3. Pair of Linear Equations in Two Variables Mathematics Exercise - 3.3 class 10 Maths in English - CBSE Study

Exercise 3.3

---

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

Solution Q1:

(i) x + y = 14  ............ (i)

       x – y = 4    ............ (ii)

Using substitution method–

From Equation (ii)

x – y = 4      

x = 4 + y

Now putting value of x as 4 + y in equation (i)  

x + y = 14

Or  (4 + y) + y = 14

Or  4 + 2y = 14

Or  2y = 14 - 4

Or  2y = 10

Now putting value of y in equation (ii)

   x = 4 + y

Or  x = 4 + 5 = 9

Therefore solution of given pair of linear equation is –

So x = 9, and y = 5 Answer

Solution: (iii) 3x – y = 3....... (i)  

9x – 3y = 9....... (ii)

Using substitution method–

From Equation (i)

3x – y = 3

Or 3x – 3 = y

Or y = 3x – 3

Now putting the value of y in equation (ii)

9x – 3y = 9

Or 9x – 3(3x – 3) = 9

Or 9x – 9x + 9 = 9

Or 9 = 9

Or x = 0 and y = 3x – 3 Answer

Therefore solution of given pair of linear equation is –

x = 0 and y = 3x – 3

Q2. Solve 2x + 3y = 11 and 2x – 4y = – 24 and hence find the value of ‘m’ for which y = mx + 3.

Solution:

2x + 3y = 11 ............. (i)

2x – 4y = –24 ........... (ii)

From equation (i)

2x + 3y = 11

⇒ 2x = 11 – 3y

Now, getting the value of m we put the value of x and y in y = mx + 3

Q3. 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 be x and the second be y.|

According to question,

Situation-I

x – y = 26 ............. (i)

Situation-II

x = 3y     ............. (ii)

Now, putting x = 3y in equation (i)  

   x – y = 26

⇒ 3y – y = 26

⇒ 2y = 26

⇒ y = 13

Now, y = 13 putting in equation (ii)

   x = 3y

⇒ x = 3 × 13

      = 39

Therefore, the first number is 39 and the second number is 13.

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

Solution:

Let the larger angle of two supplementary angles be x.

And the smaller is y.

Therefore, Situation-I

x – y = 18°  ............... (i)

Situation-II

x + y = 180°  ........... (ii)  (The sum of two supplementary angles is 180° )

Now From equation (i)

    x – y = 18°

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

Solution:

Let the larger angle of two supplementary angles be x.

And the smaller is y.

Therefore, Situation-I

x – y = 18°  ............... (i)

Situation-II

x + y = 180°  ........... (ii)  (The sum of two supplementary angles is 180° )

Now From equation (i)

    x – y = 18°

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

Solution:

Let the larger angle of two supplementary angles be x.

And the smaller is y.

Therefore, Situation-I

x – y = 18°  ............... (i)

Situation-II

x + y = 180°  ........... (ii)  (The sum of two supplementary angles is 180° )

Now From equation (i)

    x – y = 18°

⇒ x = 18° + y

Now putting the value of x in equation (ii)

   x + y = 180°

⇒ 18° + y + y = 180°

⇒ 18° + 2y = 180°

⇒ 2y = 180° – 18°

x = 18° + y

Now putting the value of x in equation (ii)

   x + y = 180°

⇒ 18° + y + y = 180°

⇒ 18° + 2y = 180°

⇒ 2y = 180° – 18°

x = 18° + y

Now putting the value of x in equation (ii)

   x + y = 180°

⇒ 18° + y + y = 180°

⇒ 18° + 2y = 180°

⇒ 2y = 180° – 18°

⇒ 2y = 162°

⇒ y = 81°

Putting the value of y in equation (i)

⇒ x = 18° + y

⇒ x = 18° + 81°

⇒ x = 99°

Therefore, the larger angle is 99° and the smaller is 81° 

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

Solution:

Let the cost of a bat be x.  

And the cost of a ball be y.

Situation I

    7 bats + 6 balls = 3800

⇒  7x + 6y = 3800 ......... (i)

Situation II

3 bats +5 balls = 1750

⇒  3x + 5y = 1750 ......... (ii)

From equation (ii)

    3x + 5y = 1750

⇒ 3x = 1750 – 5y 

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

Situation I

x + 10y = 105 ........... (i)

Situation II

x + 15y = 155 ............ (ii)

From equation (i)

x + 10y = 105

⇒  x = 105 – 10y

Now putting the value of x in equation (ii)

x + 15y = 155

⇒(105 – 10y) + 15y = 155

⇒  105 + 5y = 155

⇒  5y = 155 –105

⇒  5y = 50

Now, putting y = 10 in equation (i)  

⇒  x = 105 – 10y

⇒  x =105 –10(10)

⇒  x = 105 –100 = 5

​The charge for 25 km = x + 25y

                  = 5 + 25(10)

                  = 5 + 250

                  = 255 Rupees

(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 be x years.

And his son present age be y years.

Situation I

Five years hence, Jacob’s age = x + 5 years

And his son age will be = y + 5 years

Therefore, x + 5 = 3(y + 5)

⇒  x + 5 = 3y + 15

⇒  x – 3y = 15– 5

⇒  x – 3y = 10  ........... (i)

Situation II

Five years ago, Jacob’s age = x – 5 years

And his son’s age  = y – 5 years   

Then,  x – 5 = 7(y – 5)

⇒  x – 5 = 7y –35

⇒  x – 7y = 5 –35

⇒  x – 7y = – 30  ........... (ii)

From equation (ii)

x – 7y = – 30

Þx = 7y –30

Now, putting the value of x  in equation (i)

x – 3y = 10 

⇒ (7y – 30) – 3y = 10

⇒ 4y = 10 + 30

⇒ 4y = 40

⇒ y = 10

Putting y = 10 in equation (ii)

⇒ x = 7(10) – 30

⇒ x = 70– 30 = 40

Therefore, Jacob’s present age 40 years and his son’s age 10 years.