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1. CHAPTER - 3

2. ,
EX:3.6-3.7

5. Using cross-multiplication method, we obtain:
1. Solve the following pairs of equations by reducing them to a pair of linear
equations:
(i) 1/2x + 1/3y = 2
1/3x + 1/2y = 13/6
EX:3.6:

6. Multiplying equation (1) by 3, we obtain:
6p + 9q = 6 (3)
1. Solve the following pairs of equations by reducing them to a pair of linear
equations:
(ii) 2/√x + 3/ √y = 2
4/√x - 9/ √y = -1
EX:3.6:
Adding equation (2) and (3), we obtain:
6p + 9q = 6
4p - 9y = -1
____________
10p +0q = 5
____________
p = 5/10 = 1/2

7. 1. Solve the following pairs of equations by reducing them to a pair of linear equations:
(iii) 4/x + 3y = 14 ;
3/x - 4y = 23
EX:3.6:

8. 1. Solve the following pairs of
equations by reducing them to
a pair of linear equations
(iv) 5 + 1 = 2
x-1 y – 2
6 - 3 = 1
x-1 y - 2
EX:3.6:

9. 1. Solve the following pairs of
equations by reducing them to
a pair of linear equations
(v) 7x – 2y = 5
xy
8x +7y = 15.
xy
EX:3.6:
-2p+7q=5 (x by 7))
7p+8q=15 (x by 2)
_________
-14p+49q=35
14p+16q=30
____________
65q=65 ; q=1
-2p+7(1)=5
-2p=5-7
-2p=-2
p=1

10. 1. Solve the following pairs of
equations by reducing them to
a pair of linear equations
(vi) 6x + 3y = 6 xy
2x + 4y = 5 xy
EX:3.6:
3p+6q=6
4p+2q=5 (x by 3)
_________
3p+6q=6
12p+6q=15
____________
-9p=-9 ; p=1
3(1)+6q=6
6q=6-3
6q=3
q=1/2.

11. 1. Solve the following pairs of
equations by reducing them to
a pair of linear equations
(vii) 10 + 2 = 4
x+ y x - y
15 - 5 = -2
x+ y x - y
EX:3.6:
10p+2q=4(x by5)
15p-5q=-2 (x by 2)
_________
50p+10q=20
30p-10q=-4
____________
80p=16 ;
10p+2q=4
10(1/5)+2q=4
2q=4-2
q=2/2=1
q=1
p=16/80
p=1/5

12. 1. Solve the following pairs of
equations by reducing them to
a pair of linear equations
EX:3.6:

13. 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 :
(i) Let the speed of Ritu in still water and the speed of
stream be x km/h and y km/h respectively.
Speed of Ritu while rowing upstream = (x-y) km/h
Speed of Ritu while rowing downstream = (x + y) km /h
Distance = speed x time
EX:3.6:
According to the question,

14. Adding equations (1) and (2), we
obtain:
2x = 12
x = 6
Putting the value of x in equation (1),
we obtain:
x + y = 10
6 + y = 10
y = 10 - 6
y = 4
x + y = 10
x - y = 2
____________
2x +0y = 12
____________
x = 12/2 = 6
EX:3.6:
Thus, Ritu’s speed in still water is 6 km/h and the
speed of the current is 4 km/h.

15. Solution :
Number of days taken by women to finish the work = x
Number of days taken by men to finish the work = y
Work done by women in one day = 1/x
Work done by women in one day = 1/y
As per the question given,
2.
(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 woman alone to finish the work,
and also that taken by 1 man alone.
EX:3.6:

16. 2.
(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 woman alone to finish the work,
and also that taken by 1 man alone.
EX:3.6:

17. iii) Let the speed of train and bus
be u km/h and v km/h respectively.
(iii) Roohi travels 300 km to her home
partly by train and partly 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.
EX:3.6:

18. (iii) Roohi travels 300 km to
her home partly by train and
partly 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.
EX:3.6:

19. EX : 3.7

20. 1. The ages of two friends Ani and Biju differ by 3 years. Ani’s father Dharam is twice as old as
Ani and Biju is twice as old as his sister Cathy. The ages of Cathy and Dharam differ by 30
years. Find the ages of Ani and Biju.
Solution :
The age difference between Ani and Biju is 3 yrs.
Either Biju is 3 years older than that of Ani or Ani is 3 years older than
Biju.
From both the cases we find out that Ani’s father’s age is 30 yrs more
than that of Cathy’s age.
Let the ages of Ani and Biju be A and B respectively.
Therefore, the age of Dharam = 2 x A = 2A yrs.
And the age of Biju sister Ann B/2 yrs
By using the information that is given,
EX:3.7:

21. Case (ii)
When Biju is older than Ani,
B – A = 3 – – – – – – – – – (1)
2A−B/2=30
4A – B = 60 – – – – – – – – – (2)
Adding the equation (1) and (2) we get,
3A = 63
A = 21
Therefore, the age of Ani is 21 yrs
And the age of Biju is 21 + 3 = 24 yrs.
Case (i)
When Ani is older than that of Biju by 3 yrs then
A – B = 3 – – – – – – – – (1)
2A−B/2 = 30
4A – B = 60 – – – – – – – – – – – (2)
By subtracting the equations (1) and (2) we get,
3A = 60 – 3 = 57
A= 57/3 = 19
Therefore, the age of Ani = 19 yrs
And the age of Biju is 19 – 3 = 16 yrs.
1. The ages of two friends Ani and Biju differ by 3 years. Ani’s father Dharam is twice as old as
Ani and Biju is twice as old as his sister Cathy. The ages of Cathy and Dharam differ by 30
years. Find the ages of Ani and Biju.
EX:3.7:

22. 2. One says, “Give me a hundred, friend! I shall then become twice as rich as you”.
The other replies, “If you give me ten, I shall be six times as rich as you”. Tell me what is the amount
of their (respective) capital? [From the Bijaganita of Bhaskara II] [Hint : x + 100 = 2(y – 100), y + 10 = 6(x – 10)].
When equation (2) is multiplied by 2 we get,
12A – 2B = 140 – – – – – – – (3)
When equation (1) is subtracted from
equation (3) we get,
11A = 140 + 300
11A = 440
=> A = 40
Using A =40 in equation (1) we get,
40 – 2B = -300
40 + 300 = 2B
2B = 340
B = 170
Therefore, Sangam had Rs 40 and Reuben
had Rs 170 with them.
EX:3.7:
Solution :
Let Sangam have Rs A with him and Reuben
have Rs B with him.
Using the information that is given we get,
A + 100 = 2(B – 100) A + 100 = 2B – 200
Or A – 2B = -300 – – – – – – – (1)
And
6(A – 10) = ( B + 10 )
Or 6A – 60 = B + 10
Or 6A – B = 70 – – – – – – (2)

23. Solution :
Let the speed of the train be x km/h and the
time taken by train to travel the given distance be t hours
and the distance to travel be d km.
Or, d = xt ... (1)
3. A train covered a certain distance at a uniform speed. If the train would have been 10 km/h
faster, it would have taken 2 hours less than the scheduled time. And, if the train were slower by
10 km/h; it would have taken 3 hours more than the scheduled time. Find the distance covered
by the train.
EX:3.7:
According to the question,

24. By using equation (1), we obtain:
3x - 10t = 30 ... (3)
Adding equations (2) and (3),
we obtain:
x = 50
Substituting the value of x in equation (2),
we obtain:
(-2) x (50) + 10t = 20
-100 + 10t = 20
10t = 120
t = 12
3. A train covered a certain distance at a uniform speed. If the train would have been 10 km/h
faster, it would have taken 2 hours less than the scheduled time. And, if the train were slower by
10 km/h; it would have taken 3 hours more than the scheduled time. Find the distance covered
by the train.
EX:3.7:
From equation (1), we obtain:
d = xt = 50 x 12 = 600
d = 600
Thus, the distance covered by the train
is 600 km.
-2x + 10t = 20
3x - 10t = 30
____________
x = 50

25. EX:3.7:
4. The students of a class are made to stand in rows. If 3 students are extra in a row, there
would be 1 row less. If 3 students are less in a row, there would be 2 rows more. Find the
number of students in the class.
Solution:
Let the number of rows be A and
the number of students in a row be B.
Total number of students :
Number of rows x Number of students in a row = AB
Using the information, that is given,
First Condition:
Total number of students = (A – 1) ( B + 3)
Or AB = ( A – 1 )(B + 3) = AB – B + 3A – 3
Or 3A – B – 3 = 0
Or 3A – Y = 3 – – – – – – – – – – – – – (1)

26. Second condition:
Total Number of students = (A + 2 ) ( B – 3 )
Or AB = AB + 2B – 3A – 6
Or 3A – 2B = -6 – – – – – – – – – (2)
When equation (2) is subtracted from (1)
(3A – B) – (3A – 2B) = 3 – (-6)
-B + 2B = 3 + 6B = 9
By using the equation (1) we get,
3A – 9 =3
3A = 9+3 = 12
A = 4
Number of rows, A = 4
Number of students in a row, B = 9
Number of total students in a class => AB => 4 x 9 = 36
EX:3.7:
4. The students of a class
are made to stand in
rows. If 3 students are
extra in a row, there
would be 1 row less. If 3
students are less in a
row, there would be 2
rows more. Find the
number of students in the
class.

27. 5. In a ∆ABC, ∠ C = 3 ∠ B = 2 (∠A + ∠ B).
Find the three angles.
Solution:
EX:3.7:

28. x 2 1 0
y 5 0 -5
6. Draw the graphs of the equations 5x – y = 5 and 3x – y = 3.
Determine the co-ordinates of the vertices of the triangle formed by these lines and the y axis.
Solution:
Given,
5x – y = 5
=> y =5x – 5
Also given, 3x – y = 3
y = 3x – 3
x 2 1 0
y 3 0 -3
Its solution table will be.
EX:3.7:

29. The graphical representation of these lines will be as follows:
From the above graph we can see that the
triangle formed is ∆ABC by the lines and
the y axis.
Also the coordinates of the vertices are
A(1,0) , C(0,-5) and B(0,-3).
x 2 1 0
y 5 0 -5
x 2 1 0
y 3 0 -3

30. (I)px + qy = p - q ... (1)
qx - py = p + q ... (2)
multiplying equation (1) by p and equation (2) by q, we
obtain:
p2x + pqy = p2 - pq ... (3)
q2x - pqy = pq + q2 ... (4)
adding equations (3) and (4), we obtain:
p2x + q2x = p2 + q2
(p2 + q2)x = p2 + q2
x = p2 + q2 x = 1.
p2 + q2
Substituting the value of x in equation (1), we obtain:
p(1) + qy = p – q
p+qy-p+q = 0
qy = -q
y = -1
7. Solve the following pair of
linear equations :
I)px + qy = p - q ... (1)
qx - py = p + q ... (2)
EX:3.7:

31. 7. Solve the following pair of linear equations
(ii)
ax + by = c ... (1)
bx + ay = 1 + c ... (2)
multiplying equation (1) by a and equation
(2) by b,
we obtain:
a2x + aby = ac ... (3)
b2x + aby = b + bc ... (4)
subtracting equation (4) from equation (3),
(a2 - b2)x = ac - bc - b
substituting the value of x in equation (1),
we obtain:
EX:3.7:

32. Solution :
bx - ay = 0 ... (1)
... (2)
multiplying equation (1) and (2)
by b and a respectively, we obtain:
b2x - aby = 0 ... (3)
a2x + aby = a3 + ab2 ... (4)
EX:3.7:
7. Solve the following pair of linear equations
adding equations (3) and (4), we obtain:
b2x + a2x = a3 + ab2
x(b2 + a2) = a(a2 + b2)
x = a
substituting the value of x in equation (1),
we obtain:
b(a) - ay = 0
ab - ay = 0
ay = ab
y = b
ax + by = a2 + b2
ax + by = a2 + b2

33. (iv)
(a - b)x + (a + b)y = a2 - 2ab - b2 ... (1)
(a + b)(x + y) = a2 + b2 ... (2)
subtracting equation (2) from (1), we obtain:
(a - b)x - (a + b)x = (a2 - 2ab - b2) - (a2 + b2)
(a - b - a - b)x = -2ab - 2b2
-2bx = -2b(a + b)
x = a + b
substituting the value of x in equation (1),
we obtain:
(a - b)(a + b) + (a + b)y = a2 - 2ab - b2
a2 - b2 + (a + b)y = a2 - 2ab - b2
(a + b)y = -2ab
EX:3.7:
7. Solve the following pair of linear equations

34. (v)
152x - 378y = -74 ... (1)
-378x + 152y = -604 ... (2)
adding the equations (1) and (2), we obtain:
-226x - 226y = -678
subtracting the equation (2) from equation (1), we obtain:
530x - 530y = 530
adding equations (3) and (4), we obtain:
2x = 4
x = 2
substituting the value of x in equation (3), we obtain:
y = 1
EX:3.7:
7. Solve the following pair of linear equations

35. 8. ABCD is a cyclic quadrilateral (see Fig. 3.7). Find the
angles of the cyclic quadrilateral.
Solution:
It is known that the sum of the opposite
angles of a cyclic quadrilateral is 180o
Thus, we have
EX:3.7:

36. PAIR OF LINEAR EQUATIONS
COMPLETED