1. Solve the following pair of linear equations by the substitution method.

(i) x + y = 14

x – y = 4

(ii) s – t = 3

s/3 + t/2 = 6

(iii) 3x – y = 3

9x − 3y = 9

(iv)0.2x + 0.3y = 1.3

0.4x + 0.5y = 2.3

(v) √2x+ √3y = 0

√3x – √8y = 0

(vi) 3x/2 – 5y/3 = -2

x/3 + y/2 = 13/6

Answer:

(i) x + y = 14 …(1)

x – y = 4 … (2)

x = 4 + y from equation (2)

Putting this in equation (1), we get

4 + y + y = 14

⇒ 2y = 10⇒ y = 5

Putting value of y in equation (1), we get

x + 5 = 14

⇒ x = 14 – 5 = 9

Therefore, x = 9 and y = 5

(ii) s – t = 3 … (1)

…(2)

Using equation (1), we can say that s = 3 + t

Putting this in equation (2), we get

⇒ 2t + 6 + 3t =36

⇒ 5t + 6 = 36

⇒ 5t = 30⇒ t = 6

Putting value of t in equation (1), we get

s – 6 = 3⇒ s = 3 + 6 = 9

Therefore, t = 6 and s = 9

(iii) 3x – y = 3 … (i)

9x − 3y = 9 … (ii)

From equation (i),we get,
y =3x − 3 … (iii)

putting value of y from equation(iii) to equation(ii)
9x-3(3x-3)=9
=9x-9x+9=9
=9=9

This is always true,and pair of these equations have infinite possible solutions.
Therefore one possible solutions is x=1 and y=0

(iv) 0.2x + 0.3y = 1.3 … (1)

0.4x + 0.5y = 2.3 … (2)

Using equation (1), we can say that

0.2x = 1.3 − 0.3y

⇒ x = 

Putting this in equation (2), we get

0.4x(6.5-1.5y)+ 0.5y = 2.3

⇒ 2.6 − 0.6y + 0.5y = 2.3

⇒ −0.1y = −0.3 ⇒ y = 3

Putting value of y in (1), we get

0.2x + 0.3 (3) = 1.3

⇒ 0.2x + 0.9 = 1.3

⇒ 0.2x = 0.4 ⇒ x = 2

Therefore, x = 2 and y = 3

(v) ……….(1)

……….(2)

Using equation (1), we can say that

x = 

Putting this in equation (2), we get

⇒ 

⇒ ⇒ y = 0

Putting value of y in (1), we get x = 0

Therefore, x = 0 and y = 0

(vi)… (1)

… (2)

Using equation (2), we can say that

⇒ x = 

Putting this in equation (1), we get

⇒

⇒

⇒

⇒⇒ y = 3

Putting value of y in equation (2), we get

⇒

⇒ 

⇒ x = 2

Therefore, x = 2 and y = 3

2. Solve 2x + 3y = 11 and 2x − 4y = −24 and hence find the value of ‘m’ for which

y = mx + 3.

Answer:

2x + 3y = 11 … (1)

2x − 4y = −24 … (2)

Using equation (2), we can say that

2x = −24 + 4y

⇒ x = −12 + 2y

Putting this in equation (1), we get

2 (−12 + 2y) + 3y = 11

⇒ −24 + 4y + 3y = 11

⇒ 7y = 35 ⇒ y = 5

Putting value of y in equation (1), we get

2x + 3 (5) = 11

⇒ 2x + 15 = 11

⇒ 2x = 11 – 15 = −4⇒ x = −2

Therefore, x = −2 and y = 5

Putting values of x and y in y = mx + 3, we get

5 = m (−2) + 3

⇒ 5 = −2m + 3

⇒ −2m = 2 ⇒ m = −1

Form a pair of linear equations for the following problems and find their solution by substitution method.

3. (i) The difference between two numbers is 26 and one number is three times the other. Find them.

(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.

(iii)The coach of a cricket team buys 7 bats and 6 balls for Rs 3800. Later, she buys 3 bats and 5 balls for Rs 1750. Find the cost of each bat and each ball.

(iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is Rs 105 and for a journey of 15 km, the charge paid is Rs 155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km?

(v) A fraction becomes 9/11, if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and denominator it becomes 5/6 . Find the fraction.

(vi) Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages?

Answer:

(i) Let first number be x and second number be y.

According to given conditions, we have

x – y = 26 (assuming x > y) … (1)

x = 3y(x > y)… (2)

Putting equation (2) in (1), we get

3y – y = 26

⇒ 2y = 26

⇒ y = 13

Putting value of y in equation (2), we get

x = 3y = 

Therefore, two numbers are 13 and 39.

(ii) Let smaller angle =x and let larger angle =y

According to given conditions, we have

y = x + 18 … (1)

Also, (Sum of supplementary angles) … (2)

Putting (1) in equation (2), we get

x + x + 18 = 180

⇒ 2x = 180 – 18 = 162

⇒ x=162/2
x=81

Putting value of x in equation (1), we get

y = x + 18 = 81 + 18 = 

Therefore, two angles are .

(iii) Let cost of each bat = Rs x and let cost of each ball = Rs y

According to given conditions, we have

7x + 6y = 3800 … (1)

And,3x + 5y = 1750 … (2)

Using equation (1), we can say that 7x+6y=3800
= 6y = 3800-7x

And,

Putting value of y from equation(1) to equation (2)

Therefore, cost of each bat = Rs 500 and cost of each ball = Rs 50

(iv) Let fixed charge = Rs x and let charge for every km = Rs y

According to given conditions, we have

x + 10y = 105… (1)

x + 15y = 155… (2)

Using equation (1), we can say that

x = 105 − 10y

Putting this in equation (2), we get

105 − 10y + 15y = 155

⇒ 5y = 50 ⇒ y = 10

Putting value of y in equation (1), we get

x + 10 (10) = 105

⇒ x = 105 – 100 = 5

Therefore, fixed charge = Rs 5 and charge per km = Rs 10

To travel distance of 25 Km, person will have to pay = Rs (x + 25y)

= Rs (5 + 25 × 10)

= Rs (5 + 250) = Rs 255

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

According to given conditions,
Fraction becomes 9/11 , if 2 is added in both numerator and denominator

=11x+22 =9y+18 (by cross multiplication)
=11x =9y – 4

And, if 3 is added in both numerator and denominator

=6x+18 =5y+15 …..(ii) (by cross multiplication)

putting value of x from equation (i) to equation (ii)

Putting value of y in equation (i), we get

(vi) Let present age of Jacob = x years

Let present age of Jacob’s son = y years

According to given conditions, we have

(x + 5) = 3 (y + 5) … (1)

And, (x − 5) = 7 (y − 5) … (2)

From equation (1), we can say that

x + 5 = 3y + 15

⇒ x = 10 + 3y

Putting value of x in equation (2) we get

10 + 3y – 5 = 7y − 35

⇒ −4y = −40

⇒ y = 10 years

Putting value of y in equation (1), we get

x + 5 = 3 (10 + 5)

⇒ x = 45 – 5 = 40 years

Therefore, present age of Jacob = 40 years and, present age of Jacob’s son = 10 years

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