3. Pair of Linear Equations in Two Variables Class 10 Maths – SEBA Assam Board Textbook Solutions
SEBA Chapter 3. Pair of Linear Equations in Two Variables– Class 10 Mathematics Textbook Solutions | HSLC Assam Board
Chapter 3. Pair of Linear Equations in Two Variables
EXERCISE 3.3
1. Solve the following pair of linear equations by the substitution method .
(i) 
Solution : We have , 
And 
[ From 
]
Putting the value of 
in eq.
, we get 
Therefore, the solution is 
and 
.
(ii) 
Solution : We have , 
And 
[ From ]
Putting the value of 
in equation , we get 
Therefore, the solution is 
and 
(iii) 
Solution : We have , 
And 
[ From ]
Which is a false statement . Therefore , the equation do not have a common solution . So, the two rails will not cross each other .
(iv) 
Solution : We have , 
And 
[ From ]
Putting the value of in equation , we get 
Therefore, the solution is 
and 
.
(v) 
Solution : We have, 
And 
[ From ]
Putting the value of in equation
, we have 
Therefore, the solution is 
and 
.
(vi) 
Solution : We have , 
And 
[ From (i) ] 
Putting the value of in equation , we have 
Therefore, the solution is 
and 
2. Solve 
and
and hence find the value of 
for which 
.
Solution : We have , 
and 
Putting in equation , we get 
Therefore, the value of m is – 1 .
3. From 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 .
Solution : Let 
and 
be the two numbers .
A/Q , 
And 
[From (i)]
Putting the value of in equation , we have 
Therefore , the two numbers are 39 and 13 .
(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees . Find them .
Solution : Let 
and be the two angles .
A/Q , 
And 
[ From ]
Putting the value of in equation , we get 
Therefore, the angles are 
and 
respectively .
(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 .
Solution : Let and 
be the cost of one bat and one ball respectively .
A/Q , 
and 
[ From
]
Putting the value of
in equation , we get 
Therefore , the cost of one bat and one ball are Rs 500 and Rs 50 respectively .
(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 ?
Solution : Let, and be the fixed charge and the charge per km respectively .
A/Q , 
And 
[ From ]
Putting 
in equation 
we get 
Therefore, a person have to pay for travelling a distance of 25 km 
Rs. ( 
) Rs.( 
) Rs.( 
) Rs. 
(v) 29. A fraction becomes 
, if 2 is added to both the numerator and the denominator . If 3 is added to both the numerator and the denominator it becomes
. Find the fraction .
Solution : Let, and are the numerator and the denominator of the fraction and the fraction is 
.
A/Q , 
And , 
Putting 
in equation , we ge 
Therefore , the fraction is 
.
(vi) Five years hence, the age of Jacob will be three times that of his son . Five years ago , Jocob’s age was seven times that of his son . What are their present ages ?
Solution : Let and be the present age of the Jacob and his son respectively .
Five years hence , the age of Jacob and his son will be 
and 
years respectively .
Five years ago , the age of Jacob and his son will be 
and 
years respectively .
A/Q , 
And 
[ From ]
Putting the value of in equation , we get 
Therefore , 40 yrs and 10 yrs are the present age of the Jacob and his son respectively .
EXERCISE 3.4
1. Solve the following pair of linear equations by the elimination method and the substitution method :
(i) 
Solution : Using elimination method :
And 
Putting the value of in equation , we get 
Therefore , the solution is 
and 
.
Using substitution method :
And 
[From (i)]
Putting the value of in equation , we ge 
Therefore , the solution is and .
(ii) 
Solution : Using elimination method :
And 
Putting the value of in equation , we have 
Therefore, the solution is and 
.
Using substitution method :
And 
Putting the value of in equation , we get 
Therefore, the solution is and .
(iii) 
Solution : Using elimination method :
We have ,
And 
Putting the value of in equation 
, we get 
Therefore, the solution is 
and 
Using substitution method : We have ,
And
Putting the value of in equation we get 
Therefore, the solution is 
and .
(iv) 
Solution : Using elimination method :
We have , 
and 
Putting 
in equation , we have 
Therefore, the solution is 
and 
Using substitution method : We have
We have ,
and
[ From (i)]
Putting the value of in equation , we get 
Therefore, the solution is and
2. Form the pair of linear equations in the following problem , and find their solutions (if they exist) by the elimination method :
(i) If we add 1 to the numerator and subtract 1 from the denominator , a fraction reduces to 1 . It becomes 
if we only add 1 to the denominator . What is the fraction ?
Solution : Let and be the numerator and denominator of the fraction .
The fraction is .
A/Q , 
And 
Putting the value of in equation , we get 
Therefore , the fraction is 
.
(ii) Five years ago , Nuri was thrice as old as Sonu . Ten years later, Nuri will be twice as old as Sonu . How old are Nuri and Sonu ?
Solution : Let and be the present age of the Nuri and Sonu respectively .
Five years ago , the age of Nuri and Sonu will be and years respectively .
Five years later , the age of Nuri and Sonu will be and years respectively .
A/Q , 
And 
Putting the value of in equation , we get
Therefore , 35 yrs and 15 yrs be the present age of the Nuri and Sonu respectively .
(iii) The sum of the digits of a two-digit number is 9 . Also , nine times this number is twice the number obtained by reversing the order of the digits . Find the number .
Solution : Let and be the ten’s and the unit’s digit of the number respectively .
The number is 
.
When the digits are reversed , then the
number is 
.
A/Q , 
Putting the value of in equation , we get 
Therefore , the number is 18 .
[ 
]
(iv) Meena went to a bank to withdraw Rs 2000 . She asked the cashier to give her Rs 50 and Rs 100 notes only . Meena got 25 notes in all . Find how many notes of Rs 50 and Rs 100 she received .
Solution : Let and be the number of notes of Rs 50 and Rs 100 respectively .
A/Q , 
Putting the value of in equation , we get 
Therefore , 10 and 15 be the number of notes of Rs 50 and Rs 100 respectively .
(v) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter . Saritha paid Rs 27 for a book kept for seven days , while Susy paid Rs 21 for the book she kept for five days . Find the fixed charge and the charge for each extra day .
Solution : let, and be the fixed charge and the charges for each extra day .
A/Q , 
And 
Putting the value of in equation , we get 
Therefore, the fixed charge is Rs 13 and the charge for each extra day is Rs 3 .
EXERCISE 3.5
1. Which of the following pairs of linear equations has unique solution , no solution , or infinitely many solutions . In case there is a unique solution , find it by it by using cross multiplication method .
(i) 
Solution : Here, 
So , 
Therefore, the pair of linear equations has no solution .
(ii) 
Solution: Here, 
So, 
Therefore, the pair of linear equations has unique solution .
Using cross-multiplication method :
We know that , 
and 
Therefore, the solution is and .
(iii) 
Solution : We have , 
Here, 
So , 
Therefore, the pair of linear equations has infinitely many solutions .
(iv) 
Solution : We have, 
and 
Here, 
So ,
Therefore, the pair of linear equations has unique solution .
Using cross-multiplication method :
We know that ,
and 
Therefore, the solution is 
and .
2. (i) For which values of 
and 
does the following pair of linear equations have an infinite number of solutions ?
Solution : We have, 
Here, 
Since, the pair of linear equation have an infinitely number of solutions .
...........(i)
and 
[ From (i)]
Putting 
in Eq. (i) we get, 
Therefore, the value of is 
and the value of is 
.
(ii) For which value of 
will the following pair of linear equations have no solution ?
Solution : We have , 
and 
Here, 
, 
,
,
,
,
A/Q , 
Hence, for k 
, the given equation has no solution .
3. Solve the following pair of linear equations by the substitution and cross-multiplication methods :
Solution : Using substitution method : We have ,
and 
[ From ]
Putting the value of 
in equation , we get 
Therefore, the solution is 
and 
.
Using cross-multiplication method : We have ,
We know that , 
and 
Therefore, the solution is 
and .
4. Form the pair of linear equations in the following problems and find their solutions (if they exist) by any algebraic method :
(i) A part of monthly hostel charges is fixed and the remaining depends on the number of days one has taken food in the mess . When a student A takes food for 20 days she has to pay Rs 1000 as hostel charges whereas a student B , who takes food for 26 days , pay Rs 1180 as hostel charges . Find the fixed charges and the cost of food per day .
Solution : Let and be the fixed charges and the cost of food per day .
For student A : 
For student B : 
Putting the value of in equation , we get 
Therefore, Rs 400 and Rs 30 be the fixed charges and the cost of food per day respectively .
(ii) A fraction becomes 
when 1 is subtracted from the numerator and it becomes 
when 8 is added to its denominator . Find the fraction .
Solution : Let and be the numerator and denominator of the fraction respectively .
The fraction is .
A/Q , 
And 
Putting the value of in equation , we get 
Therefore, the fraction is 
.
(iii) Yash scored 40 marks in a test, getting 3 marks for each right answer and losing 1 mark for each wrong answer . Had 4 marks been awarded for each correct answer and 2 marks been deducted for each incorrect answer, then Yash would have scored 50 marks . How many questions were there in the test ?
Solution : let and are the number of right answers and wrong answers respectively .
A/Q, 
and 
Putting 
in equation , we get 
Therefore, the number of question in the test is 20 (15+5 = 20) .
(iv) Places A and B are 100km apart on a highway . One car starts from A and another from B at the same time . If the cars travel in the same direction at different speeds , they meet in 5 hours . If they travel towards each other , they meet in 1 hour . What are the speeds of the two cars ?
Solution : let and (in km/h) be the speed of two cars start from A and B respectively .
A/Q , when the two cars travel in the same direction at different speed , then
Again , when the two cars travel towards each other , they meet in 1 hour , then
Thus, the speeds of the two cars are 60 km/h and 40 km/h respectively .
(v) The area of a rectangle gets reduced by 9 square units , if its length is reduced by 5 units and breadth is increased by 3 units . If we increase the length by 3 units and the breadth by 2 units , the area increases by 67 square units . Find the dimensions of the rectangle .
Solution : let, and are the length and breadth of the rectangle respectively.
Therefore , the area of rectangle 
.
A/Q , 
and 
Fromi , we get 
Required the length and breadth are 17 unit and 9 unit respectively.
EXERCISE 3.6
1. Solve the following pairs of equations by reducing them to a pair of linear equations :
(i) 
Solution : Let 
and 
We have, 
and 
Putting the value of 
in equation , we get
and 
Therefore, the solution is 
and 
.
(ii) 
, 
Solution : Let 
and 
Therefore , 
and 
From we get 
and 
Therefore, the solution is and 
(iii) 
Solution : Let
and 
Putting 
in equation , we get 
Hence, 
and 
are the required solution of the given pair of equations .
(iv) 
Solution: Let, 
and 
We have, 
and 
Putting 
in equation , we get 
Therefore, 
and 
Hence, 
and are the required solution of the given pair of equations .
(v) 
Solution: We have , 
and 
Let, 
and 
So, 
and 
Putting in equation , we have 
and 
Hence, 
and are the required solution of the given pair of equations .
(vi) 
Solution : We have , 
and 
Let, 
and
Now , 
and 
From (i) , we get , 
[ Putting 
]
and 
(vii) 
Solution : We have , 
and 
Let, 
and 
and 
[ From ]
Putting 
in , we get 
and 
Hence, 
and 
are the required solution of the given pair of equations .
(viii) 
, 
Solution : We have , and
Let, 
and 
and 
Putting 
in 
equation , we get 
and 
Hence, and are the required solution of the given pair of equations .
2. Formulate the following problems as a pair of equations , and hence find their solutions :
(i) Ritu can row downstream 20 km in 2 hours , and upstream 4 km in 2 hours . Find her speed of rowing in still water and the speed of the current .
Solution: Let and (in km/h) are the speed of rowing and current water respectively .
Then the speed of downstream of water 
km/h and upstream of water 
km/h
A/Q , 
and 
Therefore, 6 km/h and 4 km/h are the speed of rowing and current water respectively .
(ii) 2 women and 5 men can together finish an embroidery work in 4 days, while 3 women and 6 men can finish it in 3 days . Find the time taken by 1 women alone to finish the work and also that taken by 1 man alone .
Solution : let time taken by 1 woman and 1 man to finish the work and days respectively .
A/Q , 
and 
Let, an
and 
and 
Putting the value of 
in equation , we get 
and 
Thus time taken by 1 woman and 1 man to finish the work 18 days and 36 days respectively .
(iii) Roohi travels 300km to her home party by train and party by bus . She takes 4 hours if she travels 60 km by train and the remaining by bus . If she travels 100 km by train and the remaining by bus ,she takes 10 minutes longer . Find the speed of the train and the bus separately .
Solution : Let and (in km/h) be the speed of the train and the bus respectively .
A/Q , 
and 
Let, and
and 
Putting the value of
in equation , we get 
and 
Therefore, 60 km/h and 80 km/h be the speed of the train and the bus respectively .
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