Exercise 3.3

(i)

(i) x + y = 5 and 2x – 3y = 4

Solution

Elimination method:

x + y = ....(1)

2x - 3y = ....(2)

Multiplying equation (1) by 2

[x + y = 5] ×

2x + 2y = ....(3)

By subtracting equation (2) from equation (3)

(2x + 2 y) - (2x - 3y) = 10 -

2x + 2y - 2x + 3y =

y = 6

y = 6556

Substituting y = 65 in equation (1)

x + 65 =

x = - 65

x = (25 - 6)5

x = 195519

Substitution method:

x + y = 5 ....(1)

2x - 3y = 4 ....(2)

By solving equation (1)

x + y = 5

y = 5 - ....(3)

Substituting y = 5 - x in equation (2)

2x - (5 - x) =

2x - 15 + 3x =

x = 4 + 15

x = 195517

Substituting x = 19/5 in equation (3)

y = - 195

y = (25 -19)5

y = 6556

Thus, x = 195 and y = 65

(ii)

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

Solution

Elimination method:

3x + 4y = ....(1)

2x - 2y = ....(2)

Multiplying equation (2) by 2

[2x - 2y = 2] ×

4x - = 4 ....(3)

By adding equation (1) and equation (3)

(3x + 4y) + (4x - 4 y) = + 4

3x + 4y + 4x - 4y =

x = 14

x = 147714

x =

Substituting x = 2 in equation (2)

2 × - 2y = 2

- 2y = 2

2y = 4 -

y = 2242

y =

Substitution method:

3x + 4y = ....(1)

2x - = 2 ....(2)

By solving equation (1)

3x + 4y =

4y = 10 - x

y = (10 - 3x)4 ....(3)

Substituting y = (10 - 3x)4 in equation (2)

2x - [(10 - 3x)4] = 2

(4x -10 + 3x)2 =

7x - = 4

7x = 4 +

x = 147714

x =

Substituting x = 2 in equation (3)

y = (10 - 3 × 2)4

y = (10 - 6)4

y = 44

y =

Thus, x = 2 and y = 1

(iii)

(iii) 3x – 5y – 4 = 0 and 9x = 2y + 7

Solution

Elimination method:

3x - 5y - = 0 ....(1)

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

Multiplying equation (1) by 3

[3x - 5y - 4 = 0] ×

9x - 15y - = 0 ....(3)

By solving equation (2)

9x - - 7 = 0 ....(4)

By subtracting equation (4) from equation (3)

(9x - 15y - 12) - (9x - 2y - ) = 0

9x - 15y - 12 - 9x + 2y + = 0

-y - = 0

y = −513513

Substituting y = −513 in equation (2)

9x = (- 5/13) + 7

9x = (-10 + 91)13

x = 8113 × 199

x = 913139

Substitution method:

3x - 5y - = 0 ....(1)

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

By solving equation (1)

3x - 5y - = 0

5y = 3x -

y = (3x - 4)5 ....(3)

Substituting y = (3x - 4)5 in equation (2)

9x = 2[(3x - 4)5] +

9x = (6x - 8 + 35)5

x = 6x +

45x - x = 27

x = 27

x = 2739−3927

x = 913−913 [By simplifying and reducing the fraction]

Substitute x = 913 in equation (3)

y = 3(9/13) - 45

y = (27 - 52)13 × 15

y = - 2513 × 15

y = −513513

Thus, y = −513 and x = 913

(iv)

(iv) x2 + 2y3 = -1 and x - y3 = 3

Solution

Elimination method:

x2 + 2y3 = - ....(1)

x - y3 = ....(2)

Multiplying equation (1) by 6 and equation (2) by 3

[x2 + 2y3 = -1] ×

3x + 4y = ...(3)

[x - y3 = 3] ×

3x - y = ....(4)

By subtracting equation (4) from equation (3)

(3x + 4y) - (3x - y) = - 6 -

3x + 4y - 3x + y = -

y = - 15

y = −155155

y =

Substitute y = - 3 in equation (2)

x - −33 =

x + = 3

x = 3 -

x =

Substitution method:

x2 + 2y3 = ...(1)

x - y3 = ...(2)

By solving equation (2)

x - y3 =

x = y3 +

x = (y + 9) 3 ....(3)

Substituting x = (y + 9)3 in equation (1)

[(y + 9)/3]2 + 2y/3 =

(y + 9 + 4y)6 = - 1

5y + 9 =

5y = - 6 -

y = −155155

y =

Substituting y = -3 in equation (3)

x = (- 3 + 9)3

x = 63

x =

Thus, x = 2 and y = -3

(i)

(i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes 12 if we only add 1 to the denominator. What is the fraction?

Solution

Let the numerator = x

And the denominator = y

Then the fraction = xyyx

When 1 is added to the numerator and 1 is subtracted from the denominator, the fraction reduces to 1 as shown below.

(x + 1)(y - 1) =

x + 1 = y -

x - y + 1 + = 0

x - y + 2 = ...(1)

When 1 is added to the denominator, the fraction becomes 12 as shown below.

x(y + 1) = 12−12

2x = y +

2x - y = 0 ...(2)

By subtracting equation (2) from equation (1)

(x - y + 2) - (2x - y - 1) =

x - y + 2 - 2x + y + = 0

-x + = 0

x =

Substitute x = 3 in equation (1)

3 - y + = 0

y =

Equations are x - y + 2 = 0 and 2x - y -1 = 0 where the numerator of the fraction is x, and denominator is y.

The fraction is 3553.

(ii)

(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

Assuming the present age of Nuri as x years and Sonu as y years, two linear equations can be formed for the given situation.

Let the present age of Nuri = x years

And the present age of Sonu = y years

5 years ago,

Nuri’s age = (x - 5)(y - 5) years

Sonu’s age = (y - 5)(x - 5) years

x - 5 = (y - 5)

x - 5 = 3y -

x - 3y - + 15 = 0

x - 3y + 10 = ...(1)

10 years later,

Nuri’s age = (x + 10)(y + 10) years

Sonu’s age = (y + 10)(x + 10) years

x + 10 = (y + 10)

x + 10 = 2y +

x - 2y + 10 - = 0

x - 2y - = 0 ...(2)

By subtracting equation (2) from equation (1)

(x - 3y +10) - ( x - 2y -) = 0

x - 3y +10 - x + 2y + = 0

- y + = 0

y =

Now, substitute y = 20 in equation (1)

x - 3 × + 10 = 0

x - + 10 = 0

x - = 0

x =

Linear equations are x - 3y +10 = 0 and x - 2 y -10 = 0 where the present age of Nuri is x and Sonu is y

The age of Sonu is 20 years and age of Nuri is 50 years.

(iii)

(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

A two-digit number’s form is 10y + x where y and x are ten’s and one’s digit respectively.

Let the one’s place = x

And the ten’s place = y

Then the number = 10y10x + x

Sum of the digits of the number;

x + y = ...(1)

By reversing the order of the digits, the number = 10x + y

Hence,

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

90y + x = x + 2 y

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

11x - y = 0

11( x - y ) = 0

x - = 0 ...(2)

By subtracting equation (2) from equation (1)

( x + y ) - ( x - 8y ) = - 0

x + y - x + 8y =

y = 9

y =

Substitute y = 1 in equation (1)

x + 1 =

x = 9 -

x =

Equations are x + y = 9 and x - 8y = 0 where y and x are ten’s and one’s digit respectively.

The two-digit number is 18.

(iv)

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

Solution

Assuming the number of notes of ₹ 50 as x and ₹ 100 as y, two linear equations can be formed for the given situation.

Let number of notes of ₹ 50 =

Number of notes of ₹ 100 =

Meena got 25 notes in all;

x + y = ...(1)

Meena withdrew ₹ 2000;

50x + y = 2000

50(x + y) = 2000

x + 2y = 200050

x + 2y = ...(2)

By subtracting equation (1) from equation (2)

(x + 2y) - (x + y) = 40 -

x + 2y - x - y =

y =

Substituting, y = 15 in equation (1)

x + 15 =

x =

Equations are x + y = 25 and x + 2y = 40 where number of ₹ 50 and ₹ 100 notes are x and y respectively.

The number of ₹ 50 notes is 10 and the number of ₹ 100 notes is 15.

(v)

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

Solution

Assuming fixed charges as ₹ x and the additional charges for each extra day as ₹ y, two linear equations can be formed for the given situation.

Let the fixed charge =

Charge per extra day =

Saritha paid ₹ 27 for a book kept for 7 days;

We know that the lending library has a fixed charge for the first three days and an additional charge for each day thereafter.

Therefore,

x + (7 - ) y = 27

x + 4y = ...(1)

Susy paid ₹ 21 for a book kept for 5 days;

x + (5 - 3) y =

x + 2y = ...(2)

By subtracting equation (2) from equation (1)

(x + 4 y) - (x + 2 y) = 27 -

x + 4y - x - 2y =

y = 6

y = 62

y =

Substituting y = 3 in equation (1)

x + 4 × = 27

x + = 27

x = 27 -

x =

Equations are x + 2y = 21 and x + 4y = 27 where fixed charge is ₹ x and charge for each extra day is ₹ y.

The fixed charge is ₹ 15 and the charge for each extra day is ₹ 3.