), we obtain

32-16y+15y =27
-y =-5

Y = 5 … (iv)

Substituting this value in equation (ii), we obtain

3x+10=4
x=-2
Hence, x=-2,y=5

Cross multiplication method

8x + 5y = 9 … (1)

3x + 2y = 4 … (2)

⇒

⇒

⇒ 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:

4. (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.

Answer:

(i)Let fixed monthly charge = Rs x and let charge of food for one day = Rs y

According to given conditions,

x + 20y = 1000 … (1),

and x + 26y = 1180 … (2)

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

6y = 180

⇒ y = 30

Putting value of y in (1), we get

x + 20 (30) = 1000

⇒ x = 1000 – 600 = 400

Therefore, fixed monthly charges = Rs 400 and, charges of food for one day = Rs 30

(ii) Let numerator = x and let denominator = y

According to given conditions,

⇒ 3x – 3 = y … (1) 4x = y + 8 … (1)

⇒ 3x – y = 3 … (1) 4x – y = 8 … (2)

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

4x – y − (3x − y) = 8 – 3

⇒ x = 5

Putting value of x in (1), we get

3 (5) – y = 3

⇒ 15 – y = 3

⇒ y = 12

Therefore, numerator = 5 and, denominator = 12

(iii) Let number of correct answers = x and let number of wrong answers = y

According to given conditions,

3x – y = 40 … (1)

And, 4x − 2y = 50 … (2)

From equation (1), y = 3x − 40

Putting this in (2), we get

4x – 2 (3x − 40) = 50

⇒ 4x − 6x + 80 = 50

⇒ −2x = −30

⇒ x = 15

Putting value of x in (1), we get

3 (15) – y = 40

⇒ 45 – y = 40

⇒ y = 45 – 40 = 5

Therefore, number of correct answers = x = 15and number of wrong answers = y = 5

Total questions = x + y = 15 + 5 = 20

(iv)Let speed of car which starts from part A = x km/hr

Let speed of car which starts from part B = y km/hr

According to given conditions,5(x-y) =100(Assuming x > y)

(Assuming x > y)

⇒ 5x − 5y = 100

⇒ x – y = 20 … (1)

And,1(x+y) =100

⇒ x + y = 100 … (2)

Adding (1) and (2), we get

2x = 120

⇒ x = 60 km/hr

Putting value of x in (1), we get

60 – y = 20

⇒ y = 60 – 20 = 40 km/hr

Therefore, speed of car starting from point A = 60 km/hr

And, Speed of car starting from point B = 40 km/hr

(v) Let length of rectangle = x units and Let breadth of rectangle = y units

Area =xy square units. According to given conditions,

xy – 9 = (x − 5) (y + 3)

⇒ xy – 9 = xy + 3x − 5y – 15

⇒ 3x − 5y = 6 … (1)

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

⇒ xy + 67 = xy + 2x + 3y + 6

⇒ 2x + 3y = 61 … (2)

By cross-multiplication method,we obtain,

Therefore, length = 17 units and, breadth = 9 units

CHAPTER NAME	OLD NCERT	NEW NCERT
Real Numbers	EXERCISE 1.1
	EXERCISE 1.2	1.1	CLICK HERE
	EXERCISE 1.3	1.2	CLICK HERE
	EXERCISE 1.4
Polynomials	EXERCISE 2.1	2.1	CLICK HERE
	EXERCISE 2.2	2.2	CLICK HERE
	EXERCISE 2.3
	EXERCISE 2.4
Pair of Linear Equations in Two Variables	EXERCISE 3.1
	EXERCISE 3.2	3.1	CLICK HERE
	EXERCISE 3.3	3.2	CLICK HERE
	EXERCISE 3.4	3.3	CLICK HERE
	EXERCISE 3.5
	EXERCISE 3.6
	EXERCISE 3.7
Quadratic Equations	EXERCISE 4.1	4.1	CLICK HERE
	EXERCISE 4.2	4.2	CLICK HERE
	EXERCISE 4.3
	EXERCISE 4.4	4.3	CLICK HERE
Arithmetic Progressions	EXERCISE 5.1	5.1	CLICK HERE
	EXERCISE 5.2	5.2	CLICK HERE
	EXERCISE 5.3	5.3	CLICK HERE
	EXERCISE 5.4	5.4 (Optional)	CLICK HERE
Triangles	EXERCISE 6.1	6.1	CLICK HERE
	EXERCISE 6.2	6.2	CLICK HERE
	EXERCISE 6.3	6.3	CLICK HERE
	EXERCISE 6.4
	EXERCISE 6.5
	EXERCISE 6.6
Coordinate Geometry	EXERCISE 7.1	7.1	CLICK HERE
	EXERCISE 7.2	7.2	CLICK HERE
	EXERCISE 7.3
	EXERCISE 7.4
Introduction to Trigonometry	EXERCISE 8.1	8.1	CLICK HERE
	EXERCISE 8.2	8.2	CLICK HERE
	EXERCISE 8.3
	EXERCISE 8.4	8.3	CLICK HERE
Some Applications of Trigonometry	EXERCISE 9.1	9.1	CLICK HERE
Circles	EXERCISE 10.1	10.1	CLICK HERE
	EXERCISE 10.2	10.2	CLICK HERE
Construction
Areas Related to Circles	EXERCISE 12.1
	EXERCISE 12.2	11.1	CLICK HERE
	EXERCISE 12.3
Surface Areas and Volumes	EXERCISE 13.1	12.1	CLICK HERE
	EXERCISE 13.2	12.2	CLICK HERE
	EXERCISE 13.3
	EXERCISE 13.4
	EXERCISE 13.5
Statistics	EXERCISE 14.1	13.1	CLICK HERE
	EXERCISE 14.2	13.2	CLICK HERE
	EXERCISE 14.3	13.3	CLICK HERE
	EXERCISE 14.4
Probability	EXERCISE 15.1	14.1	CLICK HERE
	EXERCISE 15.2