PAIR OF LINEAR EQUATIONS
IN TWO VARIABLES
3. The length of a rectangle is 3 units more than half of its breadth. Represent the situation in the form of
a linear equation in two variables in standard form.
4. Represent the following equations in the form of linear equations in two variables.
2x = 5
y = 3
5. Mohan bought 2 kg of apples and 3 kg of mangoes for Rs. 220. Khushi bought 1 kg of apples and 2 kg of
mangoes for Rs. 135. Represent both the situations algebraically.
6. Check whether the following pair of equations represents parallel lines or not.
2x + 4y = 7
3x + 6y = 21
7. Write the conditions for the pair of equations; mx + ny = p, kx + sy = t to have
(i) no solution
(ii) a unique solution.
8. Find the value of m such that the following pair of equations has a unique solution
x + my = 3
x + y = 2
9. Find the value of m, so that the following pair of linear equations has infinitely many solutions.
x + 4y = 5
2x + 8y = m
10. The present age of Ram’s father is 2 years more than seven times Ram’s present age. Also, the sum of
their present ages is 32 years. Express the situation algebraically.
Time Tracker:
11. Use elimination method to find all the possible solutions of the following pair of linear equations (if theyexist):
3x + 2y = 6
6x + 4y = 9
12. Solve the following pair of equations by substitution method.
2x + y = 4
7x + 5y = 10
OR
Solve the following pair of linear equations by substitution method:
2x + 3y = 10
7x – 3y = 8
Time Tracker:
13. Solve the following system of equations graphically:
x + 2y = 8
x – y = –1
14. Solve the following equations using cross–multiplication method.
2x + 3y = 7
3x + 2y = 8
15. Use elimination method to find all the possible solutions of the following pair of linear equations.
3x – 2y = – 2
x + y = – 4
16. Find the value of k such that the following pair of linear equations has infinitely many solutions
kx + 2y = k + 1
3x + (k – 1)y = 4
OR
Solve the following pair of equations by reducing it to a pair of linear equations.
2 – y
3 = –1
x
3 + y
4 = 7 (x 0, y 0)
x
Time Tracker:
17. Johnson travels 880 km to his home partly by train and partly by car. He takes 9 hrs, if he travels 180 km
by train and the remaining by car. He takes 12 minutes more, if he travels 360 km by train and the
remaining by car. Find the speed of the train and the car separately.
18. Check whether the given pair of equations has no solution, unique solution or infinitely many solutions.
If it has a solution, find it using elimination method and verify your answer graphically.
x – 2y = 2
4x – 2y = 5
OR
In a rectangle, if the length is increased by 2 units and the breadth is reduced by 4 units, the area gets
reduced by 26 units. If the length is reduced by 3 units and the breadth is reduced by 6 units, the area
gets reduced by 111 units. Find the area of the original rectangle.
Time Tracker:
______________________________________________________________________
SOLUTIONS
1 + y
1 = 4 … (i)
1. We have x
3
3
1 – y
1 = 3 … (ii)
and x
2
2
1 = u and y
1 = v in (i) and (ii), we get
Put x
u + 3
v = 4 or u + v = 12
3
[½]
and
u – 2
v = 3 or u – v = 6
2
[½]
2. The given equations are n + m = 16 … (i)
and n – m = 15 … (ii)
(i) + (ii) 2n = 31
[½]
31
n = 2
[½]
3. Let the length of rectangle be x units and breadth be y units.
According to the question
y + 3
x = 2
[½]
y +
6
x = 2
2x = y + 6
2x – y = 6
[½]
This is the required representation.
4. 2x = 5
2x + 0.y = 5
[½]
and y = 3
0.x + y = 3
[½]
5. Let price of 1 kg of apples be Rs. x and price of 1 kg of mangoes be Rs. y
According to the question
2x + 3y = 220
[½]
x + 2y = 135
[½]
6. Here a
= 2, b
= 4, c
= 7
1
1
1
a
= 3, b
= 6, c
= 21
[½]
2
2
2
a
2
b
4
2 ,
c = 21
7 = 3
1
Clearly, 3
= , 6
= = 3
1
1
1
a
b
c
2
2
2
a =
b
c
Here
1
1
1
a
b
c
2
2
2
The given pair of equations represents parallel lines.
[½]
7. The given equations are
mx + ny = p
kx + sy = t
(i) The condition for the equations to have no solution is
m = s
n t
p
k
[½]
(ii) The condition for the equations to have a unique solution is
m s
n
k
[½]
8. We have
x + my = 3 ……(i)
x + y = 2 ……(ii)
The given pair of equations has a unique solution
1 1
m
if 1
[½]
i.e. m 1
Therefore, the given pair of equations has a unique solution for all values of m except 1.
[½]
9. The given equations are
x + 4y = 5 ……(i)
2x + 8y = m ……(ii)
For infinitely many solutions
____________________________________________________________
4 = m
5
2
[½]
1 = m
5
i.e. 2
m = 10
[½]
10. Let Ram’s present age = x years
and present age of his father = y years
According to the first condition
y = 7x + 2
[½]
According to the second condition
x + y = 32
[½]
11. The given equations are
3x + 2y = 6 ……(i)
6x + 4y = 9 ……(ii)
Multiply equation (i) by 2 and equation (ii) by 1, we get
6x + 4y = 12 ……(iii)
and 6x + 4y = 9 ……(iv)
[1]
Subtract equation (iv) from equation (iii)
6x + 4y – 6x – 4y = 12 – 9
0 = 3, which is not possible.
Therefore, the pair of equations has no solution.
[1]
12. The given equations are
2x + y = 4 … (i)
7x + 5y = 10 … (ii)
Now, (i) y = 4 – 2x … (iii)
[½]
Substituting the value of y in equation (ii), we get
7x + 5(4 – 2x) = 10
[½]
7x + 20 – 10x = 10
–3 x + 20 = 10
–3 x = – 10
10
x = 3
[½]
Substituting the value of x in (iii), we get
____________________________________________________________
_________________________________________________________________________________
10
y = 4 – 2 × 3
20 = – 3
8
y = 4 – 3
[½]
-
10 , y = 3
8
The solution x = 3
OR
The given equations are
2x + 3y = 10 ….(i)
7x – 3y = 8 ….(ii)
From (i), 3y = 10 – 2x
Substituting the value of 3y in (ii), we get
7x – (10 – 2x) = 8
[½]
7x – 10 + 2x = 8
9x = 18
x = 2
[½]
Put the value of x in (i), we get
2(2) + 3y = 10
[½]
3y = 6
y = 2
[½]
x = 2, y = 2 is the solution of the given pair of equations.
13. The given equations are
x + 2y = 8 … (i)
x – y = – 1 … (ii)
8 -
x
equation (i) y = 2
8 - = 2
2
6 = 3
when x = 2, y = 2
8+ = 2
2
10 = 5
when x = – 2, y = 2
In tabular form
x
2 –2
y
3 5
[½]
equation (ii) y = x + 1
…
_________________________________________________________________________________
when x = 1, y = 1 + 1 = 2
when x = 4, y = 4 + 1 = 5
In tabular form
x
1 4
2 5
y
[½]
Plot the points, say A(2, 3), B(–2, 5) and C(1, 2), D(4, 5) on the graph and draw lines AB and CD
representing the lines x + 2y = 8 and x – y = – 1.
Y
D (4, 5)
•
•
5
B (–2, 5)
4
•
3
A (2, 3)
•
2
C (1, 2)
1
X’
X
1 3 5
2
4 6 7 9
–5
–3
–2
–1
8
10
–4
–1
–2
–3
–4
–5
Y’
[1½]
Now lines intersect at point (2, 3).
Therefore x =2 and y = 3 is the required solution of the given equations.
[½]
14. The given equations are
2x + 3y = 7 2x + 3y – 7 = 0
3x + 2y = 8 3x + 2y – 8 = 0
[½]
Now by cross–multiplication method,
x
y
1
)
[1]
- = )
-
-
- = )
-
-
-
(
3
)
(
8
)
(
2
)
(
7
(
7
)
(
3
)
(
8
)
(
2
(
2
)
(
2
)
(
3
)
(
3
x
y
1
14
- = 16
24
+
- = 9
21
+
4
-
x
y
1
=
=
5
-
-
-
10
5
____________________________________________________________
8
_________________________________________________________________________________
- = 2
10
x
1
i.e. 10
[½]
- = 5
- x = 5
-
y
1
- = 1
5
and 5
[½]
- = 5
- y = 5
-
Therefore x = 2 and y = 1
Which is the required solution of the given equations.
[½]
15. The given equations are
3x – 2y = – 2 … (i)
and x + y = – 4 … (ii)
Multiply (ii) by 3, we get
3x + 3y = – 12 … (iii)
[½]
Subtract equation (iii) from (i)
3x – 2y = – 2
3x + 3y = – 12
– – +
– 5y = 10
10
y = 5
[1]
- = – 2
Put the value of y in (ii)
x – 2 = – 4
[½]
x = – 4 + 2 = – 2
x = – 2
[½]
The solution of the given equations is
x = – 2 and y = – 2
[½]
16. The given equations are kx + 2y = k + 1
and 3x + (k – 1)y = 4
Here a
= k, b
= 2, c
= k + 1
1
1
1
a
= 3, b
= k – 1, c
= 4
[½]
2
2
2
Since the given pair of equations has infinitely many solutions
k = 1
2
k +
1
3
[½]
- = 4
k
k = 1
2
3
k
-
k(k – 1) = 2 × 3
k
– k = 6
2
k
– k – 6 = 0
2
k
– 3k + 2k – 6 = 0
[½]
2
k(k – 3) + 2(k – 3) = 0
(k + 2) (k – 3) = 0
k = – 2 or 3
[½]
k
1
2 +
Also, 4
=
k
-
1
(k – 1) (k + 1) = 2 × 4
k
– 1 = 8
2
k
= 8 + 1 = 9
2
k = 9= ±3
for k = –3
- ,which is not true
3
2
=
3
-
4
k = 3 is the required value.
[1]
OR
We have
2 – y
3 = –1 …(i)
x
3 + y
4 = 7 …(ii)
x
1 = u and y
1 = v in equations (i) and (ii), we get
Putting x
2u – 3v = – 1
….(iii)
3u + 4v = 7 ….(iv)
[1]
Multiplying (iii) by 3 and (iv) by 2, we get
6u – 9v = – 3
……(v)
6u + 8v = 14 …..(vi)
Now, subtracting (vi) from (v), we get
– 9v – 8v = – 3 – 14
or – 17v = – 17
or v = 1
Substituting the value of v in (iii)
2u – 3(1) = – 1
____________________________________________________________
__________
or 2u = – 1 + 3
or u = 1
[1]
1 or x = u
1
Now, u = x
1 = 1
x = 1
x = 1
1 or y = v
1
and v = y
1 = 1
y = 1
y = 1
x = 1, y = 1 is the solution of the given pair of equations.
[1]
17. Let the speed of train be t km/hr and the speed of car be c km/hr
Total distance to be covered = 880 km
dis
tan
ce
We know, time = speed
[½]
When distance covered by train = 180 km,
180 hours
time taken by train to cover this distance = t
and distance traveled by car = 880 – 180 = 700 km
700 hours
Time taken by car to cover this distance = c
According to the question
180 + c
700 = 9 ……(i)
t
[2]
Again, when distance covered by train = 360 km
360 hours
Time taken by train to cover the distance = t
and distance covered by car = 880 – 360 = 520 km
520 hours
Time taken by car to travel this distance = c
360 + c
520 = 5
46 ……(ii)
t
[2]
Multiplying equation (i) by 2
360 +
1400
= 18 ……(iii)
[½]
t
c
Subtracting (iii) from (ii)
520 - = 5
1400
46 –18
c
c
- = 5
-
880
44
c
c = 100 km/hr
[½]
Substituting the value of c in equation (ii)
360 +
520
46
=
t
100
5
520
360 =
46
– 100
t
5
t = 90 km/hr
[½]
18. Here a
= 1, b
= – 2, c
= 2
1
1
1
a
= 4, b
= – 2, c
= 5
[½]
2
2
2
a
1
b
-
2
c = 5
2
= , 2
= = 1,
=
1
Now, 4
1
1
-
a
b
c
2
2
2
a
b
Clearly,
[½]
1
1
a
b
2
2
The given system of equations has a unique solution
[½]
Now, the given equations are
x – 2y = 2 … (i)
4x – 2y = 5 … (ii)
Subtract (ii) from (i)
– 3x = – 3
x = 1
[½]
Substituting value of x in (i), we get
1 – 2y = 2
– 2y = 1
-
1
y = 2
[½]
- is the solution of the given pair of equations
1
x = 1 and y = 2
____________________________________________________________
Verification:
We have the given equations
x – 2y = 2 4x – 2y = 5
– 2y = 2 – x – 2y = 5 – 4x
x -
2
4 -
x
5
y = 2
[½]
y = 2
[½]
3 = 1.5
when x = 0, y = – 1 when x = 2, y = 2
15 = 7.5
x = 4, y = 1 x = 5, y = 2
In tabular form In tabular form
x
2 5
x
0 4
y
1.5 7.5
y
–1 1
[½]
[½]
Plot points (0, –1), (4, 1) and (2, 1.5), (5, 7.5) on the graph and draw lines by joining the points.
Y
8
(5, 7.5)
7
6
5
4
3
2
(2, 1.5)
1
(4, 1)
X’
X
–7
2
3 5
4
6 7
8
–6
–5
–3
–2
–1
–4
–1
(0, –1)
–2
1
–3
–4
–5
[1½]
Y’
Therefore (1, –0.5) is the solution of the given equations. Hence verified.
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___________________________________________________________________
OR
Let the length of rectangle = x units
Let the breadth of rectangle = y units
Therefore the area of rectangle = xy square units
[½]
According to the first condition
(x + 2) (y – 4) = x y – 26
[1]
xy + 2y – 4x – 8 = xy – 26
– 4x + 2y = – 18
2x – y = 9 … (i)
[½]
According to the second condition
(x – 3) (y – 6) = xy – 111
[1]
xy – 3y – 6x + 18 = xy – 111
6x + 3y = 129
2x + y = 43 … (ii)
[½]
Adding equations (i) and (ii), we get
4x = 52
x = 13
[1]
Substituting value of x in equation (i), we get
2 × 13 – y = 9
– y = – 17 y = 17
[1]
Area = xy = 13 × 17 = 221 sq. units.
[½]
____________________________________________________________
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