3 PAIR OF LINEAR EQUATIONS IN TWO VARIABLES CHAPTER

CHAPTER
3
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PAIR OF LINEAR EQUATIONS
IN TWO VARIABLES
Points to Remember :
1. A pair of linear equations in two variables x and y can be represented as follows :
a1x + b1y + c1 = 0; a2x + b2y + c2 = 0,
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where a1, a2, b1, b2, c1, c2 are real numbers such that a12  b12  0, a22  b22  0.
2. Graphically, a pair of linear equations a1x + b1y + c1 = 0, a2x + b2y + c2 = 0 in two variables represents a pair
of straight lines which are :
(i) Intersecting, if
a1 b1

a2 b2
here, the equations have a unique solution, and pair of equations is said to be consistent.
(ii) parallel, if
a1 b1 c1
 
a2 b2 c2
B
here, the equations have No solution, and pair of equations is said to be inconsistent.
(iii) Coincident, if
a1 b1 c1
 
a2 b2 c2
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here, the equations have infinitely many solutions, and pair of equations is said to be consistent.
3. A pair of linear equations in two variables can be solved by the :
(i) Graphical method
(ii) Algebraic Methods; which are of three types :
(a) Substitution method
(b) Elimination method
(c) Cross-multiplication method
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ILLUSTRATIVE EXAMPLES
Example 1. Draw the graph of linear equation 2x + 3y = 7.
Solution.
7  2x
3
Give atleast two suitable values to x to find the corresponding value of y.
2x + 3y = 7  3y = 7 – 2x  y 
7  2(2) 3
 1
3
3
7  2(5) 3

 1
If x  5, y 
3
3
If x  2, y 
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PAIR OF LINEAR EQUATIONS IN TWO VARIABLES
25
Representing it in a tabular form, we get
x 2 5
y 1 1
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We now plot the points (2, 1) and (5, –1) on graph paper to obtain a straight line.
2x
+3
y=
Y-axis
7
4
3
2
(2, 1)
1
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X-axis
–3
–2
0
–1
1
–1
–2
–3
–4
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2
3
B
4
5
6
(5, –1)
Example 2. Aftab tells his daughter, ‘‘Seven years ago, I was seven times as old as you were then. Also, three
years from now, I shall be three times as old as you will be. Represent this situation algebraically
and graphically.
[NCERT]
Solution. Let the present age of the daughter = x years.
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7 years ago daughter’s age = (x – 7) years
3 years from now, daughter’s age = (x + 3) years
Let the present age of father = y years
7 years ago father’s age = ( y – 7) years
3 years from now, fathers age = (y + 3) years.
According to given question, two algebraic equations are :
y – 7 = 7 (x – 7)  y = 7x – 42
and y + 3 = 3 (x + 3)  y = 3x + 6
...(1)
...(2)
Required table for y = 7x – 42
x
y
26
12 18
42 84
PAIR OF LINEAR EQUATIONS IN TWO VARIABLES
MATHEMATICS–X
and, required table for y = 3x + 6 is
x
6
12
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y 24 42
3x
+
(18, 84)
84
Age of Father 
y=
7x –
y=
96
6
42
Y
72
60
48
(12, 42)
36
(6, 24)
24
12
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6
B
12 18 24 30
Age of daughter
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From graph, we observe that :
Present age of father is 42 years and, present age of his daughter is 12 years.
Example 3. Use a single graph paper and draw the graph of the following equations :
2y – x = 8; 5y – x = 14; y – 2x = 1.
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Obtain the vertices of the triangle so obtained.
Solution.
For equation 2y – x = 8
We have, 2y – x = 8  x = 2y – 8
When y = 2, we have x = 2 (2) – 8 = 4 – 8 = – 4
when y = 3, we have x = 2(3) – 8 = 6 – 8 = – 2
Thus, we have the following table :
x –4 –2
y 2 3
For equation 5y – x = 14
We have, 5y – x = 14  x = 5y – 14
when y = 3, we have x = 5(3) – 14 = 15 – 14 = 1
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PAIR OF LINEAR EQUATIONS IN TWO VARIABLES
27
when y = 4, we have x = 5 (4) – 14 = 20 – 14 = 6
Thus, we have the following table :
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x 1 6
y 3 4
For equation y – 2x = 1
We have, y – 2x = 1  y = 2x + 1
when x = – 1, we have y = 2 (–1) + 1 = –2 + 1 = – 1
when x = 0, we have y = 2 (0) + 1 = 0 + 1 = 1
Thus, we have the following table :
x 1 0
y 1 1
The graph for the given equations is shown below :
y–
2x
=1
Y
6
5
(2,5)
4
(–2,3)
3
2
1
–7
–6
–5
–4
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–3
–2
0
–1
–1
–2
–3
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1
B
2
8
5y–x =
(1,3)
(–4, 2)
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x=
2y–
3
4
(6, 4)
5
6
7
14
X
8
–4
From graph, we observe that the vertices of the triangle are (–4, 2), (1, 3) and (2, 5).
Example 4. Draw the graphs of the equations :
x – y + 1 = 0; 3x + 2y – 12 = 0.
Determine the coordinates of the vertices of the triangle formed by these lines and the x-axis and
shade the triangular region. Also, find area of this triangular region.
Solution. For equation x – y + 1 = 0
We have, x – y + 1 = 0  y = x + 1
when, x = 0, we have y = 0 + 1 = 1
when, x = –1, we have y = –1 + 1 = 0
Thus, we have the following table :
x 0 1
y 1 0
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PAIR OF LINEAR EQUATIONS IN TWO VARIABLES
MATHEMATICS–X
For equation 3x + 2y – 12 = 0
we have, 3x + 2y – 12 = 0  y 
12  3x
2
when x = 0, we have y 
12  3(0) 12

6
2
2
when x = 4, we have y 
12  3(4) 12  12

0
2
2
Thus, we have the following table :
x 0 4
y 6 0
The graph for the given equations is shown below.
+2
3x
Y
=0
12
y–
B
7
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5
4
A (2, 3)
3
2
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(0, 1) 1
–2
–1
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0
1=
+
y
x–
(0, 6)
6
C (–1, 0)
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0
1
2
3
B (4, 0)
X
4
5
6
7
From graph, we observe that the vertices of ABC are A(2, 3), B(4, 0) and C(–1, 0).
1
Also, Area of ABC   5  3 sq. units = 7.5 sq. units.
2
Example 5. Find the values of a and b so that the following system of linear equations has an infinite number
of solutions : 2x – 3y = 7
(a + b)x – (a + b – 3) y = 4a + b
[NCERT, CBSE 2002]
Solution.
Here,
a1
c
2 b1
3
7

, 
and 1 
a2 a  b b2 (a  b  3)
c2 4a  b
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29
a1 b1 c1
For infinite solutions, we have a  b  c
2
2
2
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2
3
7


a  b a  b  3 4a  b

Taking Ist two terms :
Taking last two terms :
2
3

a b a b 3
3
7

a  b  3 4a  b

3a + 3b = 2a + 2b – 6

a+b=–6
...(1)
12a + 3b = 7a + 7b – 21

5a – 4b = – 21
From equation (1) we have a = – 6 – b
Using the value of a in equation (2), we get
5(–6  b)  4b  21

–30 – 5b – 4b = – 21  –9b = 9  b = – 1
and a = – 6 – (–1) = – 6 + 1= – 5
Hence, a = – 5 and b = – 1 Ans.
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
B
...(2)
Example 6. Use the method of substitution to solve the following system of equations :
4x +
Solution.
1
8
y= ;
3
3
1
3
5
x+ y = –
2
4
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Consider, 4 x  1 y  8
3
3

12x + y = 8
and,
1
3
5
x y  
2
4
2
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
2 x  3 y  10
...(1)
...(2)
Now, we will solve equation (1) and (2) by substitution method.
From (1), we get y = 8 – 12 x.
Now, substituting the value of y in eqn. (2), we get
2x + 3 (8 – 12x) = –10

2x + 24 – 36 x = –10

– 34x = – 34  x = 1
Now, substitute x = 1 in y = 8 – 12 x, we get
y = 8 – 12 (1) = 8 – 12 = – 4
Thus x = 1, y = – 4 is the required solution.
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PAIR OF LINEAR EQUATIONS IN TWO VARIABLES
MATHEMATICS–X
Example 7. Solve the following system of linear equations by elimination method (equating the coefficients).
6 (ax + by) = 3a + 2b
6 (bx – ay) = 3b – 2a
[CBSE 2004, 2006 outside]
Solution. Given equations :
6 (ax + by) = 3a + 2b
...(1)
6 (bx – ay) = 3b – 2a
...(2)
multiplying eqn. (1) by a and equation (2) by b, and adding, we get
6a2x + 6aby = 3a2 + 2ab
6b2x – 6aby = 3b2 – 2ab
2
2
2
2
6 (a + b ) x = 3 (a + b )

x
3 (a 2  b 2 ) 1

6 (a 2  b 2 ) 2
Putting x 
1
in eqn. (1), we get
2

1
6a   6by  3a  2b
2
3a + 6by = 3a + 2b

y

1
1
is the required solution.
x= , y=
2
3
2b 1

6b 3
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Example 8. Solve the following system of linear equations by using the method of cross-multiplication :
ax + by = a – b
bx – ay = a + b
[CBSE 2000, 2005]
Solution. The given system of equations may be written as
ax + by – (a – b) = 0
bx – ay – (a + b) = 0
Using cross-multiplication method, we get
x
–y
1
=
=
b – (a – b)
a – (a – b)
a b
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–a – (a + b)
b – (a + b)
b
–a

x
y
1

 2
b  [(a  b)]  ( a)  [(a  b)] a  [ ( a  b)]  b  [ ( a  b)] a  b 2

x
y
1


b ( a  b )  a ( a  b )  a ( a  b )  b ( a  b )  ( a 2  b 2 )
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PAIR OF LINEAR EQUATIONS IN TWO VARIABLES
31

x
y
1


b 2  a 2  a 2  b 2  ( a 2  b 2 )

x
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( a 2  b 2 )
(a 2  b2 )
 1 and y 
 1
2
2
( a  b )
( a 2  b 2 )
Thus, x =1 , y = – 1 is the required solution.
Example 9. Solve for x and y :
2
1
–5
+
=
x + 2y 2x - y
9
9
6
+
= –4
x + 2y 2x - y
Solution.
Let
1
1
 u and
 v.
x  2y
2x  y
Putting in the given equations, we get
2u  v  
5
9
B
9 u  6 v  4
multiplying eqn. (1) by 6, and eqn. (2) by 1, we get
12u  6v 
10
3
9 u  6 v  4
Subtracting eqn. (4) from eqn. (3), we get
12u – 9u =

3u 
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...(1)
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10
4
3
2
2
u
3
9
...(2)
...(3)
...(4)
Putting u  2 in eqn. (2), we get
9
2
9   6v  4  v  1
9
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Now, u 

2
, v  1
9
x + 2y =
9
2
and 2x – y = – 1
multiplying eqn. (5) by 1, and eqn. (6) by – 2, we get
x  2y 
9
2
–4x + 2y = 2
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PAIR OF LINEAR EQUATIONS IN TWO VARIABLES
...(5)
...(6)
...(7)
...(8)
MATHEMATICS–X
Subtracting eqn. (8) from eqn. (7), we get
5x 
9
5
1
2  x 
2
2
2
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1
in eqn. (6), we get
2
1–y=–1 y=2
Putting x 
1
, y = 2 is the required solution.
2
Example 10. 2 tables and 3 chairs together cost Rs. 2000 whereas 3 tables and 2 chairs together cost Rs.
2500. Find the total cost of 1 table and 5 chairs.
Solution. Let the cost of a table be Rs. x and that of a chair be Rs. y. According to given question,
2x + 3y = 2000
...(1)
and
3x + 2y = 2500
...(2)
Adding eqn. (1) and (2), we get
5x + 5y = 4500  x + y = 900
...(3)
Subtracting eqn. (1) from eqn. (2), we get
x – y = 500
...(4)
Adding eqn. (3) and eqn. (4), we get
2x = 1400  x = 700
Using x = 700 in eqn. (3), we get
700 + y = 900  y = 200
 Cost of 1 table = Rs. 700 and cost of 1 chair = Rs. 200.
Hence, cost of 1 table and 5 chairs
= Rs. (x + 5y) = Rs. 700 + 5 (200) = Rs. 1700 Ans.
Example 11. The sum of a two-digit number and the number obtained by reversing the order of digits is 165.
If the digits differ by 3, find the number.
[CBSE 2002]
Solution. Let the unit digit be x and tens digit be y.
Then, number = 10 y + x.
Number obtained by reversing the order of the digits = 10x + y.
According to the given question, we have
(10y + x) + (10 x + y) = 165
...(1)
and, x – y = 3
...(2)
OR
y–x=3
...(3)
From eqn. (1), we get
11x + 11y = 165  x + y = 15
...(4)
Solving eqn. (2) and (4) together, we get
x = 9, y = 6
Solving eqn. (3) and (4) together, we get
x = 6, y = 9
Substituting the values of x and y, for the number 10 y + x, we get number = 69 or 96. Ans.
Thus, x =
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MATHEMATICS–X
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PAIR OF LINEAR EQUATIONS IN TWO VARIABLES
33
Example 12. A fraction is such that if the numerator is multiplied by 3 and the denominator is reduced by 3,
2
18
, but if the numerator is increased by 8 and the denominator is doubled, we get .
we get
5
11
Find the fraction.
Solution.
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x
.
y
Then, according to given equestion, we have :
Let the fraction be
3x
18
x8 2

and

y  3 11
2y
5
 11x = 6y – 18
and
5x + 40 = 4y
 11x – 6y = – 18
and 5x – 4y = – 40
multiplying eqn. (1) by 2 and eqn. (2) by 3, we get :
22x – 12 y = – 36
15x – 12y = – 120
Subtracting eqn. (4) from eqn. (3), we get
7x = 84  x = 12
Substituting x = 12 in eqn. (1), we get
11(12) – 6y = – 18
 –6y = – 150  y = 25.
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...(1)
...(2)
...(3)
...(4)
12
. Ans.
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Example 13. A man travels 370 km partly by train and partly by car. If he covers 250 km by train and the rest
by car, it takes him 4 hours. But, if he travels 130 km by train and the rest by car, he takes 18
minutes longer. Find the speed of the train and that of the car.
[CBSE 2001]
Solution. Let the speed of the train be x km/h and that of the car be y km/hr.
Case I : When he travels 250 km by train and rest by car :
In this case, we have
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Hence, the required fraction is
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Time taken by the man to travel 250 km by train  250 hrs .
x
Time taken by the man to travel (370 – 250) km = 120 km by car 
120
hrs.
y
 250 120 
 Total time taken by man to cover 370 km  

 hrs.
y 
 x
It is given that the total time taken is 4 hours.

250 120
125 60

4 

2
x
y
x
y
...(1)
Case II. When he travels 130 km by train and rest by car :
Time taken by the man to travel 130 km by train  130 hrs.
x
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PAIR OF LINEAR EQUATIONS IN TWO VARIABLES
MATHEMATICS–X
Time taken by the man to travel (370 – 130) km = 240 km by car =
240
hrs.
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In this case, total time of the journey is 4 hrs. 18 minutes.


130 240

 4 hr 18 minutes
x
y
130 240
18

4
x
y
60
130 240 43


x
y
10
Now, we will solve eqn. (1) and (2).

1
1
 u,  v, the above equations reduces to
x
y
125 u + 60 v = 2
43
130 u + 240 v =
10
multiplying eqn. (3) by 4, and eqn. (4) by 1, we get
500 u + 240 v = 8
43
130 u + 240 v =
10
subtracting eqn. (6) from eqn. (5), we get
Putting
370u  8 
43
37
1
 370u 
u 
10
10
100
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Putting u  1 in eqn. (5), we get
100
5 + 240 v = 8  240 v = 3  v 
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...(2)
1
80
B
...(3)
...(4)
...(5)
...(6)
1
1
and v 
100
80
1
1
1 1
 x  100 and y  80  x  100, y  80
Hence, the speed of train = 100 km/hr and the speed of car = 80 km/hr.
Example 14. 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, pays Rs. 1180 as hostel charges.
Find the fixed charges and the cost of food per day.
[CBSE 2000, NCERT]
Solution. Let the fixed charges be Rs. x and charges per day be Rs. y.
According to given question,
x + 20 y = 1000
...(1)
x + 26 y = 1180
...(2)
Subtracting eq. (1) from eqn. (2), we get
6y = 180  y = 30
Now, u 
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35
from eqn. (1), x + 20(30) = 1000
 x = 1000 – 600 = 400
 Fixed charges = Rs. 400 and cost of food per day = Rs. 30.
Example 15. 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.
[NCERT]
Solution. Let the length and breadth of the rectangle be x and y units respectively.
Then, Area = xy sq. units
If the length is reduced by 5 units and breadth is increased by 3 units, then area is reduced by 9 sq.
units.
 xy – 9 = (x – 5) (y + 3)
 xy – 9 = xy + 3x – 5y – 15
 3x – 5y = 6
...(1)
When the length is increased by 3 units and breadth by 2 units, the area is increased by 67 sq.
units.
 xy + 67 = (x + 3) (y + 2)
 xy + 67 = xy + 2x + 3y + 6
 2x + 3y = 61
...(2)
Solving Eq. (1) and eq. (2), we get
x
y
1


305  18 183  12 9  10
x

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171
 17, y 
9
19
19
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Hence, the length and breadth of the rectangle are 17 units and 9 units respectively.
PRACTICE EXERCISE
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Section - A
1. Draw the graph of the equation 2y – x = 7 and determine from the graph whether x = 3, y = 2 is a solution
or not?
Draw the graph of the following equations (2-10)
2. 3x + 2y = 5
6. 2x – 3y = 12
10. y – 4x = 12
3. x – 3y = 2
7. 3x + 5y = – 18
4. x = 4
8. 3y – 2x = 30
5. y = – 5
9. x + 3y = – 1
Solve the following system of equations graphically (11-20)
11. 3x + 2y = 6
3x – 4y = – 12
15. 2x + 3y = – 5
x – 2y = 8
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12. 5x – y = 7
x–y=–1
16. 3x + 2y = 5
5x – 3y = 2
13. x – 2y = 6
3x – 6y = 9
17. 5x + 4y + 6 = 0
2x + 7y – 3 = 0
PAIR OF LINEAR EQUATIONS IN TWO VARIABLES
14. x – 4y + 14 = 0
3x + 2y – 14 = 0
18. 4x + 2y = 1
2x + y = – 4
MATHEMATICS–X
19. 5x + y – 7 = 0
2x + 5y = 12
20. 2x + 3y = 10
x
3
y5
2
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In each of the following system of linear equations, determine graphically whether the given system has a
unique solution, no solution or infinitely many solutions (21-30)
21. 6x + 4y = 7
3x + 2y = 10
25. 2y – 3x = 21
2
–x + y = 7
3
29. 3x + y = 5
6x + 2y = 10
22. 3x – 5y + 1 = 0
4x – y + 8 = 0
26. 3x – y = 7
7x – 2y = 15
23. x – 4y = 10
24. –2x + y = 6
3x – 12y = 30
4x + 2y = 12
27. 7x – 3y = 9
x+y=4
30. –y + 3x = 7
7x – 2y – 15 = 0
28. x – y = 8
J
A
4x – 4y = 12
Show graphically that each one of the following systems of equations has infinitely many solutions (31-33)
31.
2 y  4x  6
2x  y  3
32. 3 x  2 y  8
9 x  6 y  24
33. x  2 y  6  0
2 x  4 y  12  0
B
Show graphically that each one of the following systems of equations is in-consistent (i.e. has no solution)
(34-36)
34.
x  2 y  9
6 y  3x  21
35. x  2 y  6  0
4x  8 y  5
T
I
36. 2 x  3 y  6
6 x  9 y  10
Solve graphically each of the following systems of linear equations. Also find the co-ordinates of the points
where the lines meet y-axis. (37-39)
37.
2x  5 y  4  0
2 x  y  8  0 [CBSE 2005]
39.
3x  y  5  0
2 x  y  5  0 [CBSE 2002(C)]
M
A
38. 3 x  2 y  12
5 x  2 y  4 [CBSE 2006(C)]
Solve graphically each of the following systems of linear equations. Also find the co-ordinates of the points
where the lines meet x-axis. (40-42)
40.
2 x  3 y  4
x  2y  5
41. 2 x  y  6
x  2y  2  0
42. x  y  2  0
3x  2 y  9  0
Solve graphically the following system of equations. Shade the region bounded by these lines and x-axis.
Find the area of the shaded region. (43-45)
43.
2x  y  6
2 x  y  2  0 [CBSE 2002]
45.
x  y 1  0
3x  2 y  12  0 [CBSE 2002]
MATHEMATICS–X
44. x  y  5
2x  y  2  0
PAIR OF LINEAR EQUATIONS IN TWO VARIABLES
37
Solve graphically the following system of equations. Shade the region bounded by these lines and y-axis.
Find the area of the shaded region. (46-48).
46.
x y 3
3x  2 y  4
47. 4 x  5 y  20  0
3 x  5 y  15  0
J
A
48. x  y  1
2 x  y  8 [CBSE 2001]
Solve the following system of linear equations graphically. Also, find the vertices of the triangle formed by
these lines and x-axis. (49-51)
49.
5 x  6 y  30  0
5 x  4 y  20  0 [CBSE 2004]
50. 3 x  y  9  0
3 x  4 y  6  0 [CBSE 2006]
51. 4x – y – 8 = 0
2x – 3y + 6 = 0
J
A
Solve the following system of linear equations graphically. Also, find the vertices of the triangle formed by
these lines and y-axis (52-54) :
52. 3x + y – 11 = 0
53. 2x + 3y = 12
x – y – 1 = 0 [CBSE 2000 (C)]
54. 3x + y – 5 = 0
2x – y – 5 = 0 [CBSE 2002 (C)]
x–y=1
[CBSE 2001]
B
Determine graphically the co-ordinates of the vertices of a triangle, the equations of whose sides
are (55-60):
55.
T
I
yx
3y  x
x  y  8 [CBSE 2000]
58.
x y 5
x y  5
59. 2 x  y  5
x  2 y  7
M
A
x0
x3
56. 2 x  3 y  6  0
2 x  3 y  18  0
y2
57. 2 y  x  8
5 y  x  14
y  2x  1
60. x  y  3
2 x  5 y  12
y0
Solve the following system of equations using the method of substitution (61-69)
61. 2x + 3y = 4
5x + 8y = 9
64.
67.
38
4 x  3 y  1
3x  4 y  18
3
2
3
2x  y 
2
x  2y 
62. x + 2y = – 1
2x – 3y = 12
63. 2x – 7y = 1
4x + 3y = – 15
65. 0.4 x  0.3 y  1.7
0.7 x  0.2 y  0.8
66. x  y  11
3 4
5x y
  7
6 3
68. 6 x  5 y  11
9 x  10 y  21
69. 2 x  5 y  3
7x  2 y  4
PAIR OF LINEAR EQUATIONS IN TWO VARIABLES
MATHEMATICS–X
Solve the following system of equations using the method of elimination (by equating the coefficients)
(70-78) :
70.
x y  7
12 x  5 y  7
71. 7 x  8 y  11
8x  7 y  7
72. 3 x  5 y  7
11x  13 y  9
73.
6 x  24 y  5
2 x  3 y  2
74. 3x  2 y  11
x  5y  8
75. 2 x  7 y  3
5 x  3 y  13
76. x + y = a + b
ax – by = a2 – b2
77. 3 (bx + ay) = a – 6b
3 (ax – by) = – (6a + b) [CBSE 2004]
J
A
Solve the following system of equations using the method of cross-multiplication (79-87):
79. 2x + 3y = 17
3x – 2y = 6
82.
85.
5 x  2 y  5
3x  y  2
x y
  ab
a b
x
y
 2 2
2
a
b
80. 3x + 2y = – 25
2x + y = – 10
2
y 1  0
3
6x  y  2  0
83. 4 x 
81. 2x + y = 35
3x + 4y = 65
84. 2 x  3 y  10  0
3x  4 y  2  0
B
86. ax  by  a  b
bx  ay  a  b [CBSE 2000, 2005]
T
I
Solve the following system of equations (88-110) :
88. 1.5 x  0.1y  6.2
3x  0.4 y  11.2
89.
x
y
3 
4
6
x
 y  2
2
1 1
  1
2x y
1 1

 8, x  0, y  0
x 2y
M
A
91.
93.
95.
4
 3 y  14
x
3
 4 y  23 [CBSE 2004(C)]
x
x 1

2
x 1

3
y 1
8
3
y 1
9
2
MATHEMATICS–X
J
A
78. ax + by = a – b
bx – ay = a + b [CBSE 2000]
87. x  y  2
a b
ax  by  a 2  b 2 [CBSE 2005]
90. 101x  99 y  499
99 x  101y  501
92.
2x y
 2
a b
x y
 4
a b
6
6
y
8
3x   5 [CBSE 2007]
y
94. x 
96. 47 x  31y  63
31x  47 y  15 [CBSE 2006]
[CBSE 2007]
PAIR OF LINEAR EQUATIONS IN TWO VARIABLES
39
97.
99.
102.
105.
107.
x y
  a b
a b
ax  by  a 3  b3
x y
2
xy
x y
6
xy
[CBSE 2005]
100.
1
1

 12
5x 6 y
1
3

8
3x 7 y
6
7

3
x y x y
1
1

2( x  y ) 3( x  y )
1
5
3


2( x  2 y ) 3(3x  2 y ) 2
5
3
61


4( x  2 y ) 5(3x  2 y ) 60
a b
 0
x y
106.
22

x y
55

x y
T
I
[CBSE 2006 (C)]
J
A
15
5
x y
45
 14
x y
J
A
104. x  y  0.9
11
2
x y
1
1
3


3x  y 3x  y 4
1
1
1


2(3 x  y ) 2(3x  y )
8
B
108.
mx  ny  m 2  n 2
x  y  2m
101.
2
103. ax  by  a
ay  bx  b 2
ab2 a 2 b

 a 2  b 2 ; x, y  0
x
y
109.
98. 3(2 x  y )  7 xy
3( x  3 y )  11xy
a 2 b2

0
x
y
a 2 b b2 a

 a  b; x, y  0 [CBSE 2006 (C)]
x
y
110.
x y
 0
a b
( a  b) x  ( a  b) y  a 2  b 2
In each of the following systems of equations determine whether the system has a unique solution, no
solution or infinitely many solutions. In case there is a unique solution, find it (111-113) :
M
A
111.
3x  5 y  20
6 x  10 y  40
112. x  2 y  8
5 x  10 y  10
113. 3 x  y  1  0
2x  3 y  8  0
Find the value of k for which the following system of equations has a unique solution (114-116) :
114.
x  ky  2
3 x  2 y  5
115. 4 x  ky  5  0
2x  3 y  7  0
116. 10 x  3ky  1  0
5x  3 y  2  0
Find the value of k for which the following system of equations have infinitely many solutions (117-122) :
118. 2 x  3 y  4
119. 4 x  3 y  3
(2k  3) x  (2k  1) y  4(k  1)
(k  2) x  6 y  3k  2
117.
4x  5 y  3
kx  15 y  9
120.
2x  3 y  2
(k  2) x  (2k  1)  2(k  1)
40
121. x  (k  1) y  5
(k  1) x  9 y  8k  1 [CBSE 2003]
PAIR OF LINEAR EQUATIONS IN TWO VARIABLES
MATHEMATICS–X
122.
2 x  (k  2) y  k
6 x  (2k  1) y  2k  5
J
A
Find the value of k for which the following system of equations has no solution (123-127) :
124. x  2 y  2
2 x  ky  5
123.
kx  5 y  2
6x  2 y  7
126.
3x  y  1
(2k  1) x  (k  1) y  2k  1 [CBSE 2000]
125. 6 x  ky  9  0
3x  2 y  1  0
127. (2 k  1) x  2 y  2  0
( k 2  1) x  ( k  2) y  5  0
J
A
Find the values of a and b for which the following system of equations has infinitely many solutions
(128-130):
128.
(2a  1) x  3 y  5  0
3 x  (b  1) y  2
130.
3x  (a  1) y  2b  1
5 x  (1  2a) y  3b
129. 2 x  3 y  7
(a  b) x  (2a  b) y  3(a  b  1) [CBSE 2002]
Section - B (Word Problems)
B
Problems related to Ages (131-136) :
131. Two years ago, a father was five times as old as his son. Two years later, his age will be 8 more than three
times the age of the son. Find the present age of the father and son.
[CBSE 2004]
132. Father’s age is three times the sum of ages of his two children. After 5 years his age will be twice the sum
of ages of two children. Find the age of father.
133. A father is three times as old as his son. In 12 years, he will be twice as old as his son. Find the present
age of the father and the son.
134. Four years ago a mother was four times as old as her daughter. Six years later, the mother will be two-anda-half times as old as her daughter. Determine the present ages of mother and her daughter.
135. Five years ago, I was thrice as old as my son. Five years hence I shall be twice as old as my son. How old
are we now?
136. The present age of a father is three years more than three times the age of the son. Three years hence
father’s age will be 10 years more than twice the age of the son. Determine their present ages.
T
I
M
A
Problems related to Articles and their costs (137-140) :
137. 3 bags and 4 pens together cost Rs. 79 whereas 4 bags and 3 pens together cost Rs. 324. Find the total
cost of 1 bag and 10 pens.
138. 4 chairs and 3 tables cost Rs. 2100 and 5 chairs and 2 tables cost Rs. 1750. Find the cost of a chair and a
table seperately.
139. Two audio cassetes and three video cassettes cost Rs. 340. But three audio cassettes and two video
cassettes cost Rs. 260. Find the price of an audio cassette and that of a video cassette.
140. On selling a T.V. at 5% gain and a fridge at 10% gain, a shopkeeper gains Rs. 2000. But if he sells the T.V.
at 10% gain and the fridge at 5% loss, he gains Rs. 1500 on the transaction. Find the actual prices of T.V.
and fridge.
MATHEMATICS–X
PAIR OF LINEAR EQUATIONS IN TWO VARIABLES
41
Problems related to Numbers (141-148) :
141. The sum of two numbers is 26 and three times one of them exceeds five times the other by 6. Find the
numbers.
142. Find two numbers such that one-third of the first number added to one-fifth of the second number gives
20 and one-sixth of the first added to one-twelfth of the second gives 9.
143. The sum of the digits of a two digit number is 12. The number obtained by reversing the order of the
digits of the given number exceeds the given number by 18. Find the two digit number. [CBSE 2006]
144. In a two digit number, the ten’s digit is three times the unit’s digit. When the number is decreased by 54,
the digits are reversed. Find the number.
145. A two-digit number is such that the product of the digits is 20. If 9 is added to the number, the digits
interchange their places. Find the number.
[CBSE 2005]
146. A two-digit number is four times the sum of its digits and twice the product of the digits. Find the number.
[CBSE 2005]
147. The sum of a two digit number and the number obtained by reversing the order of its digits is 99. If the
digits differ by 3, find the number.
[CBSE 2002]
148. A two-digit number is obtained by multiplying the sum of the digits by 8. Also it is obtained by multiplying the difference of the digits by 14 and adding 2. Find the number.
B
J
A
J
A
Problems related to Fractions (149-152) :
149. The sum of the numerator and denominator of a fraction is 12. If the denominator is increased by 3, the
1
fraction becomes . Find the fraction.
[CBSE 2006(C)]
2
1
150. A fraction becomes if 1 is subtracted from both its numerator and denominator. If 1 is added to both
3
1
the numerator and denominator, it becomes . Find the fraction.
2
151. The denominator of a fraction is 4 more than twice the numerator. When both the numerator and
denominator are decreased by 6, then the denominator becomes 12 times the numerator. Determine the
fraction.
[CBSE 2001 (C)]
5
152. If the numerator of a certain fraction is increased by 2 and denominator by 1, the fraction becomes ,
8
1
and if the numerator and denominator are each diminished by 1, the fraction becomes . Find the
2
fraction.
T
I
M
A
Problems related to Speed, Distance and Time (153-158) :
153. Points A and B are 70 km apart on a highway. A car starts from A and another car starts from B at the same
time. If they travel in the same direction, they meet in 7 hours, but if they travel towards each other they
meet in one hour. What are their speeds?
[CBSE 2002]
154. A boat goes 12 km upstream and 40 km downstream in 8 hours. It can go 16 km upstream and 32 km
downstream in the same time. Find the speed of the boat in still water and the speed of the stream.
155. A sailor goes 8 km downstream in 40 minutes and returns in 1 hour. Determine the speed of the sailor in
still water and the speed of the current.
156. A man travelled 300 km by train and 200 km by taxi, it took him 5 hours 30 minutes. But if he travels 260
km by train and 240 km by taxi he takes 6 minutes longer. Find the speed of the train and that of the taxi.
157. Sumit travels 600 km to his home partly by train and partly by car. He takes 8 hours if he travels 120 km
by train and the rest by car. He takes 20 minutes longer if he travels 200 km by train and the rest by car.
Find the speed of the train and the car.
42
PAIR OF LINEAR EQUATIONS IN TWO VARIABLES
MATHEMATICS–X
158. A boat can go 20 km upstream and 30 km downstream in 3 hours. It can go 10 km upstream and 20 km
2
downstream in 1 hours. Find the speed of the boat in still water and also the speed of the stream.
3
Problems related to Area and Perimeter (159-160):
159. If in a rectangle, the length is increased and breadth reduced each by 2 units, the area is reduced by 28
sq. units. If, however the length is reduced by 1 unit and the breadth increased by 2 units, the area
increases by 33 sq. units. Find the area of the rectangle.
160. The area of a rectangle gets reduced by 80 sq. units if its length is reduced by 5 units and the breadth is
increased by 2 units. If we increase the length by 10 units and decrease the breadth by 5 units, the area
is increased by 50 sq. units. Find the length and breadth of the rectangle.
Problems related to Fixed Salary, Fare etc. (161-163) :
J
A
J
A
161. The taxi charges in a city comprise of a fixed charge together with the charge for the distance covered.
For a journey of 10 km the charge paid is Rs. 75 and for a journey of 15 km the charge paid is Rs. 110. What
will a person have to pay for travelling a distance of 25 km?
[CBSE 2000]
162. A railway half ticket costs half the full fare and the reservation charge is the same on half ticket as on full
ticket. One reserved first class ticket from Mumbai to Ahmedabad costs Rs. 216 and one full and one half
reserved first class tickets cost Rs. 327. What is the basic first class full fare and what is the reservation
charge?
163. A part of the monthly expenditure of a family is constant and the remaining varies with the price of wheat.
When the rate of wheat is Rs. 250 per quintal, the total monthly expenditure is Rs. 1,000 and when it is Rs.
240 per quintal, the total monthly expenditure of the family is Rs. 980. Find the total monthly expenditure
of the family when the cost of wheat is Rs. 350 per quintal.
Problems related to Geometry (164-165) :
T
I
B
164. In a ABC, C = 3B and 3B = 2 (A + B). Find the angles of the triangle.
165. In a cyclic quadrilateral ABCD, A = (2x + 4)°, B = (y + 3)°, C= (2y + 10)° and D = (4x – 5)°. Find the
angles of the cyclic quadrilateral.
Miscellaneous Problems (166-180) :
166. A man sold a chair and a table together for Rs. 1520 thereby making a profit of 25% on the chair and 10%
on the table. By selling them together for Rs. 1535 he would have made a profit of 10% on the chair and
25% on the table. Find the cost price of each.
167. The incomes of A and B are in the ratio of 8 : 7 and their expenditures are in the ratio 19 : 16. If each saves
Rs. 1250, find their incomes.
168. Students of a class are made to stand in a rows. If one student is extra in a row, there would be 2 rows less.
If one student is less in a row, there would be 3 rows more. Find the number of students in the class.
169. Riya has only 50 p and 25 p coins in her purse. If in all she has 210 coins of the total value of Rs. 82.50,
find the number of coins of each type.
170. A person invested some amount at the rate of 12% simple interest and some other amount at 10% simple
interest. He received yearly interest of Rs. 130. If he had interchanged the amounts invested, he would
have received Rs. 4 as more interest. Find the amount invested in each case.
171. 8 women and 12 girls can finish a piece of work in 10 days, while 6 women and 8 girls can finish it in 14
days. Find the time taken by one woman alone and that by one girl alone to finish the same work.
172. In a school, there are only two sections of X students A and B. If 10 students are sent from A to B, the
number of students in each room is same. If 20 students are sent from B to A the number of students in
A is double the number of students in B. Find the number of students in each room.
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A
MATHEMATICS–X
PAIR OF LINEAR EQUATIONS IN TWO VARIABLES
43
173. A number consists of two digits. When it is divided by the sum of the digits, the quotient is 8. The sum
of the reciprocals of digits is nine times the product of the reciprocals of the digits. Find number.
174. Bhavya 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 she would have scored 50 marks. How many questions were there in the test?
175. A and B each have certain number of oranges. A says to B, ‘‘If you give me 10 of your oranges, I will have
twice the number of oranges left with you. B replies, ‘‘If you give 10 of your oranges, I will have the same
number of oranges as left with you.’’ Find the number of oranges with A and B seperately.
176. The age of father 8 years back was five times that of his son. After 8 years, his age will be 8 years more
than twice the age of his son. Find their present ages.
177. It takes 12 hours to fill a swimming pool using two pipes. If the larger pipe is used for 4 hours and the
smaller pipe for 9 hours, only half the pool is filled. How long would it take for each pipe alone to fill the
pool?
J
A
J
A
178. A takes 3 hours more than B to walk 30 km. But if A doubles his pace, he is ahead of B by 1
1
hours. Find
2
their speeds of walking.
179. A number consists of two digits. When it is divided by the sum of the digits, the quotient is 6 with no
remainder. When the number is divided by 9, the digits are reversed. Find the number.
180. A man lent a part of money at 10% p.a. and the rest at 15% p.a. His annual income is Rs. 1900. If he had
interchanged the rate of interest on two sums, he would have earned Rs. 200 more. Find the amount lent
in each case.
B
HINTS TO SELECTED QUESTIONS
90. Add and subtract given equations to get the value of (x + y) and (x – y).
T
I
95. Take LCM and them simplify the equations.
99.
104.
x y
x
y
1 1
1
1
2 

 2    2. Now, let  a,  b etc.
xy
xy xy
y x
x
y
11
x y 1
11
 2
  x y 
x y
11
2
2
M
A
2
3
7


a  b 2a  b 3 (a  b  1)
Now, equate first two and last two expressions, and get two linear equations in a and b. Solve them
now.
129. For infinitely many solutions,
134. Let present age of mother be x years and present age of daughter be y years. Then,
1
x – 4 = 4 (y – 4) and x  6  2 ( y  6) .
2
140. Let actual prices of TV and fridge be Rs. x and Rs. y respectively. Then,
5
10
10
5
x
y  2000 and
x
y  1500
100
100
100
100
44
PAIR OF LINEAR EQUATIONS IN TWO VARIABLES
MATHEMATICS–X
146. Let unit digit be x and tens digit be y.
Then, number = 10 y + x.
here, 10 y + x = 4 (x + y) and 10y + x = 2xy.
147. Let unit digit be x and ten’s digit be y.
Then, number = 10 y + x.
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Number obtained by reversing order of digits = 10 x + y.
here, (10 y + x) + (10 x + y) = 99 and x – y = ± 3.
151. Let the fraction be
x
.
y
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Then, y = 2x + 4 and y – 6 = 12 (x – 6).
158. Let speed of boat in still water be x km/hr and the speed of the stream be y km/hr.
Then,
20
30
10
20
2

 3 and

1
x y x y
x y x y
3
162. Let basic first class full fare be Rs. x and the reservation charge be Rs. y.
x

Then, x  y  216 and ( x  y )    y   327
2

B
168. Let number of students be x and the number of rows be y. Then, number of students in each row  x .
y
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here, Total number of students = No. of rows × No. of students in each row
x 
x 
x    1 ( y  2) and x    1  ( y  3) .
y 
y 
171. Let one woman along can finish the work in x days and one girl along can finish it in y days. Then,

One woman’s one day’s work  1
x
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and, one girl’s one day’s work 
1
y
8 12 1
6 8 1
here, x  y  10 and x  y  14 .
175. Let no. of oranges with A and B be x and y respectively. Then, x + 10 = 2 (y – 10) and y + 10 = x – 10.
177. Let larger and smaller pipes can fill the pool along in x and y hours respectively. Then,
4 9 1
1 1 1
  and  
x y 2
x y 12
178. Let A’s speed = x km/hr and B’s speed = y km/hr.
30 30
1 1 1

3  
x
y
y x 10
Again, A’s speed = 2x km/hr and B’s speed = y km/hr.
Then,
MATHEMATICS–X
...(1)
PAIR OF LINEAR EQUATIONS IN TWO VARIABLES
45
Then,
Put
30 30 3
2 1 1

   
y 2x 2
y x 10
...(2)
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1
1
 a,  b and solve now..
x
y
180. Let the amount lent bet Rs. x and Rs. y.
Then, 10 x  15 y  1900 and 15 x  10 y  1900  200  2100.
100
100
100
100
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MULTIPLE CHOICE QUESTIONS
Mark the correct alternative in each of the following :
1. The system of equations 3x – 2y = 7 and 7x + 8y = 12 has :
(a) unique solution
(b) no solution
(c) infinitely many solutions
(d) can’t determine
2. The value of x and y satisfying the system of equations 3x + y = – 1 and 2x – 3y = – 8 is:
(a) x =1, y = – 2
(b) x = –1, y = 2
B
(c) x = 2, y = – 3
(d) none of these
3. The value of x and y satisfying the system of equations 2x – 3y + 13 = 0 and 3x – 2y + 12 = 0 is :
(a) x = 2, y = 3
(b) x = – 2, y = – 3
(c) x = – 2, y = 3
(d) none of these
4. The value of x and y satisfying the system of equations x  y  2 and ax – by = a2 – b2 is :
a b
(a) x = a, y = b
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(b) x = – a, y = – b
(c) x = a2, y = b2
(d) none of these
38
21
57
6

 9 is

 5 and
x y x y
x y x y
(c) x = 11, y = 8
(d) none of these
5. The value of x and y satisfying the systems of equations
(a) x = 11, y = – 8
(b) x = – 11, y = 8
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6. The value of k for whcih the system of equations 2x + ky = 1, 3x – 5y = 7 has a unique solution, is :
(a) k   10
(b) k   10
(c) k = 0
(d) k  0
3
3
7. The value of k for which the system of equations 3x – 2y = 8, 6x – ky = 16 has infinitely many solutions,
is
(a) k = 1
(b) k = 2
(c) k = 3
(d) k = 4
8. The value of k for which the system of equations x + 2y = 3 and 5x + ky + 7 = 0 has no solution, is :
(a) 1
(b) 10
(c) 3
(d) 6
9. If the system of equations 3x + y = 1, (2k – 1) x + (k – 1) y = 2k + 1 is inconsistent, then the value of k is:
(a) 1
(b) 0
(c) –1
(d) 2
10. If 2x – 3y = 7 and (a + b) x – (a + b – 3) y = 4a + b represent coincident lines, then a and b satisfy the
equation:
(a) 5a + b = 0
46
(b) a – 5b = 0
(c) a + 5b = 0
PAIR OF LINEAR EQUATIONS IN TWO VARIABLES
(d) 5a – b = 0
MATHEMATICS–X
VERY SHORT ANSWER TYPE QUESTIONS (1 MARK QUESTIONS)
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1. Form a pair of linear equations for the following problem :
Fourteen students of class X took part in a quiz and the number of boys is 2 more than the number of
girls.
2. Is x = 2 and y = – 1 a solution of the pair of linear equations x + 2y = 0 and 3x + 4y = 20?
3. How many solutions do two linear equations in two variables have, if their graph are parallel?
4. What is the minimum and maximum number of solutions that a system of simultaneous linear equations
can have, if it is consistent system?
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5. How many solutions will the following pair of linear equations have?
7x – 4y + 11 = 0
2x – 9y + 15 = 0
6. The graph of y = – 3 is a straight line parallel to which axis?
7. Given a linear equation 3x – 5y = 15. Write another linear equation in two variables, such that the
geometric representation of the pair so formed is coincident lines.
B
8. For what value of k, will the equations 2x – y + 8 = 0 and 4x – ky + 16 = 0 represent coincident lines?
9. For what value of k, will the equations 4x + my = 8 and 3x – 5y + 7 = 0 represent parallel lines?
10. For what value of k, will the equations 2x + ky = 7 and 3x + 9y = 13 may have a unique solution?
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11. Find a point on y-axis satisfying 3x – 4y = 12.
12. At what point does the line 4x + 5y = 20 intersects x-axis?
13. Give an equation of a line which passes through the origin.
14. If a system of equation is inconsistent, then what type of graph the equations will have?
15. What are the coordinates of the point of intersection of two lines 3x + 2y = 0 and 2x – y = 0?
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16. What are the points of intersection of the line x  y  3  0 with x-axis and with y-axis?
a b
17. If x = – y, x = – 3, and x-axis form a triangle as shown, find the co-ordinates of the three vertices of the
triangle.
X
MATHEMATICS–X
Y
y =–x
–3 –2 –1
X
O
Y
PAIR OF LINEAR EQUATIONS IN TWO VARIABLES
47
18. What is the area of the triangle formed by the lines 2x + y = 6, 2x – y + 2 = 0 and x-axis in the figure given
below ?
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Y
(1, 4)
4
3
2
1
–4
–3
–2
0
–1
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X
1
2
3
–1
4
5
6
6
y=
2x+
2x–
y+
2=
0
X
–2
–3
–4
Y
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B
19. For what value of k, the equations 2x – ky + 3 = 0 and 3x + 2y – 1 = 0 has no solution?
20. For what value of k, the equations kx + 3y = k – 3 and 12 x + ky = k has infinitely many solutions?
PRACTICE TEST
M.M : 30
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General Instructions :
Time : 1 hour
Q. 1-4 carry 2 marks, Q. 5-8 carry 3 marks and Q. 9-10 carry 5 marks each.
1. Solve for x and y :
6
4 x   15
y
4
7 y0
y
2. Solve for x and y using cross-multiplication method :
x y  ab
ax – by = a2 – b2
3. For what value of k will the following system of linear equations has no solution?
3x + y = 1
(2k – 1) x + (k – 1) y = 2k + 1
3x 
48
PAIR OF LINEAR EQUATIONS IN TWO VARIABLES
MATHEMATICS–X
4. Show that the following system of equations has an infinite number of solutions :
3x + 5y – 8 = 0
9x + 15y = 24
5. Solve for x and y :
37 x + 53 y = 320
53 x + 37 y = 400
6. In a two digit number, the units digit is twice the ten’s digit. If 27 is added to the number, the digits
interchange their places. Find the number.
7. Solve graphically the system of equations :
x + y = 3 ; 3x – 2y = 4
9. On the same axes, draw the graph of the following equations :
x – 5y + 14 = 0
x – 2y + 8 = 0
2x – y + 1 = 0
Hence obtain the vertices of the triangle so formed.
B
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8. The larger of two supplementary angles exceeds the smaller by 18°. Find them.
10. A man travels 600 km partly by train and partly by car. If he covers 400 km by train and the rest by the car,
it takes him 6 hours and 30 minutes. But, if he travels 200 km by train and the rest by the car, he takes half
an hour longer. Find the speed of the train and that of the car.
ANSWERS OF PRACTICE EXERCISE
1.
15.
19.
23.
26.
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No
11. x = 0, y = 3
12. x = 2, y = 3 13. No solution
x = 2, y = – 3
16. x = 1, y = 1 17. x = – 2, y = 1
x = 1, y = 2 20. Infinitely many solutions
21. No solution
Infinitely many solutions
24. Unique solution
Unique solution
27. Unique solution
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29. Infinitely many soluions
38.
41.
44.
46.
48.
50.
52.
54.
57.
60.
63.
30. Unique solution
14. x = 2, y = 4
18. No solutoin
22. Unique solutoin
25. Infinitely many soluions
28. No solution
 4
37. x  3, y  2;  0,  , (0,8)
 5
x = 2, y = 3; (0, 6), (0, –2)
39. x = 2, y = –1; (0, 5), (0, –5)
40. x = 1, y = 2; (–2, 0), (5, 0)
x = 2, y = 2; (–2, 0), (3, 0)
42. x = 1, y = 3; (–2, 0), (3, 0)
43. x = 1, y = 4; Area = 8 sq. units
x = 1, y = 4; Area = 12 sq. units
45. x = 2, y = 3; Area = 7.5 sq. units
x = 2, y = 1; Area = 5 sq. units
47. x = 5, y = 0; Area = 17.5 sq. units
x = 3, y = 2; Area = 13.5 sq. units
49. x = 0, y = 5; (–6, 0), (0, 5), (4, 0)
x = – 2, y = 3; (–2, 3), (–3, 0), (2, 0)
51. x = 3, y = 4; (3, 4), (2, 0), (–3, 0)
x = 3, y = 2; (3, 2), (0, –1), (0, 11)
53. x = 3, y = 2; (0, 4), (0, –1), (3, 2)
x = 2, y = – 1; (2, –1), (0, 5), (0, –5)
55. (0, 0), (4, 4), (6, 2)
56. (3, 4), (0, 2), (6, 2)
(–4, 2), (1, 3), (2, 5)
58. (0, 5), (0, –5), (5, 0)
59. (3, –5), (3, –11), (–1, –3)
(6, 0), (3, 0), (1, 2)
61. x = 5, y = – 2
62. x = 3, y = – 2
x = – 3, y = – 1
64. x = – 2, y = – 3
65. x = 2, y = 3
MATHEMATICS–X
PAIR OF LINEAR EQUATIONS IN TWO VARIABLES
49
67. x 
66. x = 6, y = 36
1
1
,y
2
2
1
9
68. x  , y 
3
5
69.
x
2
1
,y
3
3
70. x = 4, y = 3
7
13
71. x   , y 
5
5
72.
x
68
25
,y
47
47
1
1
73. x   , y 
2
3
74. x = 3, y = – 1
75. x = 2, y = 1
76. x = a, y = b
77. x = – 2, y 
78. x = 1, y = – 1
79. x = 4, y = 3
81. x = 15, y = 5
82. x  
1
x  , y  2
5
102.
100. x 
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x
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117. k = 12
120. k = 4
123. k = –5
50
1
7
,y
24
4
86. x = 1, y = – 1
89. x = 20, y = 12
83. x 
92. x = 2a, y = – 2b
B
95. x 
23
83
,y
5
5
98. x = 0, y = 0 or x  1, y 
3
2
101. x = 8, y = 3
104. x  3.2, y  2.3
109. x  m  n, y  m  n
112. No solution
110. x = a, y = – b
113. Unique solution, x = – 1, y = 2
115. k  6
116. k  2
118. k = 2
119. k 
121. k = 2
124. k = 4
5
2
122. k = 5
125. k = – 4
4
17
11
128. a  , b 
3
4
5
a = 5, b = 1
130. a = 8, b = 5
Father’s age = 42 years, son’s age = 10 years
132. 45 years
Father’s age = 36 years, son’s age = 12 years
Mother’s age = 44 years, daughter’s age = 14 years
35 years, 15 years
136. Father’s age = 33 years, son’s age = 10 years.
Rs. 155
138. Rs. 150, Rs. 500
139. Rs. 20 and Rs. 100
Rs. 20,000 and Rs. 10,000
141. 17 and 9
142. 24 and 60
126. k = 2
129.
131.
133.
134.
135.
137.
140.
107. x = a, y = b
97. x  a 2 , y  b 2
1
1
x , y
2
4
5
1
x
,y
4
4
1
5
,y
2
4
108. x = a2, y = b2
111. infinitely many solutions
2
114. k  
3
105.
106. x  1, y  1
94. x  3, y  2
96. x = 2, y = – 1
99.
89
89
,y 
4080
1512
a 2  b 2  ab
ab
,y
103. x 
a b
ab
1
1
91. x  , y 
6
4
90. x = 2, y = 3
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1
25
,y
11
11
85. x = a2, y = b2
88. x = 4, y = 2
84. x = – 2, y = 2
87. x = a, y = b
93.
1
3
80. x = 5, y = – 20
127. k  
PAIR OF LINEAR EQUATIONS IN TWO VARIABLES
MATHEMATICS–X
143. 57
146. 36
149.
144. 93
147. 63 or 36
5
7
150.
8
15
155. 10 km/hr and 2 km/hr
158. 20 km/hr, 10 km/hr
161. Rs. 180
152.
145. 45
148. 72
3
7
151.
153. 40 km/hr and 30 km/hr
154. 6 km/hr and 2 km/hr
156. 100 km/hr, 80 km/hr
157. 60 km/hr, 80 km/hr
159. 253 sq. units
160. 40 units and 30 units
162. Fare = Rs. 210, Reservation charge = Rs. 6
163. Rs. 1200
164. 120°, 40° and 20°
166. CP of chair = Rs. 600, CP = table = Rs. 700
167. A’s income = Rs. 6000, B’s income = Rs. 5250
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165. 127°, 110°, 53° and 70°
168. 60
169. 25 p coins = 90, 50 p coins = 120
170. Rs. 500, Rs. 700
171. 140 days, 280 days
173. 72
174. 20
176. 16 years and 48 years
177. 20 hours and 30 hours
178. A’s speed =
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7
18
10
km/hr , B’s speed = 5 km/hr
3
179. 54
B
172. 100, 80
175. 70, 50
180. Rs. 10000 at 10% and Rs. 6000 at 15%
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ANSWERS OF MULTIPLE CHOICE QUESTIONS
1. (a)
6. (b)
2. (b)
7. (d)
3. (c)
8. (b)
4. (a)
9. (d)
5. (c)
10. (b)
ANSWERS OF VERY SHORT ANSWER TYPE QUESTIONS
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1. Let x and y denote the number of boys and girls respectively. Then, x = y + 2 and x + y = 14.
2. No
3. No solution
4. One and infinite
5. unique
7. 6x – 10y = 30
8. k = 2
9. k  20
3
10. All real numbers except 6.
11. (0, –3)
12. (5, 0)
13. y = mx
14. parallel lines
15. x = 0, y = 0
16. (–3a, 0), (0, –3b)
18. 8 sq. units
19.
4
3
6. x-axis
17. (0, 0), (–3, 0), (–3, 3)
20. 6
ANSWERS OF PRACTICE TEST
1. x = 3, y = 2
2. x = a, y = b
3. k = 2
5. x = 6.5, y = 1.5
7. x = 2, y = 1
8. 99°, 81°
9. (2, 5), (–4, 2), (1, 3)
10. 100 km/hr, 80 km/hr
MATHEMATICS–X
PAIR OF LINEAR EQUATIONS IN TWO VARIABLES
6. 36
51