Exercise 3.2

(i)

(i) x + y = 14; x – y = 4

Solution

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

x - y = ...(2)

By solving the equation (1)

y = 1 - x ...(3)

Substitute y = 14 - x in equation (2), we get

x - ( - x) = 4

x - 14 = 4

2x = 4 +

2x =

x =

Substituting x = 9 in equation (3), we get

y = - 9

y =

Thus, x = 9, y = 5

(ii)

(ii)s – t = 3; s3 + t2 = 6

Solution

s - t = ...(1)

s3 + t2 = ...(2)

By solving equation (1)

s - t = 3

s = + t ...(3)

Substitute s = 3 + t in equation (2), we get

(3 + t)3 + t2 =

(6 + 2t + 3t)6=

6 + t = 6 ×

5t = - 6

t = 305530

t =

Substituting t = 6 in equation (3), we get

s = 3 +

s =

Thus, s = 9, t = 6

(iii)

(iii) 3x – y = 3; 9x – 3y = 9

Solution

3x - y = ...(1)

9x - 3y = ...(2)

By solving the equation (1)

3x - y = 3

y = - 3 ...(3)

Substitute y = 3x - 3 in equation (2), we get

9x - 3(3x - 3) =

9x - x + = 9

9 =

This shows that the lines are coincident having infinitely many solutions.

x can take any value. i.e. Infinitely many Solutions.

(iv)

(iv) 0.2x + 0.3y = 1.3; 0.4x + 0.5y = 2.3

Solution

0.2x + 0.3y = ...(1)

0.4x + 0.5 y = ...(2)

Multiply both the equations (1) and (2) by 10, to remove the decimal number and making it easier for calculation.

[0.2x + 0.3y = 1.3] ×

⇒ 2x + y = ...(3)

(0.4x + 0.5y = ) × (10)

⇒ 4x + y = ...(4)

By solving the equation (3)

2x + 3y =

3y = 13 -

y = (13 - 2x)3 ...(5)

Substitute y = (13 - 2x)3 in equation (4), we get

4x + [(13 - 2x)3] =

(12x + 65 - 10x)3 =

2x + = 23 ×

2x = 69 -

x = 42 =

Substituting x = 2 in equation (5), we get

y = (13 - 2 × 2)3

y = 93 =

Thus, x = 2, y = 3

(v)

(v) 2x + 3y = 0; 3x - 8y = 0.

Solution

2x + 3y = ...(1)

3x - 8y = ...(2)

By solving equation (1)

2x +3y = 0

3y = - 2x

y = - √2x3 ...(3)

Substitute y = - √2x3 in equation (2), we get

3x - 8- √2x3 =

3x + √16x3 = 0

3√3x + 4x3 = 0

x(33 + ) = 0

x =

Substituting x = 0 in equation (3), we get

y = - √2 × 0 3

y =

Thus, x = 0, y = 0

(vi)

(vi) 3x2 - 5y3 = -2; x3 + y2 = 136

Solution

3x2 - 5y3 = ...(1)

x3 + y2 = 136−136 ...(2)

Multiply both the equations (1) and (2) by 6, to remove the denominatores and making it easier for calculation.

[3x2 - 5y3 = - 2] ×

9x - 10y = ...(3)

[x3 + y2 = 136] ×

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

By solving equation (3)

9x - 10y =

10y = 9x +

y = (9x + 12)10 ...(5)

Substituting y = (9x + 12)10 in equation (4), we get

2x + 3 [(9x + 12)10] =

(20x + 27x + 36)10 =

x = 130 -

x = 9447 =

Substituting x = 2 in equation (5), we get

y = (9 × 2 + 12)10

y = 30103410

y =

Thus, x = 2, y = 3.

(i)

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

Solution

Let the first (larger) number = x

And the second number = y

The difference between two numbers is .

x - y = ....(1)

One number is three times the other

x = ....(2)

Substituting x = 3y in equation (1), we get

3y - = 26

y = 26

y =

Substituting y = 13 in equation (2)

x = 3 ×

x =

The two numbers are 39 and 13.

(ii)

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

Solution

Let the larger angle = x

And the smaller angle = y

Since the angles are supplementary,

x + y = ....(1)

Larger angle exceeds the smaller by

x = y + ....(2)

Substituting x = y + 18 in equation (1), we get y + 18 + y =

2y = 180 -

y = 16221562

y =

Substituting y = 81 in equation (2), we get

x = 81 +

x =

The angles are 99° and 81°.

(iii)

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

Solution

Let the cost of 1 bat = ₹ x

And the cost of 1 ball = ₹ y

7x + 6y = ....(1)

3x + 5y = ....(2)

By solving equation (1),

7x + 6y =

6y = 3800 - x

y = (3800 - 7x)6 ....(3)

Substituting y = (3800 - 7x) 6 in equation (2), we get

3x + 5 [(3800 - 7x)6] =

(18x + 19000 - 35x)6 =

-17x + 19000 = 1750 ×

17x = 19000 - 1050012980

x = 850017750017

x =

Substituting x = 500 in equation (3), we get

y = (3800 - 7 × 500)6

y = 30063406

y =

Cost of 1 bat is ₹ 500

Cost of 1 ball is ₹ 50

(iv)

(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 ₹ 105 and for a journey of 15 km, the charge paid is ₹ 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?

Solution

Let the fixed charge = ₹ x

And charge per km = ₹ y

Charge for a distance of 1015 km

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

Charge for a distance of 1510 km

x + 15y = ....(2)

By solving equation (1),

x +10y =

x = 105 - ....(3)

Substituting x = 105 - 10y in equation (2), we get

105 - 10y + 15y =

y = 155 - 105

y = 505−505

y =

Substituting y = 10 in equation (3)

x = 105 - 10 ×

x = 105 -

x =

Now, charge for a distance of 25 km = x + 25y

= 5 + × 10

= 5 + 250 =

Fixed charge = ₹ 5

Charge per km = ₹ 10

Charge for 25 km = ₹ 255

(v)

(v) A fraction becomes 911 , if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes 56.Find the fraction.

Solution

Let the numerator = x

And denominator = y

Then the fraction is xyyx

When 2 is added to both numerator and denominator,

(x + 2)(y + 2) = 911119

11(x + 2) = ( y + 2)

11x + = 9y +

11x - y + 22 - 18 =

11x - + = 0 ...(1)

When 3 is added to both numerator and denominator

(x + 3)(y + 3) = 56−56

(x + 3) = ( y + 3)

6x + = 5y +

6x - 5y + 18 - = 0

- 5y + = 0 ....(2)

5y = 6x +

y = (6x + 3)5 ....(3)

Substituting y = (6x + 3)5 in equation (1)

11x - [(6x + 3)5] + = 0

[55x - 9(6x + 3) + 20] 5 =

55x - x - 27 + = 0

x - = 0

x =

Substituting x = 7 in equation (3) we get,

y = (6 × 7 + 3)5

y = (42 + 3)5

y = 455−455

y =

Thus, the fraction is 79.

(vi)

(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?

Solution

Let the present age of Jacob = x years

And present age of his son = y years

5 years from now,

Jacob’s age = (x + 5) (y + 5) years

Son’s age = (y + 5)(x + 5) years

(x + 5) = (y + 5)

x + 5 = 3y +

x - 3y + - 15 = 0

x - - = 0 ...(1)

5 years ago, Jacob’s age = (x - 5) years

Son’s age = (y - 5) years

(x - 5) = (y - 5)

x - 5 = 7y -

x - 7y - 5 + = 0

x - 7y + = 0 ...(2)

7y = x +

y = (x + 30)7 ....(3)

Substituting y = (x + 30)7 in equation (1)

x - 3[(x + 30)7] - = 0

[7x - 3(x + 30) - 70]7 = 0

7x - 3x - - = 0

4x - = 0

x = 16044160

x =

Substituting x = 40 in equation (3)

y = (40 + 30)7

y = 707−707

y =

Thus, the present age of Jacob is 40 years and his son is 10 years.