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Revision Question Bank
Pair of Linear Equations in Two Variables
1.
Draw the graph of the equations x – y +1 = 0 and 3 x + 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.
Solution :
We have
x–y+1=0
3x + 2y – 12 = 0
Graph of the equation x – y + 1 = 0
 y = x + 1.
When x = 0, we have y = 1
When x = 1, we have y = 0
Thus, we have the following table
x
0
–1
y
1
0
Plotting the points A(0, 1) and B(–1, 0) on the graph paper on a suitable scale and drawing a line joining
them.
Graph of the equation 3x + 2y – 12 = 0
 2y = 12 – 3x
When x = 0, we have y = 6
When x = 4, we have y = 0
Thus, we have the following table
x
0
4
y
6
0
Plotting the points C(0, 6) and D(4, 0) on the same graph paper and joining them, we obtain the
following graph : –
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It is evident from the graph that the two lines intersect at point P(2, 3). The area enclosed the lines and
x-axis is shown in the shaded region.
2.
Solve the following system of linear equations
8v–3u = 5uv and 6v– 5u=–2uv.
Solution :
Clearly, the given equations are not linear in the variables u and v but can be reduced into linear
equations by an appropriate substitution.
If we put u = 0 in either of the two equations, we get v = 0.
Thus, u = 0 and v = 0 form one solution of the given system of equations.
To find the other solutions we assume that.
Since, u  0 and v  0 Therefore, uv  0 .
On dividing each of the given equations by uv, we get
and
3 3
 5
u v
..(i)
6 5
  2
u v
..(ii)
on putting
1
1
= y, the above equations become
 x and
u
v
8x – 3v = 5
...(iii)
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and
6x – 5y = –2
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..(iv)
On multiplying Eq. (iii) by 3 and Eq. (iv) by 4, we get
24x – 9y = 15
...(v)
and 24x – 20y = – 8
...(vi)
On subtracting Eq. (vi) from Eq. (v), we get
11y = 23
 y
23
11
On putting y =
8x 
23
in Eq. (iii), we get
11
69
5
11
 8x  5 
 8x 
60
55  69
 8x 
11
11
124
124 31
x

11
8 11 22
Now, x 
31
22
1 31
22

u
u 22
31
and
y
23
11
1 23
11

v
v 11
23
Hence, the given system of equations has two solutions given by
(i) u = 0 and v = 0
(ii) u 
3.
22
11
and v 
31
23
Solve for x and y by using the method of elimination
0.4x + 0.3y =1.7and 0.7x– 0.2y = 0.8.
Solve for x and y by using the method of elimination
0.4x + 0.3y =1.7and 0.7x– 0.2y = 0.8.
Solution :
Given equations are 0.4x + 03y = 1.7
and
0.7x - 0.2y = 0.8
... (i)
... (ii)
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On multiplying each one of the equations by 10, we get
and
4x + 3y= 17
... (iii)
7x– 2y =8
...(iv)
On multiplying Eq. (iii) by 2 and Eq. (iv) by 3, we get
and
8x + 6y = 34
... (v)
21x – 6y = 24
...(vi)
On adding Eqs. (v) and (vi), we get
29x = 58  x =
58
=2
29
On putting x = 2 in Eq. (iv), we get
7 × 2 – 2y = 8  2y= 14
–8
 2y = 6  y = 3
 x = 2 and y = 3
4.
Solve the following system of linear equations by the method of cross-multiplication
2x– y – 3 = 0 and 4x + y– 3 = 0.
Solution :
The given system of equations is
2x
and
–y–3=0
4x + 3y –3 = 0
By cross-multiplication method, we have

=

x
y

 1   3  1   3 2   3  4   3
1
2  1  4   1 
x
y
1


3  3 6  12 2  4
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x y 1


6 6 6

 x
6
6
 1 and y    1
6
6
Hence, the solution of the given system of equations is x = 1 and y = –1.
5.
Solve the following system of equations in x and y
(a – b)x + (a + b)y =a2– 2ab– b2 and(a + b)(x + y) = a2 + b2.
Solve the following system of equations in x and y
(a – b)x + (a + b)y =a2– 2ab– b2 and(a + b)(x + y) = a2 + b2.
Solution :
The given system of equations may be written as
(a – b)x + (a + b)y – (a2 –2ab –b2) = 0
and
(a + b)x + (a + b) y – (a2 + b) = 0
By cross-multiplication method, we have
x
 a  b  a  b  a  b   a2  2ab  b2

=
=
2
2



y
 a  b   a  b  a  b    a2  2ab  b2

2
2


1
 a  ba  b   a  b 
2

x
  a  b a  b   a  b  a2  2ab  b2


y
  a  b a  b   a  b  a2  2ab  b2

=


=



2
2
2
2




1
 a  ba  b   a  b 
2
x
 a  b[ a2  b2   a2  2ab  b2 ]
y
 a  b a  2ab  b2  a  b a2  b2

2



1
 a  ba  b  a  b 
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x
 a  b 2ab  b2



=
y
a  a b  3ab  b  a3  ab2  a 2b  b3
=
1
  a  b  2b
3
2
2
x

2b  a  b
2
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3
y
1

2
4ab 2b  a  b

2b  a  b 
2
 x
2b  a  b 
and y 
ab
4ab2
2ab

2b  a  b a  b
Hence, the solution of the given system of equations is
x = a + b and y = 
6.
2ab
.
ab
A Lending library has a fixed charge for the first three days and an additional charge for each day,
thereafter and 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 fixed charge for first 3 days = Rs x
and additional charge for each day = Rs y
According to the question,
x + 4y = 27
and
x + 2y = 21
...(i)
... (ii)
On subtracting Eq. (ii) from Eq. (i), we get
x  4y  27
x  2y  21
  
2y  6
 y
6
3
2
On substituting the value of y in Eq, (i), we get
x + 4 × 3 = 27  x +12 = 27
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x = 15
Hence, charge for first 3 days = Rs 15 and extra charge per day = Rs 3.
7.
For the following system of equations, determine the value of k for which the given system of equations
has a unique solution. 2x + 3y– 5 = 0 and kx– 6y– 8 = 0
Solution :
The given system of equations is
2x + 3y
and
kx
–5=0
– 6y – 8 = 0
It is of the form a1x +b1y + c1 = 0
and
a2x + b2y + c2 = 0
where,
a1 = 2,b1 =3,c1 = –5
and
a2 = k, b2 =-6, c2 = – 8
For a unique solution, we must have
a1 b1

a2 b2
2 3

 k   4 i.e.,
k 6
Hence, the given system of equations will have a unique solution for all real values of k other than – 4.
8.
Five years ago, Ramesh was thrice of Shyam's age. Ten years later, Ramesh will be twice of Shyam's age.
How old are Ramesh and Shyam?
Solution :
Let the present age of Ramesh be x years
2nd the present age of Shyam be y years.
Five years ago, Ramesh’s age = (x – 5)years
Shyam’s age = (y – 5) years
Using the given information, we have
x – 5 = 3(y – 5)

x – 5 = 3y – 15

x – 3y = – 10
…(i)
Also, ten years, hence, Ramesh’s age = (x + 10) years
Shyam’s age = (y + 10) years
Using the given information, we have
(x + 10) = 2(y + 10)
 x + 10 = 2y + 20
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 x – 2y = 10
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..(ii)
Subtracting equation (ii) from (i), we get
–3y + 2y = – 10 – 10
– y = – 20
 y = 20
Putting y = 20 in equation (i), we get : –
x – 3(20) = – 10
x – 60 = – 10
x = – 10 + 60
x = 50
Hence, present age of Ramesh is 50 years 2nd present age of shyam is 20 years.
9.
Father's age is 3 times the sum of ages of his two children. After 5 yr, his age will be twice the sum of
ages of the two children. Find the age of father.
Solution :
Let father's age be x year and let sum of ages of his two children by y year.
Condition I
x = 3y
...(i)
After 5 yr, age of father = (x + 5)
After 5 yr, sum of ages of two children =(y +10)
Condition II
x + 5 = 2(y+10)
...(ii)
On putting the value of x from Eq. (i) in Eq. (ii), we get
3y + 5 = 2(y+10)
 3y+5 = 2y + 20
 3y – 2y = 20 – 5  y=15
On putting the value of y in Eq. (i), we get
x = 3(15) = 45
Hence, age of father is 45 yr.
10. Draw the graph of the pair of equations 2x + y=4 and 2x - y=4. Write the vertices of the triangle formed
by these lines and the y-axis. Find the area of this triangle.
Solution :
The given pair of linear equations 2x + y =4 and 2x – y = 4.
Table for line 2x + y = 4  y =4 – 2x
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x
0
2
y = 4 – 2x
4
0
Points
A(0, 4)
B(2, 0)
Now, we plot all these points on a graph paper and join them to get a line AB.
and table for line 2x – y = 4 
x
y = 2x – 4
Points
y = 2x – 4
0
–4
C(0, 4)
2
0
B(2, 0)
Now, we plot all these points on a graph paper and join them to get a line BC.
Graphical representation of both lines are given below
The vertices formed by a triangle are A (0, 4), B (2, 0) and C(0,–4).
Here,
AC = OA + OC = 4 + 4 = 8
and
OB =2
1
 Area of  ABC × base ×height
2
=
1
× AC × OB
2
=
1
× 8 × 2 = 8 sq units.
2
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Chapter Test {Polynomial}
M:M: 40
M: Time: 40
Each Question carry 4 marks
1.
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 me 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 separately.
[4]
Solution :
Let A has x oranges and B has y oranges. According to the question,
x + 10 = 2(y – 10)
 x + 10 = 2y – 20
 x – 2y = – 20 – 10
 x – 2y = – 30
..(i)
 x – y = 20
On subtracting Eq. (ii) from Eq. (i), we get
x  2y  30
x  y  20
  
 y  50
 y = 50
On putting y = 50 in Eq. (ii), we get
x – 50 = 20
 x = 20 + 50
 x = 70
Hence, A has 70 oranges and B has 50 oranges.
2.
Draw the graph of the equations 4x – y = 4 and 4x + y = 12. Determine the vertices of the triangle formed
by the line representing these equations and the X-axis. Shade the triangular region, so formed.
[4]
Solution :
The given equations are
4x – x = 4
...(i)
and
...(ii)
4x + y –12
From Eq. (i), we have
4x – y = 4  y = 4x – 4
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Put x = 0, then y = 4 × 0 – 4 = – 4
Put x = 1, then y = 4 – 4 = 0
Put x
= 2, then y =4 × 2 – 4 = 4
x
0
1
2
y
–4
0
4
From Eq. (ii), we have
4x + y = 12  y = 12 – 4x
Put
x = 3, then y = 12 – 4 × 3 = 0
Put x = 3, then y = 12 – 4 × 2 = 4
Put x = 4, then y = 12 – 4 × 4 = – 4
x
3
2
4
y
0
4
–4
Now, plotting the points (0, –4), (1, 0), (2, 4), (3, 0) and (4, –4) on the graph paper with a suitable scale and
drawing lines joining them equation wise, we obtain the graph of the lines represented by the equations 4x
– y = 4 and 4x + y = 12,
From the graph, we observe that points A(2, 4), B (1, 0) and C(3, 0) are the three vertices of the triangle
formed by the given lines and the X-axis.
3.
An express train takes 1 h less than a passenger train to travel 132 km between Mysore and Bangalore
(without taking into consideration the time for which they stop at intermediate stations). If average
speed of the express train is 11 km/h more than average speed of the passenger train, then find the
speed of the express train.
[4]
Solution :
Let average speed of the passenger train = xkm/h and average speed of the express train
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= (x+11)km/h
Time taken to cover 132 km by passenger train

132
h
x
and time taken to cover 132 km by express train
=
132
h
x  11
According to the question,

132 x  1  132x
132 132

1
x
x  11
1
x  x  11
 132x + 1452 – 132x = x2 +11X
 x2+11x – 1452 = 0
 x2 +44x – 33x – 1452 = 0
[by splitting middle term]
 (x – 33)(x +44) = 0
 x – 33 = 0 and
 x = 33
and
x + 44= 0
x = – 44 [impossible]
Hence, the speed of express tram is 33+11 i.e., 44 km/h.
4.
A train travels 288 km at a uniform speed. If the speed had been 4 km/h more, if would have taken 1 h
less for the same journey. Find the speed of the train.
[4]
Solution :
Let the speed of the train be x km/h.
Time taken to cover 288 km at x km/h =
288
h
x
Time taken to cover 288 km at (x + 4) km/h
=
288
 x  4
According to the question,
288
288

=1
x
 x  4
 288 (x + 4) – 288x = x (x + 4)
 288x + 1152 – 288x = x2 +4x
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 x2+4x – 1152 = 0
 x2 +36x – 32x – 1152 = 0
[by splitting middle term]
 x (x + 36) – 32 (x + 36) = 0
 (x + 36)(x – 32) = 0
 x + 36=0 or x – 32 = 0
 x = – 36 or x = 32
 x = 32
[
x   36 , as speed cannot be negative]
Hence, the speed of the train is 32 km/h.
5.
Solve for x and y
[4]
15
22

5
xy xy
40
55

 13
xy xy
x  y and x   y
Solution :
On putting
1
1
 u and
 v , then given equations become
xy
xy
15u + 22v = 5
and
..(i)
40u + 55y = 13
..(ii)
On multiplying Eq. (i) by 5 and Eq. (ii) by 2 and then subtract, we get
75u  110v  25
80u  110v  26
 

 5u  1
 u
1
5
On putting u 
15 
1
in Eq. (i), we get
5
1
 22v  5
5
 22 v = 5 – 3
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 v
2
1
 v
22
11
1
1
1
1
 and

xy 5
x  y 11

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


 x–y=5
 x + y = 11
u
1
1
and v  
5
11 
..(iii)
..(iv)
On solving Equations. (iii) and (iv), we get
x = 8,y = 3
 x = 8 and y = 3 is the required solution.
6.
Find the value of k, in which system equations x + 3y = 2 and 2x + ky = 8 has no solution.
[4]
Solution :
Given lines are x + 3y = 2 and 2x + ky = 8
On comparing with ax + by = c, we get
a1 =1, b1 = 3, c1 =2
and
a2 = 2, b2 = k, c2 = 8
Condition for no solution,
a1 b1 c1


a2 b2 c2

1 3 2
 
2 k 8
 k=6
7.
Show graphically that each one of the following systems of equations is inconsistent (i.e., has no
solution).
[4]
3x – 4y – 1 = 0
8
2x  y  5  0
3
Solution :
We have,
3x – 4 y – 1 = 0
and
8
2x  y  5  0
3
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Table for equation 3x – 4y – 1 = 0
 y
3x  1
4
x
y
3x  1
4
Points
0
3
–1
–1/4
2
–1
 1 
A  0, 
 4 
B(3, 2)
C(–1, –1)
Now, plot all the points on a graph paper and join them to get a line AB.
Table for equation 2x –
8
y+5=0
3
 6x – 8y +15 = 0  y =
6x  15
8
x
y
6x  15
8
Points
0
–2.5
1.5
1.875
0
3
D(0, 1875)
E(–2.5, 0)
F(15, 3)
Now, we plot all the points on a graph paper and join them to get a line EF.
We find the lines represented by equations 3x – 4y – 1 = 0 and 2x –
8
+ 5 = 0 are parallel. So, the
3
two lines have no common point. Hence, the given system of equations is inconsistent.
8. Solve for x and y
ax + by – a + b = 0
and
bx – ay – a – b = 0.
[4]
Solution :
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Given,
and
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ax + by – a + b = 0
bx – ay – a – b = 0.
The given equations may be written as
ax + by = a – b
and
...(i)
bx – ay = a + b
...(ii)
On multiplying Eq. (i) by a and Eq. (ii) by b, we get
a 2x + aby = a2 - ab
and
...(iii)
b 2x – aby = ab + bz
...(iv)
On adding Eqs. (iii) and (iv), we get
(a2 +b2)x = a 2 +b 2
 x
a2  b2
1
a2  b2
On putting x = 1 in Eq. (i), we get
a × 1+ by = a – b
 a = by = a – b
 by = a – b – a
 by = – b
 y=–1
 x = 1 and y = – 1 is the required solution.
9. Find the value of k for which the system of equations kx – y =2, 6x – 2y = 3 has (i) a unique solution (ii)
no solution. Is there a value of k for which the given system has infinitely many solutions?
[4]
Solution:
The given system of equations is
kx – y – 2 = 0 and 6x – 2y – 3 = 0.
This is of the form
a 1x + b1y + c1 = 0
and
a 2x + b2y + c2 = 0
where, a1 = k, b1 = – 1, c1 = – 2
and a 2 = 6, b 2 = –2, c 2 = – 3
(i) For a unique solution, we must have
a1 b1

a2 b2

k 1

6 2
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
k 1

6 2

k 3
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(ii) For no solution, we must have
a1 b1 c1


a2 b2 c2

k 1 2


6 2 3

k 1 2
 
6 2 3

k 1
 and
6 2
k 2

6 3
k = 3 and k  4
Clearly, k = 3 also satisfies the condition k  4.
Hence, the given system will have no solution when k = 3.
Further, the given system will have infinitely many solutions only when
a1 b1 c1

 .
a2 b2 c2
k 1 2


6 2 3

k 1 2
 
6 2 3
This is never possible.
Since,
1 2

2 3
Hence, there is no value of k for which the given system of equations has infinitely many solutions.
10.Two places A and Bare 100 km 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 h. If they travel
towards each other, they meet in 1 h. What are the speeds of the cars?
[4]
Solution :
Let the speed of two cars be x km/h and y km/h, respectively.
Case I
When two cars travel in the same direction, they will meet each other at P after 5
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The distance covered by a car from A = 5x km
[
distance = speed × time]
and distance covered by another car from B = 5y km
5x
–
5y = 100
 x–
y=2
Case II
When two cars travel in opposite (towards) direction, they will meet each other at Q after 1 h.
[divide both sides by 5]
...(i)
The distance covered by a car from A = x km and distance covered by another car from
B = y km
According to the question,
x + y = 100 km
...(ii)
On adding Eqs. (i) and (ii), we get
x  y  100
x  y  200
2x  120
 x
120
2
 x = 60 km/h
On putting x = 60 in Eq. (ii), we get
60 + y =100
 y =100 – 60
 y = 40 km/h
Hence, the speed of two cars are 60 km/h and 40 km/h, respectively.
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