150-NCERT-10-3-Pair of Linear Equations in Two Variables - Ex-3.3

NCERT

10th Mathematics

Exercise 3.3

Topic: 3 Pair of Linear Equations in Two Variables

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EXERCISE 3.3

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

(i) x + y = 14, x – y = 4,   (ii) s – t = 3, (s/3) + (t/2) = 6,   (iii) 3x – y = 3, 9x – 3y = 9

(iv) 0.2x + 0.3y = 1.3, 0.4x + 0.5y = 2.3,     (v) √2 x + √3 y = 0,  √3 x - √8 y = 0

(vi) (3x/2) - (5y/3) = - 2, (x/3) + (y/2) = 13/6

Explanation:

1) Take a simple equation.

2) Simply get the value of x or y with the easy steps (say).

3) Put the obtained value of x in the other equation and get the value of y.

Solution:

(i) x + y = 14, x – y = 4

1) x + y = 14 ------------ equation 1

2) x - y = 4 ------------ equation 2

3) Simplify equation 2,  we get

x - y = 4

y = x - 4 ------------ equation 3

4) Substitute the value of y from equation 3 in equation 1, we get

x + y = 14

x + (x - 4) = 14

2x - 4 = 14

2x = 14 + 4

2x = 18

x = 9 ------------ equation 4

5) Put the value of x from equation 4 in equation 3, we get

y = x - 4

y = 9 - 4 

y = 5

6) The value of x = 9 and the value of y = 5. 

(ii) s – t = 3, (s/3) + (t/2) = 6

1) s - t = 3 ------------ equation 1

2) (s/3) + (t/2) = 6 ------------ equation 2

3) Simplify equation 1, we get

s - t = 3

s = t + 3 ------------ equation 3

4) Substitute the value of s from equation 3 in equation 2, we get

(s/3) + (t/2) = 6

((t + 3)/3) + (t/2) = 6

[(2t + 6) + (3t)]/6 = 6

(5t + 6)/6 = 6

(5t + 6) = 36

5t = 36 - 6

5t = 30 

t = 6 ------------ equation 4

5) Put the value of t from equation 4 in equation 3, we get

s = t + 3

s = 6 + 3

s = 9

6) The value of s = 9 and the value of t = 6. 

(iii) 3x – y = 3, 9x – 3y = 9

1) 3x - y = 3 ------------ equation 1

2) 9x - 3y = 9 ------------ equation 2

3) Simplify equation 1, we get

3x - y = 3

y = 3x + 3 ------------ equation 3

4) Substitute the value of y from equation 3 in equation 2, we get

9x - 3y = 9

9x - 3(3x + 3) = 9

9x - 9x + 9 = 9

9 = 9

5) This shows that the lines are coincident with infinitely many solutions.

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

1) 0.2x + 0.3y = 1.3 ------------ equation 1

2) 0.4x + 0.5y = 2.3 ------------ equation 2

3) Multiplying by 10 to equation 1 and equation 2, we get,

2x + 3y = 13 ------------ equation 3

4x + 5y = 23 ------------ equation 4

4) Simplify equation 3, we get

2x + 3y = 13

2x = 13 - 3y

x = (13 - 3y)/2 ------------ equation 5

5) Substitute the value of x from equation 5 in equation 4, we get

4x + 5y = 23

4[(13 - 3y)/2] + 5y = 23

2(13 - 3y) + 5y = 23

26 - 6y + 5y = 23

26 - y = 23 

y = 26 - 23

y = 3 ------------ equation 6

5) Put the value of y from equation 6 in equation 5, we get

x = (13 - 3y)/2

x = (13 - 3(3))/2

x = (13 - 9)/2

x = 4/2

x = 2 

6) The value of x = 2 and the value of y = 3.

(v) √2 x + √3 y = 0,  √3 x - √8 y = 0

1) √2x + √3y = 0 ------------ equation 1

2) √3x - √8y = 0 ------------ equation 2

3) Simplify equation 2, we get

√3x - √8y = 0

√3x = √8y

x = √(8/3)y ------------ equation 3

4) Substitute the value of x from equation 3 in equation 1, we get

√2x + √3y = 0

√2(8/3)y + √3y = 0

[√2(8/3) + √3]y = 0

[√16/3) + √3]y = 0

y = 0 ------------ equation 4

5) Put the value of y from equation 4 in equation 3, we get

x = √(8/3)y

x = √(8/3)x0 

x = 0

6) The value of x = 0 and the value of y = 0.

(vi) (3x/2) - (5y/3) = - 2, (x/3) + (y/2) = 13/6

1) (3x/2) - (5y/3) = - 2 ------------ equation 1

2) (x/3) + (y/2) = 13/6 ------------ equation 2

3) Simplify equation 1 and equation 2, and we get,

9x - 10y = - 12 ------------ equation 3

2x + 3y = 13 ------------ equation 4

4) Simplify equation 3, we get

9x - 10y = - 12

9x = 10y - 12 

x = (10y - 12)/9 ------------ equation 5

5) Substitute the value of x from equation 5 in equation 4, we get

2x + 3y = 13

2[(10y - 12)/9] + 3y = 13

(20y - 24)/9 + 3y = 13

(20y - 24 + 27y)/9 = 13

(47y - 24)/9 = 13 

47y - 24 = 13 x 9

47y = 24 + 117

47y = 141

y = 141/47 

y = 3 ------------ equation 6

6) Put the value of y from equation 6 in equation 4, we get

2x + 3y = 13 

2x + (3 x 3) = 13

2x = 13 - 9 = 4

x = 4/2

x = 2 

7) The value of x = 2 and the value of y = 3.

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

1) 2x + 3y = 11 ------------ equation 1

2) 2x - 4y = - 24 ------------ equation 2

3) Simplify equation 2, we get

2x - 4y = - 24

2x = 4y - 24

x = (4y - 24)/2 

x = 2y - 12 ------------ equation 3

4) Substitute the value of x from equation 3 in equation 1, and we get

2x + 3y = 11

2(2y - 12) + 3y = 11

4y - 24 + 3y = 11

7y - 24 = 11

7y = 24 + 11

7y = 35

y = 35/7

y = 5 ------------ equation 4

5) Put the value of y = 5, from equation 4 in equation 3, and we get

x = 2y - 12

x = (2 x 5) - 12

x = 10 - 12  

x = - 2

6) The value of x = - 2 and the value of y = 5.

7) Put x = - 2 and y = 5 in the equation y = mx + 3, we get

y = mx + 3

5 = m(- 2) + 3

- 2m = 5 - 3

- 2m = 2

m = 2/(- 2)

m = - 1

8) Therefore, x = - 2, y = 5, and m = - 1. 

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.

(ii) The larger of two supplementary angles exceeds the smaller by 18

degrees. Find them.

(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. 

(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?

(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.

(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?

Explanation:

1) Here, let x and y be two variables.

2) Apply the given conditions and frame the equations.

3) We will get two equations from the above two conditions, then solve these

equations to get the values of x and y. 

Solution:

(i) The difference between two numbers is 26 and one number is three times

the other. Find them.

1) Let the greater number be x, and the smaller number be y.

2) According to the first condition, 

x - y = 26 -------------------------- equation 1.

3) According to the second condition, 

x = 3y -------------------------- equation 2.

4) Put the value of x from equation 2 in equation 1, we get,

x - y = 26

3y - y = 26

2y = 26

y = 26/2

y = 13 -------------------------- equation 3.

5) Put the value of y from equation 3 in equation 2, we get,

x = 3y

x = 3 x 13

x = 39 -------------------------- equation 4.

6) The numbers are 13 and 39. 

(ii) The larger of two supplementary angles exceeds the smaller by 18

degrees. Find them.

1) Let the greater angle be x, and the smaller angle be y.

2) According to the first condition, 

x = y + 18 -------------------------- equation 1.

3) We know that the sum of the supplementary angles is 180 degrees, 

x + y = 180 -------------------------- equation 2.

4) Put the value of x = y + 18 from equation 1 in equation 2, we get,

x + y = 180

(y + 18) + y = 180

2y + 18 = 180

2y = 180 - 18

2y = 162 

y = 81⁰ -------------------------- equation 3.

5) Put the value of y = 81 from equation 3 in equation 1, we get,

x = y + 18

x = 81 + 18

x = 99 -------------------------- equation 4.

6) The angles are 81⁰ and 99⁰. 

(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.

1) Let the cost of the bat be Rs x and the cost of the ball be Rs y.

2) As 7 bats and 6 balls cost Rs 3800, we have,

7x + 6y = 3800

7x = 3800 - 6y

x = (3800 - 6y)/7 ------------ equation 1

3) As 3 bats and 5 balls cost Rs 1750, we have,

3x + 5y = 1750 ------------ equation 2

4) Put the value of x = (3800 - 6y)/7 from equation 1 in equation 2, we get,

3x + 5y = 1750

3(3800 - 6y)/7 + 5y = 1750

[(11400 - 18y) + 35y]/7 = 1750

[(11400 - 18y) + 35y] = 1750 x 7

11400 + 17y = 1750 x 7

11400 + 17y = 12250

17y = 12250 - 11400

17y = 850

y = 850/17

y = 50 -------------------------- equation 3.

5) Put the value of y = 50 from equation 3 in equation 2, we get,

3x + 5y = 1750

3x + 5(50) = 1750

3x + 250 = 1750

3x = 1750 - 250 

3x = 1500

x = 1500/3 

x = 500 -------------------------- equation 4.

6) The cost of the bat is Rs 500 and the cost of the ball is Rs 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 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?

1) Let the fixed charges be Rs x and the charges per km be Rs y.

2) According to the first condition, we have,

x + 10y = 105

x = 105 - 10y ------------ equation 1

3) According to the second condition, we have,

x + 15y = 155 ------------ equation 2

4) Put the value of x = 105 - 10y from equation 1 in equation 2, we get,

x + 15y = 155

(105 - 10y) + 15y = 155

105 + 5y = 155

5y = 155 - 105

5y = 50

y = 10 -------------------------- equation 3.

5) Put the value of y = 10 from equation 3 in equation 1, we get,

x = 105 - 10y

x = 105 - 10(10)

x = 105 - 100

x = 5  -------------------------- equation 4.

6) The charges for 25 km travel will be x + 25y.

7) So the total charges for 25 km travel will be,

x + 25y = 5 + 25(10)

= 5 + 250

= 255

8)  the fixed charges will be Rs 5, the charges per km are Rs 10 and the travel charges for 25 km will be Rs 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.

1) Let the numerator be x and the denominator be y.

2) According to the first condition, we have,

(x + 2)/(y + 2) = 9/11

11x + 22 = 9y +18

9y = 11x + 22 - 18

9y = 11x + 4

y = (11x + 4)/9 ------------ equation 1

3) According to the second condition, we have,

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

6x + 18 = 5y +15

6x - 5y = 15 - 18

6x - 5y = - 3 ------------ equation 2

4) Put the value of y = (11x + 4)/9 from equation 1 in equation 2, we get,

6x - 5y = - 3

6x - 5(11x + 4)/9 = - 3

[54x - 5(11x + 4)]/9 = - 3

[54x - 5(11x + 4)] = - 3 (9)

[54x - 55x - 20] = - 27 

54x - 55x = - 27 + 20

- x = - 7 

x = 7 -------------------------- equation 3.

5) Put the value of x = 7 from equation 3 in equation 1, we get,

y = (11x + 4)/9

y = (11(7) + 4)/9

y = (77 + 4)/9 

y = 81/9

y = 9  -------------------------- equation 4.

6) So, here, the numerator is 7 and the denominator is 9. So the fraction is 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?

1) Let Jacob's present age be x and his son's present age be y.

2) 5 years hence their ages will be (x + 5) and (y + 5).

3) According to the first relation given in the problem,

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

(x + 5) = 3y + 15

x = 3y + 15 - 5

x = 3y + 10  ---------------------- equation 1

4) 5 years ago, their ages were (x - 5) and (y - 5). 

5) According to the relation given in the problem,

(x - 5) = 7(y - 5)

(x - 5) = 7y - 35

x = 7y - 35 + 5

x = 7y - 30  ---------------------- equation 2

6) Put the value of x = 3y + 10 from equation 1 in equation 2, and we get,

x = 7y - 30

3y + 10 = 7y - 30

3y - 7y = - 30 - 10

     - 4y = - 40

       4y = 40 

y = 10 -------------------------- equation 3.

7) Put the value of y = 10 from equation 3 in equation 1, we get,

x = 3y + 10

x = 3(10) + 10

x = 30 + 10 

x = 40  -------------------------- equation 4.

8) So Jacob's present age is 40 years and his son's present age is 10 years.

Click here for ⇨ NCERT-10-3-Pair of Linear Equations in Two Variables - Ex-3.4

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