NCERT solutions for class 10 Maths chapter 3 Pair of Linear Equations in Two Variables (Exercise 3.3)

NCERT solutions for class 10 Maths


Chapter 3


Pair of Linear Equations in Two Variables


Exercise 3.3


Ex 3.3 Question 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)

Solution:

(i) x + y = 14      … (i)

x – y = 4        … (ii)

From equation (i),

x = 14 – y      … (iii)

Substituting value of x in equation (ii)

(14 – y) – y = 4

14 – 2y = 4

2y = 10

y = 5

Substituting the value of y in equation (iii),

x = 14 -5

x = 9

∴ x = 9 , y = 5

(ii) s – t = 3 ………(i)

(s/3) + (t/2) = 6 ……… (ii)

From equation (i),

s = t + 3 …..(iii)

Substituting the value of s in equation (ii),

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

⇒ (2(3+t) + 3t )/6 = 6

⇒ (6+2t+3t)/6 = 6

⇒ (6+5t) = 36

⇒5t = 30

⇒t = 6

Substituting the value of t in equation (iii),

s = 3 + 6 = 9

∴ s = 9 and t = 6.

(iii)  3x – y = 3        … (i)

9x – 3y = 9        … (ii)

From (i),

y = 3x – 3        … (iii)

Substituting this value in equation (ii),

9(3+y)/3 – 3y = 9

⇒9 +3y -3y = 9

⇒ 9 = 9

(iv) 0.2x + 0.3y = 1.3 ……….(i)

0.4x + 0.5y = 2.3 ……….(ii)

From  equation (i)

x = (1.3- 0.3y)/0.2 ………(iii)

Substituting the value of x in  equation (ii)

0.4(1.3-0.3y)/0.2 + 0.5y = 2.3

⇒2(1.3 – 0.3y) + 0.5y = 2.3

⇒ 2.6 – 0.6y + 0.5y = 2.3

⇒ 2.6 – 0.1 y = 2.3

⇒ 0.1 y = 0.3

⇒ y = 3

Substituting the value of y in equation (iii),

x = (1.3-0.3(3))/0.2 = (1.3-0.9)/0.2 = 0.4/0.2 = 2

∴ x = 2 and y = 3.

(v) 2 x + √3 y = 0 ……..(i)

√3 x – √8 y = 0 ……….(ii)

From  equation (i)

x = – (√3/√2)y ……….(iii)

Substituting the value of x in the equation (iii)

√3(-√3/√2)y – √8y = 0 ⇒ (-3/√2)y- √8 y = 0

⇒ y = 0

Substituting the value of y in equation (ii),

x = 0

∴ x = 0 and y = 0.

(vi) 3x/2)-(5y/3) = -2 …….(i)

(x/3) + (y/2) = 13/6 (ii)

From equation (i)

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

⇒ x = 2(-6+5y)/9 = (-12+10y)/9 ………………………(iii)

Substituting the value of x in equation (iii),

((-12+10y)/9)/3 + y/2 = 13/6

⇒y/2 = 13/6 –( (-12+10y)/27 ) + y/2 = 13/6

Substituting the value of y in equation (ii),

(3x/2) – 5(3)/3 = -2

⇒ (3x/2) – 5 = -2

⇒ x = 2

∴ x = 2 and y = 3.


Ex 3.3 Question 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…………………………..(i)

2x – 4y = -24………………………… (ii)

From equation (ii),

x = (11-3y)/2 ………………….(iii)

Substituting the value of x in equation (II),

2(11-3y)/2 – 4y = 24

11 – 3y – 4y = -24

-7y = -35

y = 5……….(iv)

Substituting the value of y in equation (iii),

x = (11-3×5)/2 = -4/2 = -2

Hence, x = -2, y = 5

Also,

y = mx + 3

5 = -2m +3

-2m = 2

m = -1

Hence, the value of m is -1.


Ex 3.3 Question 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.

(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 ₹ 3800. Later, she buys 3 bats and 5 balls for ₹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 ₹ 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?

(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?

Solution:

Let the two numbers be x and y respectively, such that y > x.

According to the question,

y = 3x ……………… (i)

y – x = 26 …………..(ii)

Substituting the value of (i) into (ii),

3x – x = 26

x = 13 ……………. (3)

Substituting the value of x  in (i),

y = 39

∴ The numbers are 13 and 39.

(ii) Let the larger angle by xo and smaller angle be yo.

[The sum of two supplementary pair of angles is always 180o.]

According to the question,

x + y = 180o……………. (i)

x – y = 18……………..(ii)

From (i),

x = 180o – y …………. (iii)

Substituting the value of x in (ii),

180– y – y =18o

162o = 2y

y = 81o ………….. (iv)

Substituting the value of y in (iii)

x = 180o – 81o

= 99o

∴ The angles are 99º and 81º.

(iii) Let the cost a bat = x and the cost of a ball = y.

7x + 6y = 3800 ………………. (i)

3x + 5y = 1750 ………………. (ii)

From equation (i),

y = (3800-7x)/6………………..(iii)

Substituting the value of y in (ii).

3x+5(3800-7x)/6 =1750

⇒3x+ 9500/3 – 35x/6 = 1750

⇒3x- 35x/6 = 1750 – 9500/3

⇒(18x-35x)/6 = (5250 – 9500)/3

⇒-17x/6 = -4250/3

⇒-17x = -8500

x = 500 …………….. (iv)

Substituting the value of x in (iii),

y = (3800-7 ×500)/6 = 300/6 = 50

∴The cost of a bat is ₹ 500 and that of a ball is ₹ 50.

(iv) Let the fixed charge = ₹ x and per km charge = ₹ y.

x + 10y = 105 …………….. (i)

x + 15y = 155 …………….. (ii)

From equation (i),

x = 105 – 10y ………………. (iii)

Substituting the value of x in (ii),

105 – 10y + 15y = 155

5y = 50

y = 10 …………….. (iv)

Substituting the value of y in (iii)

x = 105 – 10 × 10 = 5

Fixed charge = ₹ 5

Per km charge= ₹ 10

Charge for 25 km = x + 25 y

⇒5 + 25 (10) = 5 + 250

= ₹ 255

(v) Let the fraction = x/y.

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

11x + 22 = 9y + 18

11x – 9y = -4 …………….. (i)

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

6x + 18 = 5y +15

6x – 5y = -3 ………………. (ii)

From equation (i),

x = (-4+9y)/11 …………….. (iii)

Substituting the value of x in (ii),

6(-4+9y)/11 -5y = -3

-24 + 54y – 55y = -33

-y = -9

y = 9 ………………… (iv)

Substituting the value of y in (iii),

x = (-4+9×9 )/11 = 7

∴The fraction is 7/9.

(vi) Let the age of Jacob = x and his son = y respectively.

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

x – 3y = 10 ……………….. (i)

(x – 5) = 7(y – 5)

x – 7y = -30 ……………………. (ii)

From equation (i),

x = 3y + 10 ……………………. (iii)

Substituting the value of x in (ii),

3y + 10 – 7y = -30

-4y = -40

y = 10 ………………… (iv)

Substituting the value of y in (iii),

x = 3 x 10 + 10 = 40

∴ The present age of Jacob’s and his son is 40 years and 10 years.


 

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