Lesson: Linear Equations In Two Variables
Exercise 4.1 (19 Multiple Choice Questions and
Answers)
Question: 1
The linear equation 2x – 5y = 7 has:
(a) A unique solution
(b) Two solutions
(c) Infinitely many solutions
(d) No solution
Solution:
c
Since, 2x 5y = 7 is a linear equation, we get a line
when we join all the points which satisfy the equation.
So, there are infinite numbers of points on this line
which satisfy the equation 2x – 5y = 7.
Question: 2
The equation 2x + 5y = 7 has a unique solution, if x, y
are:
(a) Natural numbers
(b) Positive real numbers
(c) Real numbers
(d) Rational numbers
Solution:
a
In case of natural numbers, there is only one solution
i.e. (1, 1) which satisfies the equation 2x + 5y = 7. But
in case of positive real numbers, real numbers or
rational numbers, there will be infinite number of
solutions for the equation.
Question: 3
If (2, 0) is a solution of the linear equation
2x + 3y = k, then the value of k is:
(a) 4
(b) 6
(c) 5
(d) 2
Solution:
a
As (2, 0) is a solution of the linear equation
2x +3y = k,
2 2 3 0 k
4 k
Question: 4
Any solution of the linear equation 2x + 0y + 9 = 0 in
two variables is of the form:
(a)
9
,
2
m
(b)
9
,
2
n
(c)
9
0,
2
(d) (−9, 0)
Solution:
a
Given linear equation is 2x + 0y + 9 = 0.
2 9 x
9
2
x
So, the value of and that of can be any real
9
2
x
number.
Question: 5
The graph of the linear equation 2x + 3y = 6 cuts the y
axis at the point:
(a) (2, 0)
(b) (9, 3)
(c) (3, 0)
(d) (0, 2)
Solution:
d
The given linear equation is 2x + 3y = 6. If the line cut
the y axis, the value of x at the point of intersection
will be zero. Hence, the point is (0, 2).
Question: 6
The equation x = 7, in two variables can be written as:
(a) 1. x + 1. y = 7
(b) 1. x + 0. y = 7
(c) 0. x + 1. y = 7
(d) 0. x + 0. y = 7
Solution:
b
The given linear equation is x = 7.
Here, the coefficient of y in the equation is zero.
Hence, the required equation is 1. x + 0. y = 7.
Question: 7
Any point on the x axis is of the form:
(a) (x, y)
(b) (0, y)
(c) (x, 0)
(d) (x, x)
Solution:
c
The y – coordinate of any point on the x – axis is zero.
Question 8
Any point on the line y = x is of the form:
(a) (a, a)
(b) (0, a)
(c) (a, 0)
(d) (a, a)
Solution:
a
The given equation y = x shows that the x –coordinate
of the point is equal to the y – coordinate of the point.
Question: 9
Any equation of x axis of the form:
(a) x = 0
(b) y = 0
(c) x + y = 0
(d) x = y
Solution:
b
The value of the y – coordinate for all the points at the
x – axis is zero.
Question: 10
The graph of y = 6 is a line
(a) parallel to x axis at a distance 6 units from the
origin
(b) parallel to y axis at a distance 6 units from the
origin
(c) making an intercept 6 on the x axis.
(d) making an intercept 6 on both the axes.
Solution:
a
The graph of y = 6 is a line parallel to the x axis at a
distance 6 units from the origin.
Question: 11
x = 5, y = 2 is a solution of the linear equation
(a) x + 2y = 7
(b) 5x + 2y = 7
(c) x + y = 7
(d) 5x + y = 7
Solution:
c
For option (a), the given equation is x + 2y = 7.
LHS = x + 2y
= 5 + 2 × 2
= 9 2
Hence, LHS RHS
For option (b), the given equation is 5x + 2y = 7.
LHS = 5x + 2y
= 5 × 5 + 2 × 2
= 29 7
Hence, LHS RHS
For option (c), the given equation is x + y = 7.
LHS = x + y
= 5 + 2
= 7
Hence, LHS RHS
For option (d), the given equation is 5x + y = 7.
LHS = 5x + y
= 5 × 5 + 2
= 27 7
Hence, LHS RHS
Question: 12
If a linear equation has solutions (−2, 2), (0, 0) and
(2, −2) then it is of the form:
(a) y x = 0
(b) x + y = 0
(c) −2x + y = 0
(d) x + 2y = 0
Solution:
b
Let us consider a linear equation ax + by + c = 0 … (i)
Since, (−2, 2), (0, 0) and (2, −2) are the solutions of
the linear equation, therefore these points satisfy
eq (i).
Corresponding to the point (−2, 2), the equation is
−2x + 2b + c = 7 … (ii)
Corresponding to the point (0, 0), the equation is
0 + 0 + c = 0 … (iii)
Corresponding to the point (2, −2), the equation is
2a 2b + c = 7 … (iv)
From eqs. (ii) and (iii),
c = 0 and −2a + 2b + c = 0
2 2 a b
2
2
b
a b
On putting a = b and c = 0 in eq. (i)
bx + by + 0 = 0
+ 0 bx by
+( ) 0 b x y
+ 0, 0 x y b
Hence, x + y = 0 is the required form of the linear
equation.
Question: 13
The positive solutions of the equation ax + by + c = 0
always lie in the:
(a) 1st quadrant
(b) 2nd quadrant
(c) 3rd quadrant
(d) 4th quadrant
Solution:
a
The first quadrant contains only positive values of x
and y. So, the positive solutions of the equation
ax + by + c = 0 always lie in the 1
st
quadrant.
Question: 14
The graph of the linear equation 2x + 3y = 6 is a line
which meets the x axis at the point:
(a) (0, 2)
(b) (2, 0)
(c) (3, 0)
(d) (0, 3)
Solution:
c
The given linear equation is 2x + 3y = 6.
At the x axis, the value of y is zero.
2 3 0 6 x
2 6 x
6
3
2
x
Hence, the coordinate on the x axis is (3, 0).
Question 15
The graph of the linear equation y = x passes
through the point:
(a)
(b)
3
0,
2
(c) (1, 1)
(d)
Solution:
c
The given equation is y = x.
In this equation the value of x is equal to the value of
y. So, at any point both the coordinates are equal.
Hence, the required point is (1, 1).
Question: 16
If we multiply or divide both sides of a linear equation
with a non-zero number, then the solution of the linear
equation:
(a) Doubles
(b) Remains the same
(c) Changes in case of multiplication only
(d) Changes in case of division only
Solution:
b
By property, if we multiply or divide both sides of a
linear equation with a non-zero number, then the
solution of the linear equation remains the same i.e.,
the solutions of the linear equation remains
unchanged.
Question: 17
How many linear equations in x and y can be satisfied
by x = 1 and y = 2?
(a) Only one
(b) Two
(c) Infinitely many
(d) Three
Solution:
c
Let the linear equation be ax + bx + c = 0.
On putting x = 1 and x = 2, in this equation we get
x + 2b + c = 0, where a, b and c, are real number.
Here, different values of a, b and c, satisfy
a + 2b + c = 0.
Hence, infinitely many linear equations in x and y can
be satisfied by x = 1 and x = 2.
Question: 18
The point of the form (a, a) always lies on:
(a) a axis
(b) y axis
(c) On the line y x
(d) On the line y + x = 0
Solution:
c
If the value of x is equal to value of y. The point of the
form (a, a) always lies on the line y = x.
Question: 19
The point of the form (a, −a) always lies on the line:
(a) x = a
(b) x = a
(c) y = x
(d) x + y = 0
Solution:
d
The given equation is x + y = 0.
This can be written as x = –y.
So, x and y are equal and of opposite sign. Hence, the
point of the form (a, –a) always lies on the line
x + y = 0
Exercise 4.2
Question: 1
Write whether the following statements are True or
False? Justify your answer:
(a) The point (0, 3) lies on the graph of the linear
equation 3x + 4y = 12.
(b) The graph of the linear equation x + 2y = 7 passes
through the point (0, 7).
(c) The graph given below represents the linear
equation x + y = 0.
(d) The graph given below represents the linear
equation x = 3
(e) The coordinates of points in the table
represent some of the solutions of the equation
x y + 2 = 0.
X
0
1
2
3
4
Y
2
3
4
5
6
(f) Every point on the graph of a linear equation in two
variables does not represent a solution of the linear
equation.
(g) The graph of every linear equation in two variables
need not be a line.
Solution:
(a) 3x + 4y = 12 …… (i)
Putting x = 0 and y = 3, we get,
3 × 0 + 4 × 3
= 0 +12
= 12
LHS = RHS
True, since (0, 3) satisfies the equation 3x + 4y = 12.
(b) LHS = x +12y
Putting x = 0, y = 7
= 0 + 2 × 7
= 12
RHS
False, since (0, 7) does not satisfies the equation.
(c) LHS = x +y
Putting x = −3 and y = 3, we get
= −3 + 3
= 0 = RHS
Again, at the point (1, 1) putting x = −1 and y = 1,
we get
−1 + 1 = 0 = RHS
True, since (–1, 1) and (–3, 3) satisfy the given
equation and the two points determine a unique line.
(d) The given linear equation is x = 3
This can be written as x + 0.y = 3
x
3
3
3
y
0
1
2
It is a line parallel to the y – axis at a distance of 3
units in the +ve direction.
True, since this graph is a line parallel to y – axis at a
distance of 3 units (to the right) from it.
(e) The coordinates of the points are (0, 2), (1, 3),
(2, 4), (3, –5), and (4, 6).
Given equation is x – y + 2 = 0.
At the point (0, 2),
0 2 2 22 = 0
It satisfies the given equation
At the point (1, 3),
1 3 2 = 3 3 0 0 = 0.
It satisfies the given equation
At the point (3, –5),
3 ( 5)+ 2 = 3+ 5+ 2 = 0 10 0
It does not satisfy the given equation.
At the point (4, 6),
4 6 2 = 6 6 0 0 0.
It satisfies the given equation.
Hence, the point (3, 5) does not satisfy the given
equation.
(f) False, since every point on the graph of the
equation represents a solution.
(g) False, since the graph of a linear equation in
two variables is always a line.
(vi) False, since every point on the graph of the
equation represents a solution.
(vii) False, since the graph of a linear equation in two
variables is always a line.
Exercise 4.3
Question: 1
Draw the graphs of linear equations y = x and
y = –x on the same cartesian plane. What do you
observe?
Solution:
y = x
x
0
1
2
y
0
1
2
y = x
y
0
1
−1
x
0
−1
1
Graphs of each equation is a line passing through
(0, 0).
Question: 2
Determine the point on the graph of the linear
equation 2x + 5y = 19, whose ordinate is
times its abscissa
1
1
2
Solution:
a
Given, 2x + 5y = 19 … (1)
As the ordinate is times its abscissa, .
1
1
2
3
2
y
x
Hence eq (1) becomes,
3
2 5 19
2
x
x
15
2 19
2
x
x
4 15
19
2
x x
19
19
2
x
x = 2
3 3 2
3
2 2
x
y
Thus, the point on the graph is (2, 3).
Question: 3
Draw the graph of the equation represented by a
straight line which is parallel to the x axis and at a
distance 3 units below it.
Solution:
Any line parallel to the x axis and at a distance of 3
units below it is given by y = –3.
Question: 4
Draw the graph of the linear equation whose solutions
are represented by the
points having the sum of the coordinates as 10 units.
Solution:
The sum of the coordinates is 10 units.
Let x and y be two coordinates, then
x + y = 10.
x
0
10
y
10
0
Question: 5
Write the linear equation such that each point on its
graph has an ordinate 3 times its abscissa.
Solution:
Let the abscissa of the point be x.
According to the question, the ordinate (y) = 3 ×
abscissa
= 3 y x
When x = 1, then y = 3 and when y = 2, then
y = 3 × 2 = 6.
x
1
2
y
3
6
Here, we find two points A(1, 3) and (2, 6).
So, we draw the graph by plotting the points and
joining the line AB.
Hence, y = 3x is the required equation such that point
on its graph has an ordinate 3 times its abscissa.
Question: 6
If the point (3, 4) lies on the graph of 3y = ax + 7, then
find the value of a.
Solution:
Given the graph,
3y = ax + 7 …… (i)
Putting the value x = 3 and y = 4 in eq. (i),
3.4 = a.3 + 7
.3 12 7 a
5
3
a
Question 7
How many solution(s) of the equation 2x + 1 = x – 3
are there on the:
(a) Number line
(b) Cartesian plane
Solution:
(a) 2x + 1 = x – 3
2 3 1 x x
4 x
x = 4 can be represented on the number line as
shown.
(b) x = −4
x = −4 has infinitely many solutions on the
cartesian plane.
Question: 8
Find the solution of the linear equation x + 2y = 8
which represents a point on:
(a) x axis
(b) y axis
Solution:
(i) x + 2y = 8
When the point is on the x axis, then putting y = 0,
we get,
x + 2 × 0 = 8
8 x
The point on the x axis is (8, 0)
When the point is on the y axis, then putting x = 0,
we get,
0+ 2y = 8
8
4
2
y
The point on the y axis is (0, 4).
Question: 9
For what value of c, the linear equation 2x + cy = 8
has equal values of x and y for its solution.
Solution:
The given linear equation is 2x + cy = 8.
Now, by the condition, and y coordinate of given
linear equation are same i.e., x = y.
Putting x = y, we get
2x + cy = 8
8 2 cx x
8 2
, 0
x
c x
x
Thus, for every value of x, we have a corresponding
value of c.
Question: 10
Let y varies directly as x. If y = 12 when x = 4, then
write a linear equation. What is the value of y when x
= 5?
Solution:
It is given that y is directly proportional to x.
So,
y x
y = kx (k = constant)
Putting y =12 and x = 4, we get
.
12
3
4
k
3 y x
if x = 5 then
x = 3 × 5 = 15
Exercise 4.4
Question: 1
Show that the points A (1, 2), B (– 1, – 16) and
C (0, – 7) lie on the graph of the linear equation
y = 9x – 7.
Solution:
y = 9x – 7
At the point A(1, 2), x = 1 and y = 2
we have,
2 = 9(1) – 7
2 2
Thus, the point A lies on the given line.
For the point B(–1, –16), x = –1, and y = –16
we have,
–16 = 9(1) – 7
16 16
At the point B lien on the given line.
For the point C(0, –7), we have
–7 = 9(0) – 7
7 7
Thus, the point C lies on the given line.
Hence, all the three points A, B and C lie on
y = 9x 7.
Question: 2
The following values of x and y satisfy a linear
equation.
x
6
6
y
2
6
Write the linear equation.
Draw the graph using the values of x and y as given in
the above table.
At what points the graph of the linear equation
(i) cuts the x axis?
(ii) cuts the y axis?
Solution:
(i) Given,
x
6
6
y
2
6
Let the linear equation be ax + by + c = 0.
Putting x = 6 and x = −2 in the above equation, we get
6a 2b + c = 0 (A)
Putting x = −6 and y = 6 in the above equation, we get
−6a 6b + c = 0 (B)
Adding equations (A) and (B), we get
4b + 2c = 0
2 c b
Putting the value of c in (A), we get
6a 2b 2b = 0
6 4 a b
2
3
b
a
Now, putting the values of and in ax + by + c = 0, we
get
2
2 0
3
b
x by b
2
2 0
3
x y
2
2
3
y x
Multiplying both the sides by 3, we get
3y = 2x + 6
2 3 6 0 x y
The graph cuts the x axis at (3, 0) and the y – axis at
(0, 2).
Question: 3
Draw the graph of the linear equation 3x + 4y = 6. At
what points, the graph cuts the x-axis and the y-axis.
Solution:
3x + 4y = 6
3 6 4 x y
6 4
3
y
x
x
0
2
y
3
2
0
The graph cuts the x axis at (2, 0) and the y axis at
.
3
0,
2
Question: 4
The linear equation that converts
Fahrenheit (F) to Celsius (C) is given by the relation
5 160
9
F
C
(a) If the temperature is 86°F, what is the temperature
in Celsius?
(b) If the temperature is 35°C, what is the temperature
in Fahrenheit?
(c) If the temperature is 0°C what is the temperature in
Fahrenheit and if the temperature is 0°F, what is the
temperature in Celsius?
(d) What is the numerical value of the temperature
which is same in both the scales?
Solution:
(a)
5 160
9
F
C
Putting f = 86, we get,
5 86 160
9
C
430 160
9
270
30
9
The temperature is 30°C.
(b)
5 160
9
F
C
5 160
9
F
C
Putting c = 35,
5 160
35
9
F
5 160 315 F
5 315 160 F
475
5
F
F = 95
The temperature is 95°F.
(c)
5 160
9
F
C
Putting C = 0,
5 160
0
9
F
5 160 0 F
5 160 F
160
5
F
F = 32
The temperature is 32°F.
5 160
9
F
C
Putting F = 0,
5 160
9
F
C
160
9
C
(d)
5 160
9
F
C
Putting F = C
5 160
9
C
C
9 5 160 C C
9 5 160 C C
4 160 C
160
4
C
40 C
The required value is 40°C.
Question: 5
If the temperature of a liquid can be measured in
Kelvin units as x °K or in Fahrenheit units as y °F, the
relation between the two systems of measurement of
temperature is given by the linear equation
9
273 32
5
y x
(a) Find the temperature of the liquid in Fahrenheit if
the temperature of the liquid is 313°K.
(b) If the temperature is 158° F, then find the
temperature in Kelvin.
Solution:
(a)
9
273 32
5
y x
Putting x = 313,
9
313 273 32
5
y
9
40 32
5
= 72 + 32
= 104
So the value is 140°F.
(b)
9
273 32
5
y x
Putting y = 158,
9
158 273 32
5
x
5
158 32 273
9
x
5
126 273
9
x
x = 343
The required value is 343°K.
Question: 6
(i) The force exerted to pull a cart is directly
proportional to the acceleration produced in the body.
Express the statement as a linear equation of two
variables and draw the graph of the same by taking the
constant mass equal to 6 kg. Read from the graph, the
force required when the acceleration produced is
(a) 5 m/sec
2
(b) 6 m/sec
2
Solution:
y = mx, where y denotes the force, x denotes the
acceleration and m denotes the constant mass.
Let the force exerted to pull a cart be F and the
acceleration be a.
F a
F ma
6 F a
The required linear equation in two variables is
F = 6a.
a
1
2
f
6
12
Representing the force on the x axis and the
acceleration on the y axis, we get the following
graph
From the graph, we find that
(a) When acceleration produced = 5m/sec
2
,
the forced required = 30 N
(b) When acceleration produced = 6m/sec
2
,
the forced required = 36 N.