# Pair of Linear Equations in two Variables Exercise 3.3

Pair of Linear Equations in two Variables Exercise 3.3

Page No: 53

1. Solve the following pair of linear equations by the substitution method.
(i) = 14  ; – = 4

(ii) – = 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) √2x+  √3y =  0 ; √3x -  √8y =  0

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

(i) x + y = 14 ... (i)

x – y = 4 ... (ii)
From equation (i), we get

x = 14 - y ... (iii)
Putting this value in equation (ii), we get

(14 - y) - y = 4

14 - 2y = 4

10 = 2y

= 5 ... (iv)

Putting this in equation (iii), we get

= 9

∴ = 9 and y = 5

(ii) – = 3 ... (i)
s/3 + t/2 = 6 ... (ii)
From equation (i), we gett + 3
Putting this value in equation (ii), we get
t+3/3 + t/2 = 6
2t + 6 + 3t = 36
5t = 30
t = 30/5 ... (iv)
Putting in equation (iii), we obtain
s = 9
∴ s = 9, t = 6

(iii) 3x - = 3 ... (i)
9x - 3y = 9 ... (ii)
From equation (i), we get
y = 3x - 3 ... (iii)
Putting this value in equation (ii), we get
9x - 3(3x - 3) = 9
9x - 9x + 9 = 9
9 = 9
This is always true.
Hence, the given pair of equations has infinite possible solutions and the relation between these variables can be given by
y = 3x - 3
Therefore, one of its possible solutions is x = 1, y = 0.

(iv) 0.2x + 0.3y = 1.3 ... (i)
0.4x + 0.5y = 2.3 ... (ii)
0.2x + 0.3y = 1.3
Solving equation (i), we get
0.2x  = 1.3 – 0.3y
Dividing by 0.2, we get
= 1.3/0.2   - 0.3/0.2
= 6.5 – 1.5 y              …(iii)
Putting the value in equation (ii), we get
0.4x + 0.5y    = 2.3
(6.5 – 1.5y) × 0.4x + 0.5y        =  2.3
2.6 – 0.6y + 0.5y       = 2.3
-0.1y  = 2.3 – 2.6
= -0.3/-0.1
= 3
Putting this value in equation (iii) we get
x = 6.5 – 1.5 y
x = 6.5 – 1.5(3)
x = 6.5  - 4.5
x = 2
∴ = 2 and y = 3 (vi) 3/2x - 5/3y = -2 ... (i)

x/3 + y/2 = 13/6 ... (ii)

From equation (i), we get

9x - 10y = -12

x = -12 + 10y/9 ... (iii)

Putting this value in equation (ii), we get 2. Solve 2x + 3y = 11 and 2x - 4y = - 24 and hence find the value of 'm' for which y =mx + 3.

2x + 3= 11                          ... (i)
Subtracting 3y both side we get
2x = 11 – 3y                   … (ii)
Putting this value in equation second we get
2x – 4y    = –  24                         … (iii)
11- 3y – 4y = -  24
7y = - 24  – 11
-7y  = - 35
y  =  - 35/-7
y  = 5
Putting this value in equation (iii) we get
2x   = 11 – 3 × 5
2x   = 11- 15
2x   = - 4
Dividing by 2 we get
x  = - 2
Putting the value of x and y
y  = mx + 3.
5     = -2m +3
2m  = 3 – 5
m    =  -2/2
m    =  -1

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.

Let larger number = x
Smaller number = y
The difference between two numbers is 26
x – y = 26
x = 26 + y
Given that one number is three times the other
So = 3y
Putting the value of x we get
26= 3y
-2y = - 2 6
y= 13
So value of x  = 3y
Putting value of y, we get
= 3 × 13 = 39
Hence the numbers are 13 and 39.

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

Let first angle = x
And second number = y
As both angles are supplementary so that sum will 180
x + y = 180
x = 180 - y                     ... (i)
Difference is 18 degree so that
x – y = 18
Putting the value of we get
180 – y – y= 18
- 2= -162
= -162/-2
= 81
Putting the value back in equation (i), we get
= 180 – 81 = 99Hence, the angles are 99º and 81º.

(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.

Let cost of each bat = Rs x
Cost of each ball = Rs y

Given that coach of a cricket team buys 7 bats and 6 balls for Rs 3800.
7x + 6= 3800
6= 3800 – 7x
Dividing by 6, we get
= (3800 – 7x)/6         … (i)

Given that she buys 3 bats and 5 balls for Rs 1750 later.
3x + 5= 1750
Putting the value of y
3x + 5 ((3800 – 7x)/6) = 1750
Multiplying by 6, we get
18x  + 19000 – 35= 10500
-17x =10500 - 19000
-17x = -8500
x = - 8500/- 17
=  500
Putting this value in equation (i) we get
= ( 3800 – 7 × 500)/6
= 300/6
= 50

Hence cost of each bat = Rs 500 and cost of each balls = 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 traveling a distance of 25 km?

Let the fixed charge  for taxi = Rs x
And variable cost per km = Rs y
Total cost = fixed charge + variable charge
Given that for a distance of 10 km, the charge paid is Rs 105
x + 10= 105 … (i)
= 105 – 10y
Given that for a journey of 15 km, the charge paid is Rs 155
x + 15= 155
Putting the value of x we get
105 – 10y + 15y= 155
5y= 155 – 105
5y= 50
Dividing by 5, we get
y = 50/5 = 10
Putting this value in equation (i) we get
x = 105 – 10 × 10
x  = 5
People have to pay for traveling a distance of 25 km
x + 25y
= 5 + 25 × 10
= 5 + 250
=255

A person have to pay Rs 255 for 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.

Let Numerator = x
Denominator = y
Fraction will = x/y
A fraction becomes 9/11, if 2 is added to both the numerator and the denominator
(x + 2)/y+2 = 9/11

By Cross multiplication, we get
11x + 22 = 9y + 18
Subtracting 22 both side, we get
11= 9y – 4
Dividing by 11, we get
= 9y – 4/11                          … (i)
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)/+3 = 5/6 … (ii)
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 side we get
6(9y – 4 )/11 = 5y  - 3
54   – 24=  55-  33
-= -9
y= 9
Putting this value of y in equation (i), we get
= 9y – 4
11                                … (i)
x = (81 – 4)/77
= 77/11
= 7

Hence our 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?

Let present age of Jacob = x year
And present Age of his son is = y year
Five years hence,
Age of Jacob will = x + 5 year
Age of his son will = y + 5year
Given that the age of Jacob will be three times that of his son
x + 5 = 3(+ 5)
Adding 5 both side, we get
= 3y + 15 - 5
= 3y + 10                    … (i)
Five years ago,
Age of Jacob will = x - 5 year
Age of his son will = y - 5 year
Jacob’s age was seven times that of his son
x – 5 = 7(y -5)
Putting the value of x from equation (i)  we get
3+ 10 – 5 = 7y – 35
3y + 5 = 7y – 35
3y – 7= -35 – 5
-4y =  - 40
= - 40/- 4
= 10 year
Putting the value of y in equation first we get
x = 3 × 10 + 10
= 40 years
Hence, Present age of Jacob = 40 years and present age of his son = 10 years.