# Pair of Linear Equations in Two VariablesExercise – 3.3

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

(i) x + y = 14, x – y = 4

Solution:

x + y = 14……………..(1)

x – y = 4……………..(2)

From equation (1),

y = 14 – x ……………….(3)

Substituting the value of equation (3) in (2),

x – (14 – x) = 4

x – 14 + x = 4

2x = 14 + 4

2x = 18

x = 9

Substituting the value of x in equation (2),

9 – y = 4

y = 9 – 4

y = 5

Hence, x = 9, y = 5.

Solution:

s - t = 3……………..(1)

s/3 + t/2 = 6 ……………..(2)

From equation (1),

s = 3 + t ……………….(3)

Substituting the value of equation (3) in (2),

5t + 6 = 36

t = 6

Substituting the value of x in equation (3),

s = 3 + 6

s = 9

Hence, s = 9, t = 6.

(iii) 3x - y = 3, 9x – 3y = 9

Solution:

3x - y = 3……………..(1)

9x – 3y = 9……………..(2)

From equation (1),

y = 3x - 3 ……………….(3)

Substituting the value of equation (3) in (2),

9x – 3(3x – 3) = 9

9x – 9x + 9 = 9

9 = 9

Hence, y = 3x – 3 where x can take any value, that is, there are infinitely many solutions.

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

Solution:

0.2x + 0.3y = 1.3……………..(1)

0.4x + 0.5y = 2.3……………..(2)

From equation (1),

Substituting the value of equation (3) in (2),

0.12x + 0.65 – 0.10x = 0.69

0.02x = 0.04

x = 2

Substituting the value of x in equation (2),

0.4×2 + 0.5y = 2.3

0.8 + 0.5y = 2.3

0.5y = 1.5

y = 3

Hence, x = 2, y = 3.

(v) √2x + √3y = 0, √3x - √2y = 0

Solution:

√2x + √3y = 0……………..(1)

√3x - √2y = 0……………..(2)

From equation (1),

Substituting the value of equation (3) in (2),

3x – 2x = 0

x = 0

Substituting the value of x in equation (2),

√3×0 - √2y = 0

y = 0

Hence, x = 0, y = 0.

Solution:

From equation (1),

Substituting the value of equation (3) in (2),

282x + 216 = 780

282x = 564

x = 2

Substituting the value of x in equation (2),

2/3 + y/2 = 13/6

y/2 = 9/6

y = 3

Hence, x = 2, y = 3.

2. Solve 2x + 3y = 11 and 2x – 4y = -24 and hence find the value of 'm' for which y = mx + 3.

Solution:

2x + 3y = 11…………………(1)

2x – 4y = -24………………(2)

From equation (1),

y = (11-2x)/3 ………………(3)

Substituting the value of equation (3) in (2),

2x – 4[(11-2x)/3] = -24

6x – 44 + 8x = -72

14x = -28

x = -2

Substituting the value of x in equation (2),

2(-2) – 4y = -24

-4 – 4y = -24

-4y = -20

y = 5

So, x = -2, y = 5.

Substituting the values ​​of x and y in y = mx + 3,

5 = -2m + 3

m = -1

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

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

Solution:

Let the first number = x

and second number = y

According to Question,

x = 3y …………(1)

The difference of both the numbers is 26, so

x  - y = 26

x = y + 26 …………………..(2)

Substituting the value of x in Equation (1),

y + 26 = 3y

3y – y = 26

2y = 26

y = 13

Substituting the value of y in equation (1),

x = 3×13 = 39

Hence, one number = 39, and the second number = 13.

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

Solution:

Let the greater angle = x

and smaller angle = y

According to Question,

x = 18 + y …………(1)

Both angles are supplementary, so

x  + y = 180 …………………..(2)

[The sum of supplementary angles is 180 degrees.]

Substituting the value of x in equation (2),

y + 18 + y = 180

2y = 180 – 18

2y = 162

y = 81

Substituting the value of y in equation (1),

x = 81 + 18

x = 99

Hence, greater angle = 99, and smaller angle = 81.

(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 cost of one bat = ₹x

and cost of one ball = ₹y

according to the first condition,

7x + 6y = 3800

according to the second condition,

3x + 5y = 1750 ………….(2)

Substituting the value of x in equation (2),

11400 – 18y + 35y = 12250

17y = 850

y = 50

Substituting the value of y in equation (1),

x = 500

Hence cost of one bat = ₹ 500 and cost of one ball = ₹ 50

(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 of taxi = ₹x

And the cost of each additional km = ₹ y

According to the first condition,

x + 10y = 105 …………………(1)si 10*

according to the second condition,

x + 15y = 155……………………(2)

From equation (1),

x + 10y = 105

x = 105 – 10y

Substituting the value of x in equation (2),

x + 15y = 155

(105 – 10y) + 15y = 155

105 + 5y = 155

5y = 50

y = 10v) 50050

2250

..y

Substituting the value of y in equation (1),

x + 10y = 105

x + 10×10 = 105

x = 105 – 100

x = 5

Hence, the fixed charge of the taxi = ₹ 5 and the cost for each additional km = ₹ 10.

for 10 km,

k x + 10y

5 + 10×10 = 105

Similarly charges for 25 km = x + 25y

= 5 + 25×10

= 5 + 250

= ₹255

(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 the denominator it becomes 5/6, Find the fraction.

Solution:

Let numerator = x

and denominator = y

Therefore, rational number = x/y

according to the first condition,

11x + 22 = 9y + 18

according to the second condition,

6x + 18 = 5y + 15

6x – 5y = -3………………(2)

Substituting the value of x in Equation (1),

54y – 24 – 55y = -33

y = 9

Substituting the value of y in equation (2),,

6x – 5×9 = -3

6x = 45 – 3

6x = 42

x = 7

Hence the fractional number = x/y = 7/9.

(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 Jacob's present age = x years

and present age of son = y years

5 years later,

Jacob's age = x + 5 years

Son's age = y + 5 years

According to Question,

x + 5 = 3(y + 5)

x + 5 = 3y + 15

x = 3y + 10………………(1)

5 years ago,

Jacob's age = x - 5 years

Son's age = x - 5 years

According to Question,

x – 5 = 7(y-5)

x – 5 = 7y – 35

x – 7y = -30…………………(2)

Substituting the value of x in equation (2),,

3y + 10 – 7y = -30

-4y = -40

y = 10

Substituting the value of y in equation (1),

x = 3×10 + 10

x = 30 + 10

x = 40

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