### NCERT Solutions For Class 10 Maths Chapter – 3 Exercise – 3.3

NCERT Solutions For Class 10 Math Chapter – 3 Linear equation in two variables Exercise  3.3

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

• X + y = 14 ; x – y = 4
• S –  t = 3 ; s/3 + t/2 = 6
• 3x – y = 3 9x – 3y = 9
• 2 x + 0.3 y = 1.3 ; 0.4x + 0.5 y = 2.3
• 2x + 3y = 0 ; 3x – 8y = 0
• 3/2x – 5/3y = -2 ; x/3 + y/2 = 13/6

Solution:-

• X + y  = 14 (i)

X – y = 4 (ii)

From

equation (I) we get

X = 14 – y  (iii)

Putting this values in equation (ii) we get

(14 – y) – y = 4

Y = 5 (iv)N

Putting this in equation  (iii) , we get

X = 9

X = 9 and y = 5

• S – t = 3 (I)

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

From equation (I) we get = t + 3

Putting this value in equation (ii), we get

T+3/3 + t/2 = 6

T = 30/5

S = 9

S = 9 , t = 6

• 3x – y  = 3

9x – 3y = 9

From equation (I) we get

Y = 3x – 3

Putting this value in equation (ii), we get

9x -3(3x – 3) = 9

9 = 9

X = 1 and y = 0

• 2x + 0.3 y = 1.3

0.4x + 0.5y = 2.3

Solving equation (I) we get

0.2x = 1.3 – 0.3y

Dividing by 0.2 we get

X = 1.3/2 – 0.3/2

X = 6.5 – 1.5y

Putting value of x in equation

0.4x + 0.5y  = 2.3

(6.5 – 1.5y) x 0.4x + 0.5y = 2.3

Y = 3

Putting value of y in this equation

X = 6.5 – 1.5y

X = 6.5 – 1.5(3)

X = 2

• √2x + √3y = 0

√3x – √8y = 0

Solution:-

X = -√3y/√2

Putting value of x in this equation

√3(-√3y/√2 ) – √8y = 0

Y = 0

Putting value of y in this equation.

X = 0

X = 0 and y = 0

• 3/2x – 5/3y = -2

X/3 + y/2 = 13/16

From equation (I) we get

9x- 10y = -12

X = -12 + 10y/9

Putting value of x in this equation

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

47y = 17 + 24

Y = 3

Putting value of yin this equation

X = -12 + 10 x 3 /9

= 2

X = 2 and y = 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

Subtracting 3y both sides we get

2x = 11 – 3y

Putting value of x in this equation

2x – 4y = -24

11 – 3y – 4y = -24

7y = 24 – 11

-7y = -35

Y = 5

Putting value of y in this equation

2x  = 11 – 3  x 5

2x = -4

X = -2

Putting value of x in this equation

Y = mx + 3.

5 = -2m + 3.

M = -2/2

M = -1

Q3. From the pair of linear equation for the following problems and find their 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.

Solution:-

X – y = 26

X = 26 + y

Given that one number is three times the other

So x = 3y

Putting the value of x we get

26y = 3y

Y = 13

So value of x =  3y

X = 3 X 1 3

X = 39

Hence the number are 13 and 39.

• The larger of two supplementary angles exceeds the smaller by 18  degrees . Find them.

Solution:-

X + y = 180

X = 180 – y

Difference is 18 degrees so that

X – y = 18

Putting the value of x we get

180 – y – y = 18

-2y = -162

Y = 81

Putting value of y in this equation

X = 180 – 81

X = 99

• 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 ball

Solution:-

7x + 6y = 3800

6y = 3800 – 7x

Divide by 6 we get

Y = (3800 – 7x)/6

3x + 5y = 1750

Putting the value of y

3x + 5 (3800 – 7x)/6) = 1750

Multiplying by 6, we get

18x + 19000 – 35x = 10500

-17x = 10500 – 19000

-17x = -8500

X = 500

Putting value of x in this equation

Y = (3800 – 7 x 500)/6

Y = 50

• 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 charge and the charge per km ? How much does a person have to pay for travelling a distance of 25 km ?

Solution:-

X + 10y = 105

X = 105 – 10y

Given that for a journey of 15 km the charge paid is Rs 155

X + 15y = 155

Putting the value of x in this equation

105 – 10y + 15y = 155

5y = 155 – 105

Y = 10

Putting value of y in this equation

X = 5

Peoples have to pay for travelling a distance of 25 km

X + 25y

= 5 + 250

= 255

A person have to pay Rs 255 for 25 km.

• A fraction becomes 9/11 If 2 is added to both the numerators and the denominators It becomes 5/6 Find the fraction.

Solution:-

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

By cross multiplication we get

11x + 22 = 9y + 18

Subtracting 22 both side , we get

11x = 9y – 4

Dividing by 11 , we get

X = 9y – 4/11

Given that 3 is added to both the numerator and the denominator it becomes 5/6.

If 3 is added to both the numerator and the denominator it becomes 5/6

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

By cross multiplication we get

6x + 18 = 5y + 15

Subtracting the value of x , we get

6(9y – 4)/11 + 18 = 5y + 15

Subtract 18 both sides we get

6(9y  -4)/11 = 5y – 3

Y= 9

Putting this value of y in this equation (I) we get

X = 9y – 4 = 11

X= (81 – 4)/77

X = 7

Hence our fractions is 7/9.

• Five years hence , the age of Jacob will be three times that of his son. Five years ago Jacob age was seven times that of his son. What are their present ages ?

Solution:-

X -3y = 10 (I)

X = 3y + 10

Jacob age was seven times that of his son

X – 5 = 7(y – 5)

Putting the value of x from equation (I) we get

3y + 10 – 5 = 7y – 35

Y =10

Putting value of y in this equation

X = 3 x 10 + 10

X = 40

NCERT Solutions For Class 10 Math Chapter – 3 Linear equation in two variables Exercise  3.3   Q1.  Solve the following…