Question 1.

Solve the following pair of equations by substitution method.

- x + y = 14, x – y = 4
- s – t = 3, \(\frac{s}{3}+\frac{t}{2}=6\)
- 3x – y = 3, 9x – 3y = 9
- 0.2x + 0.3y = 1.3, 0.4x + 0.5y = 2.3
- \(\sqrt{2} x+\sqrt{3} y=0\), \(\sqrt{3} x-\sqrt{8} y=0\)
- \(\frac{3 x}{2}-\frac{5 y}{3}=-2\), \(\frac{x}{3}+\frac{y}{2}=\frac{13}{6}\)

Solution:

1. x + y = 14

x – y = 4

The given pair of linear equation is

x + y = 14 …..(1)

x – y = 4 …..(2)

From equation (1),

y = 14 – x …(3)

Substituting this value of y in equation (2), we get

x – (14 – x) = 4

â‡’ x – 14 + x = 4

â‡’ 2x – 14 = 4

â‡’ 2x = 4 + 14

â‡’ 2x = 18

â‡’ x = \(\frac{18}{2}\) = 9

Substituting this value of x in equation (3), we get

y = 14 – 9 = 5

Therefore the solution is

x = 9, y = 5

Verification:

Substituting x = 9 and y = 5, we find that both the equations (1) and (2) are satisfied as shown below:

x + y = 9 + 5 = 14

x – y = 9 – 5 = 4

This verifies the Solution.

2. s – t = 3

\(\frac{s}{3}+\frac{t}{2}=6\)

The given pair of linear equations is

s – t = 3 …(1)

\(\frac{s}{3}+\frac{t}{2}=6\) …(2)

From equation (1),

s = t + 3 …(3)

Substituting this value of s in equation (2), we get

â‡’ 2 (t + 3) + 3t = 36

â‡’ 2t + 6 + 3t = 36

â‡’ 5t + 6 = 36

â‡’ 5t = 36 – 6

â‡’ 5t = 30

â‡’ t = \(\frac{30}{5}\) = 6

Substituting this value of t in equation (3), we get

s = 6 + 3 = 9

Therefore, the solution is

s = 9, t = 6

Verification:

Substituting s = 9 and t = 6, we find both equation (1) and (2) are satisfied as shown below:

s – t = 9 – 6 = 3

\(\frac{s}{3}+\frac{t}{2}=\frac{9}{3}+\frac{6}{2}\)

= 3 + 3 = 6

This verifies the solution.

3. 3x – y = 3

9x – 3y = 9

The given pair of linear equations is

3x – y = 3 …(1)

9x – 3y = 9 …(2)

From equation (1),

y = 3x – 3 …(3)

Substituting this value of y in equation (2), we get

9x – 3(3x – 3) = 9

â‡’ 9x – 9x + 9 = 0

â‡’ 9 = 9

Which is true. Therefore, equations (1) and (2) have infinitely many solutions.

4. 0.2x + 0.3y = 1.3

0.4x + 0.5y = 2.3

The given system of linear equation is

0.2x + 0.3y = 1.3 …(1)

0.4x + 0.5y = 2.3 …(2)

From equation (1),

0.3y = 1.3 – 0.2x

â‡’ y = \(\frac{1.3-0.2 x}{0.3}\) …(3)

Substituting this value of y in equation (2), we get

0.4x + 0.5\(\left(\frac{1.3-0.2 x}{0.3}\right)\) = 2.3

â‡’ 0.12x + 0.65 – 0.1x = 0.69

â‡’ 0.12x – 0.1x = 0.69 – 0.65

â‡’ 0.02x = 0.04

â‡’ x = \(\frac{0.04}{0.02}\) = 2

Substituting this value of x in equation (3), we get

Therefore, the solution is x = 2, y = 3.

Verification:

Substituting x = 2 and y = 3, we find that both the equations (1) and (2) are satisfied as shown below:

0.2x + 0.3y = (0.2) (2) + (0.3) (3)

= 0.4 + 0.9= 1.3

0.4x + 0.5y = (0.4) (2) + (0.5) (3)

= 0.8 + 1.5 = 2.23

This verifies the solution.

5. \(\sqrt{2} x+\sqrt{3} y=0\)

\(\sqrt{3} x-\sqrt{8} y=0\)

The given pair of linear equations is

\(\sqrt{2} x+\sqrt{3} y=0\) …..(1)

\(\sqrt{3} x-\sqrt{8} y=0\) …..(2)

From equation (2),

Substituting this value of* in equation (1), we get

Substituting this value of y in equation (3), we get

x = \(\frac{\sqrt{8}}{\sqrt{3}}(0)\) = 0

Therefore, the solution is

x = 0, y = 0.

Verification:

Substituting x = 0 and y = 0, we find that both the equations (1) and (2) are satisfied as shown below:

This verifies the solution.

6. \(\frac{3 x}{2}-\frac{5 y}{3}=-2\)

\(\frac{x}{3}+\frac{y}{2}=\frac{13}{6}\)

The given system of linear equation is

Substituting the value of x in equation (4), we get

2\(\left(\frac{10 y-12}{9}\right)\) + 3y = 13

20y – 24 + 27y = 117

47y = 117 + 24

y = \(\frac{141}{47}\)

y = 3

Substituting the value of y in equation (4), we get

2x + 3 Ã— 3 = 13

2x + 9 = 13

2x = 13 – 9

x = \(\frac{4}{2}\) = 2

Therefore, the solution is x = 2, y = 3

Verification:

Substituting x = 2 and y = 3, we find that both the equation (1) and (2) are satisfied as shown below:

This verifies the solution.

Question 2.

Solve 2x + 3y = 11 and 2x – 4y = -24 and hence find the value of â€˜mâ€™ for which y = mx + 3.

Solution:

The given pair of linear equations is

2x + 3y = 11 …….(1)

2x – 4y = -24 …….(2)

From equation (1),

3y = 11 – 2x

â‡’ y = \(\frac{11-2 x}{3}\) …….(3)

Substituting this value of y in equation (2), we get

2x – 4\(\left(\frac{11-2 x}{3}\right)\) = -24

â‡’ 6x – 44 + 8x = -72

â‡’ 14x – 44 = -72

â‡’ 14x = 44 – 72

â‡’ 14x = -28

â‡’ x = \(-\frac{28}{14}\) = -2

Substituting this value of x in equation (3), we get

Verification:

Substituting x = -2 and y = 5, we find that both the equations (1) and (2) are satisfied as shown below:

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

= -4 + 15 = 11

2x – 4y = 2(-2) – 4(5)

= -4 – 20 = -24

This verifies the solution.

Now, y = mx + 3

â‡’ 5 = m(-2) + 3

â‡’ -2m = 5 – 3

â‡’ -2m = 2

â‡’ m = \(\frac{2}{-2}\) = -1

Question 3.

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

- The difference between two numbers is 26 and one number is three times the other. Find them.
- The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.
- The coach of a cricket team buys 7 bats and 6 balls for â‚¹ 3800. Later, she buys 3 bats and 5 balls for â‚¹ 1750. Find the cost of each bat and each ball.
- 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 â‚¹ 105 and for a journey of 15 km, the charge paid is â‚¹ 155. What are the fixed charges and the charge per kilometre? How much does a person have to pay for travelling a distance of 25 km?
- A fraction becomes \(\frac{9}{11}\), if 2 is added to both the numerator and the denominator. If 3 is added to both the numerator and the denominator it becomes \(\frac{5}{6}\), find the fraction. [CBSE]
- Five years hence, the age of Jacob will be three times that of his son, five years ago, Jacobâ€™s age was seven times that of his son. What are their present age?

Solution:

1. Let the two numbers be x and y(x > y), then according to the question, the pair of linear equations formed is

x – y = 26 …(1)

x = 3y …(2)

Substituting the value of x from equation (2) in equation (1), we get

3y – y = 26

â‡’ 2y = 26

â‡’ y = \(\frac{26}{2}\)

â‡’ y = 13

Substituting this value ofy in equation (2), we get

x = 3(13) = 39

Hence, the required number are 39 and 13.

Verification:

Substituting x – 39 and y = 13, we find that both the equations (1) and (2) are satisfied as shown below:

x – y = 39 – 13 = 26

3y = 3(13) = 39 = x

This verifies the solution.

2. Let the larger and the smaller of two supplementary angles be xÂ° and yÂ° respectively.

Then, according to the question.

The pair of linear equations formed is

xÂ° = yÂ° + 18Â° …(1)

xÂ° + yÂ° = 180Â° …(2)

| âˆµ The two angles are supplementary

Substituting the value of xÂ° from equation (1) in equation (2), we get

yÂ° + 18Â°+ yÂ° = 180Â°

â‡’ 2yÂ° + 18Â° = 180Â°

â‡’ 2yÂ° = 180Â° – 18Â°

â‡’ 2yÂ° = 162Â°

â‡’ yÂ° = \(\frac{162Â°}{2}\) = 81Â°

Substituting this value ofyÂ° in equation (1), we get

xÂ° = 81Â° + 18Â° = 99Â°

Hence, the larger and the smaller of the two supplementary angles are 99Â° and 81Â° respectively.

Verification:

Substituting xÂ° = 99Â° and yÂ° = 81Â°, we find that both the equations (1) and (2) are satisfied as shown below:

yÂ° + 18Â° = 81Â°+ 18Â° = 99Â° = xÂ°

xÂ° + yÂ° = 99Â°+ 81Â° = 180Â°

This verifies the solution.

3. Let the cost of each bat and each ball be â‚¹ x and â‚¹ y respectively.

Then, according to the question, the pair of linear equations formed is

7x + 6y = 3800 …(1)

3x + 5y = 1750 …(2)

From equation (2),

5y = 1750 – 3x

y = \(\frac{1750-3 x}{5}\) …(2)

Substituting this value of y in equation (1), we get

7x + 6\(\left(\frac{1750-3 x}{5}\right)\) = 3800

â‡’ 35x + 10500 – 18x = 19000

â‡’ 17x + 10500 = 19000

â‡’ 17x = 19000 – 10500

â‡’ 17x = 8500

â‡’ x = \(\frac{8500}{17}\) = 500

Substituting this value of x in equation (3), we get

Hence, the cost of each bat is â‚¹ 500 and each that of ball is â‚¹ 50 respectively.

Verification:

Substituting x = 500 and y = 50, we find that both the equations (1) and (2) are satisfied as shown below:

7x + 6y = 7(500) + 6(50)

= 3500 + 300 = 3800

3x + 5y = 3(500) + 5(50)

= 1500 + 250 = 1750

This verifies the solution.

(iv) Let the fixed charges be â‚¹ x and the charge per kilometre be â‚¹ y.

Then, according to the question, the pair of

linear equations formed is

x + 10y = 105 …(1)

x + 15y = 155 …(2)

From equation (1)

x = 105 – 10y …(3)

Substituting this value of x in equation (2), we get

105 – 10y + 15y = 155

â‡’ 5 + 5y = 155

â‡’ 5y = 155 – 105

â‡’ 5y = 50

â‡’ y = \(\frac{50}{5}\) = 10

Substituting this value of y in equation (3), we get

x = 105 – 10(10)

= 105 – 100 = 5

Hence, the fixed charges are â‚¹ 5 and the charge per kilometre is â‚¹ 10.

Verification:

Substituting x = 5 and y = 10, we find that both the equations (1) and (2) are satisfied as shown below:

x + 10y = 5 + 10(10)

= 5 + 100 = 105

x + 15y = 5 + 15(10)

= 5 + 150 = 155

This verifies the solution.

Again, for travelling a distance of 25 km, a person will have to pay

= 5 + 10(25)

= 5 + 250 = â‚¹ 255

5. Let the fraction be \(\frac{x}{y}\).

Then, according to the question, the pair of linear equations formed is

\(\frac{x+2}{y+2}=\frac{9}{11}\)

â‡’ 11(x + 2) = 9(y + 2)

â‡’ 11x + 22 = 9y + 18

â‡’ 11x = 9y + 18 – 22

â‡’ 11x – 9y + 4 = 0 …(1)

and \(\frac{x+3}{y+3}=\frac{5}{6}\)

â‡’ 6(x + 3) = 5(y + 3)

â‡’ 6x + 18 = 5y + 15

â‡’ 6x – 5y = -3 …(2)

From equation (1),

â‡’ x = \(\frac{9 y-4}{11}\) …(3)

Substituting this value of x in equation (2), we get

â‡’ \(6\left(\frac{9 y-4}{11}\right)\) – 5y = -3

â‡’ 6(9y – 4) – 55y = -33

â‡’ 54y – 24 – 55y = -33

-y + 9 = 0

y = 9

Substituting this value of y in equation (3), we get

Hence, the required fraction is \(\frac{7}{9}\).

Verification:

Substituting x = 7 and y = 9, we find that both the equations (1) and (2) are satisfied as shown below:

This verified the solution.

6. Let the present ages of Jacob and his son be x years and y years respectively.

Then, according to the question, the pair of linear equations formed is

x + 5 = 3(y + 5)

x – 5 = 7(y – 5)

â‡’ x – 3y = 10 …(1)

x – 7y = -30 …(2)

From equation (1),

x = 3y + 10 …(3)

Substituting this value of x in equation (3), we get

3y + 10 – 7y = -30

-4y = -40

y = 10

Substituting y = 10 in equation (3), we get

x = 3(10) + 10

= 30 + 10 = 40

Hence, the present age of Jacob and his son are 40 years and 10 years respectively.

Verification:

Substituting x = 40 and y = 10, we find that both the equations (1) and (2) are satisfied as shown below:

x – 3y = 40 – 3(10)

= 40 – 30 = 10

x – 7y = 40 – 7(10)

= 40 – 70 = -30

This verifies the solution.