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