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

(i) x + y = 5, 2x – 3y = 4

(ii) 3x + 4y = 10, 2x – 2y = 2

(iii) 3x − 5y – 4 = 0, 9x = 2y + 7

(iv) x/2 + 2y/3 = -1, x – y/3 = 3

Answer:

(i) x + y = 5 … (1)

2x – 3y = 4 … (2)

Elimination method:

Multiplying equation (1) by 2, we get equation (3)

2x + 2y = 10 … (3)

2x − 3y = 4 … (2)

Subtracting equation (2) from (3), we get

5y = 6⇒ y = 6/5

Putting value of y in (1), we get

x + 6/5= 5

⇒ x = 5 − 6/5= 19/5

Therefore, x = 19/5 and y = 6/5

Substitution method:

x + y = 5 … (1)

2x − 3y = 4 … (2)

From equation (1), we get,

x = 5 − y

Putting this in equation (2), we get

2 (5 − y) − 3y = 4

⇒ 10 − 2y − 3y = 4

⇒ 5y = 6 ⇒ y = 6/5

Putting value of y in (1), we get

x = 5 − 6/5= 19/5

Therefore, x = 19/5 and y = 6/5

(ii) 3x + 4y = 10… (1)

2x – 2y = 2… (2)

Elimination method:

Multiplying equation (2) by 2, we get (3)

4x − 4y = 4 … (3)

3x + 4y = 10 … (1)

Adding (3) and (1), we get

7x = 14⇒ x = 2

Putting value of x in (1), we get

3 (2) + 4y = 10

⇒ 4y = 10 – 6 = 4

⇒ y = 1

Therefore, x = 2 and y = 1

Substitution method:

3x + 4y = 10… (1)

2x − 2y = 2… (2)

From equation (2), we get

2x = 2 + 2y

⇒ x = 1 + y … (3)

Putting this in equation (1), we get

3 (1 + y) + 4y = 10

⇒ 3 + 3y + 4y = 10

⇒ 7y = 7⇒ y = 1

Putting value of y in (3), we get x = 1 + 1 = 2

Therefore, x = 2 and y = 1

(iii) 3x − 5y – 4 = 0 … (1)

9x = 2y + 7… (2)

Elimination method:

Multiplying (1) by 3, we get (3)

9x − 15y – 12 = 0… (3)

9x − 2y – 7 = 0… (2)

Subtracting (2) from (3), we get

−13y – 5 = 0

⇒ −13y = 5

⇒ y = −5/13

Putting value of y in (1), we get

3x – 5 (−5/13)− 4 = 0

⇒ 3x = 4 − 

⇒ x = 

Therefore, x = and y = 

Substitution Method:

3x − 5y – 4 = 0 … (1)

9x = 2y + 7… (2)

From equation (1), we can say that

3x = 4 + 5y⇒ x = 

Putting this in equation (2), we get

9 − 2y = 7

⇒ 12 + 15y − 2y = 7

⇒ 13y = −5 ⇒ y = 

Putting value of y in (1), we get

3x – 5 = 4

⇒ 3x = 4 − 

⇒ x = 

Therefore, x = and y = 

(iv) … (1)

… (2)

Elimination method:

Multiplying equation (1) by 2, we get

… (3)

substracting equation (2)from (3) we get

Putting value of x in (2), we get

=x-4=-2
=x=2

Substitution method:

… (1)

… (2)

From equation (2), we can say that 

Putting this in equation (1), we get

Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method:

2. (i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes ½ if we only add 1 to the denominator. What is the fraction?

(ii) Five years ago, Nuri was thrice as old as sonu. Ten years later, Nuri will be twice as old as sonu. How old are Nuri and Sonu?

(iii) The sum of the digits of a two–digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.

(iv) Meena went to a bank to withdraw Rs 2000. She asked the cashier togive her Rs 50 and Rs 100 notes only. Meena got 25 notes in all. Find how many notes of Rs 50 and Rs 100 she received.

(v) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid Rs 27 for a book kept for seven days, while Susy paid Rs 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.

Answer:

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

According to given condition, we have

⇒ x + 1 = y – 1 and 2x = y + 1

⇒ x – y = −2 … (1) and 2x – y = 1… (2)

So, we have equations (1) and (2), multiplying equation (1) by 2 we get (3)

2x − 2y = −4… (3)

2x – y = 1… (2)

Subtracting equation (2) from (3), we get

−y = −5⇒ y = 5

Putting value of y in (1), we get

x – 5 = −2⇒ x = −2 + 5 = 3

Therefore, fraction = 

(ii) Let present age of Nuri = x years and let present age of Sonu = y years

5 years ago, age of Nuri = (x – 5) years

5 years ago, age of Sonu = (y – 5) years

According to given condition, we have

(x − 5) = 3 (y − 5)

⇒ x – 5 = 3y – 15

⇒ x − 3y = −10… (1)

10 years later from present, age of Nuri = (x + 10) years

10 years later from present, age of Sonu = (y + 10) years

According to given condition, we have

(x + 10) = 2 (y + 10)

⇒ x + 10 = 2y + 20

⇒ x − 2y = 10 … (2)

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

y = 10 − (−10) = 20 years

Putting value of y in (1), we get

x – 3 (20) = −10

⇒ x – 60 = −10

⇒ x = 50 years

Therefore, present age of Nuri = 50 years and present age of Sonu = 20 years

(iii) Let digit at ten’s place = x and Let digit at one’s place = y

According to given condition, we have

x + y = 9 … (1)

And 9 (10x + y) = 2 (10y + x)

⇒ 90x + 9y = 20y + 2x

⇒ 88x = 11y

⇒ 8x = y

⇒ 8x – y = 0 … (2)

Adding (1) and (2), we get

9x = 9⇒ x = 1

Putting value of x in (1), we get

1 + y = 9

⇒ y = 9 – 1 = 8

Therefore, number = 10x + y = 10 (1) + 8 = 10 + 8 = 18

(iv) Let number of Rs 100 notes = x and let number of Rs 50 notes = y

According to given conditions, we have

x + y = 25 … (1)

and 100x + 50y = 2000

⇒ 2x + y = 40 … (2)

Subtracting (2) from (1), we get

−x = −15⇒ x = 15

Putting value of x in (1), we get

15 + y = 25

⇒ y = 25 – 15 = 10

Therefore, number of Rs 100 notes = 15 and number of Rs 50 notes = 10

(v) Let fixed charge for 3 days = Rs x

Let additional charge for each day thereafter = Rs y

According to given condition, we have

x + 4y = 27 … (1)

x + 2y = 21 … (2)

Subtracting (2) from (1), we get

2y = 6⇒ y = 3

Putting value of y in (1), we get

x + 4 (3) = 27

⇒ x = 27 – 12 = 15

Therefore, fixed charge for 3 days = Rs 15 and additional charge for each day after 3 days = Rs 3

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