Q1. Solve the following pair of linear equations by the elimination method and the substitution method:

(i) x +y = 5 and 2x -3y = 4

(ii) 3x + 4y =10 and 2x -2y = 2

(iii) 3x -5y -4 = 0 and 9x =2y +7

(iv) x/2  +2y/3 = -1 and x – y/3 = 3

Solution. The given pair of linear  equations is

x +y=5 ……(i)

2x -3y = 4…(ii)

Multiplying the equation (i) by 2,we get equation (iii)

2x +2y=10….(iii)

Substracting equation (iii) from equation (ii)

-5y = -6

y = 6/5

Putting the value y = 6/5 in equation (i)

x + 6/5 = 5

x = 5 – 6/5 =(25 – 6)/5 = 19/5

Hence value of x is 6/5 and value of y is 19/5

(ii) The given equations are

3x + 4y =10…..(i) and 2x -2y = 2…..(ii)

Multiplying equation (i) by 2 and equation (ii) by 3,we get the equation (iii) and equation (iv)

6x + 8y =20…..(iii) and 6x -6y =6….(iv)

Substracting equation (iv) from equation (iii),we get

14y = 14

y = 1

Putting the value of y in equation (i)

3x +4×1=10

3x + 4 =10

3x = 6

x = 2

Hence the value of x is 2 and value of y is 1

(iii) The given equations are

3x -5y -4 = 0 (i) and 9x =2y +7(i.e 9x -2y -7=0)…..(ii)

Multiplying equation (i) by 3,we get equation (iii)

9x -15y -12 = 0 (iii)

Substracting equation (iii) from equation (ii)

13y+5 =0

y = -5/13

Putting this value of y in equation (i)

3x -5(-5/13) -4 = 0

3x +25/13 = 4

3x = 4 -(25/13)

3x = (52-25)/13 = 27/13

x = 9/13

Hence the value of x is 9/13 and value of y is -5/13

(iv) The given equations are x/2  +2y/3 = -1 and x – y/3 = 3

Arranging both equations in standard form

3x + 4y = -6……(i) and 3x -y = 9…..(ii)

Subtracting equation (ii) from equation (i)

5y =-15

y = -3

Putting the value of y in equation (i)

3x + 4×-3 =-6

3x -12 = -6

3x = -6 +12 = 6

x = 2

Hence the value of x is 2 and value of y is -3

Similarly, you can do other parts of the question and study exercise 3.3 for the substitution method.

Q2. Form the pair of linear equations in the following problems, and find their solutions(if they exist) by elimination method:

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

Solution. Let the numerator is x and the denominator is y

According to the first condition

(x +1)/(y -1) = 1

x + 1 = y -1

x -y = -2……(i)

According to the second condition

x/(y+1) = 1/2

2x = y +1

2x -y = 1……(ii)

Multiplying the equation (i) by 2

2x -2y = -4….(iii)

Subtracting the equation (iii) from the equation (ii)

y =5

Putting the value of y  in equation (i)

x -5 =-2

x =-2 +5 =3

Hence the required fraction is 3/5

(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?

Solution. Let the age of Nuri is x and the age of Sonu is y

According to first condition

5 years back the age of Nuri = 3× 5 years back the age of Sonu

x -5 = 3(y -5)

x -5 = 3y -15

x – 3y = -10….(i)

According to second condition

10 years later the age of Nuri = 2× 10 years later the age of Sonu

x +10 = 2(y +10)

x +10= 2y +20

x +10= 2y +20

x -2y =10…..(ii)

Substracting the equation (ii) from the equation (i)

-y = -20

y = 20

Putting the value of y in equation (i)

x -3×20 = -10

x -60 = -10

x = -10 +60 = 50

Hence the age of Nuri is 50 years and the age of Sonu is 20 years

(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.

Solution. Let once of a two-digit number is x and tense of a two-digit number is y

∴ The number = 10y +x

According to the first condition

The sum of the digits = 9

x + y =9……(i)

9× The number = 2× The number obtained by reversing its digits

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

90y + 9x = 20x + 2y

90y -2y +9x -20x =0

88y -11x =0

8y -x =0…..(ii)

Substituting x = 8y from equation (ii) to equation (i)

8y +y = 9

9y =9

y =1

Putting the value of y in equation (i)

x + 1 = 9

x = 8

Hence the required number is 10y +x =10×1+ 8 = 18

(iv) Meena went to a bank to withdraw Rs.2000. She asked the cashier to give 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.

Solution. Let the notes of Rs 50 are x and of Rs 100 are y

According to first condition

The notes of Rs 50 ×The number of Rs 50 notes +The notes of Rs 100 ×The number of Rs 100 notes = Rs 2000

50x + 100y =2000

x + 2y = 40…..(i)

According to second condition

The number of Rs 50 notes +The number of  Rs 100 notes =25

x + y =25…….(ii)

Subtracting equation (ii) from equation (i)

y = 15

Putting the value y =15 in equation (i)

x + 2×15 =40

x + 30 =40

x = 40 -30 = 10

Hence the Rs 50 notes are 10 and the Rs 100 notes are 15

(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.

Solution. Let a lending library has a fixed charge of x for 3 days and charge y for each extra day.

According to the first condition, Saritha paid Rs 27 for a book kept for  7 days

Total charge = 3 days fixed charge + charge for (7 -3) days

27 = x + 4y

x + 4y =27….(i)

According to the second condition, Susy paid Rs.21 for the book she kept for five days

Total charge = 3 days fixed charge + charge for (5 -3) days

21 = x + 2y

x + 2y =21….(ii)

Subtracting equation (ii) from equation (i)

2y =6

y =3

Putting the value of y in equation (i)

x + 4×3 = 27

x +12 = 27

x = 15

Therefore the fixed charge for 3 days is Rs 15 and the charge for an extra day is Rs 3