NCERT Solutions for Class 9 Math Chapter 4 - Linear Equation in two Variables

Answer:

Let the cost of a note book is `x and cost of a pen is `y.
Therefore a linear equation in two variables (x and y) representing the given statement is x = 2y.

Answer:

Answer:

Answer:

–2x + 3y = 6 in the form of ax + by + c = 0 is
written as
–2x + 3y – 6 = 0 [Changing 6 to LHS]
$\Rightarrow$ – 2x + 3y + (–6) = 0
On comparing the coefficients of x and y and also constant terms, we observe that
a = –2, b = 3 and c = –6

Answer:

x = 3y in the form of ax + by + c = 0 is written as
x – 3y = 0
$\Rightarrow$ 1.x + (–3)y + 0 = 0
On comparing the coefficients of x and y and also constant terms, we observe that
a = 1, b = – 3 and c = 0.

Answer:

2x = –5y in the form of
ax + by + c = 0 is written as 2x + 5y = 0
[Changing –5y to LHS]
$\Rightarrow$ 2x + 5y + 0 = 0
On comparing the coefficients of x and y and also constant terms, we observe that
a = 2, b = 5 and c = 0

Answer:

3x + 2 = 0 in the form of ax + by + c = 0 is written
as
3x + 0.y + 2 = 0
On comparing the coefficients of x and y and also constant terms, we observe that
a = 3, b = 0 and c = 2.

Answer:

y – 2 = 0 in the form of ax + by + c = 0 is written as
0.x + 1.y + (–2) = 0
On comparing the coefficients of x and y and also
constant terms, we observe that
a = 0, b = 1 and c = –2.

Answer:

5 = 2x in the form of ax + by + c = 0 is written as
5 – 2x = 0
$\Rightarrow$ –2x + 0.y + 5 = 0
On comparing the coefficients of x and y also constant terms, we observe that
a = –2, b = 0 and c = 5

Answer:

Statement (iii) is true because for every value of x, there is corresponding value of y and vice-versa. Thorough justification:
Let assume that
(i) x = 0 then y = 3 × 0 + 5
$\Rightarrow$ y = 0 + 5
$\Rightarrow$ y = 5
So, x = 0, y = 5 is a solution.
(ii) x = 1; then y = 3 × 1 + 5
$\Rightarrow$ y = 3 + 5
$\Rightarrow$ y = 8
So, x = 1, y = 8 is also a solution.
(iii) x = –2 then y = 3 (–2) + 5
$\Rightarrow$ y = –6 + 5
$\Rightarrow$ y = –1
So, x = –2, y = –1 is also a solution.
Similarly, we can find infinitely many solutions of given equation as we wish simply by putting the value of x (or y) and get the corresponding value of y (or x).

Answer:

2x + y = 7
We can write equation as:
y = 7 – 2x
When x = 0;
y = 7 – 2 × 0 $\Rightarrow$ y = 7 – 0 $\Rightarrow$ y = 7
When x = 1;
y = 7 – 2 × 1 $\Rightarrow$ y = 7 – 2 $\Rightarrow$ y = 5
When x = 2;
y = 7 – 2 × 2 $\Rightarrow$ y = 7 – 4 $\Rightarrow$ y = 3
When x = –1;
y = 7 – 2 × (–1) $\Rightarrow$ y = 7 + 2 $\Rightarrow$ y = 9
So, four of the infinitely many solutions of the equation 2x + y = 7 are (0, 7), (1, 5), (2, 3) and (–1, 9).

Answer:

$\mathrm{\pi }$x + y = 9
We can write the equation as:
y = 9 – $\mathrm{\pi }$x
When x = 0;
y = 9 – $\mathrm{\pi }$ 0 $\Rightarrow$ y = 9 – 0 $\Rightarrow$ y = 9
When x = 2;
y = 9 – $\mathrm{\pi }$ 2 $\Rightarrow$ y = 9 – 2$\mathrm{\pi }$

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Answer:

Put x = 0 and y = 2 in equation
x – 2y = 4
LHS, x – 2y = 0 – 2 × 2
= 0 – 4 = –4
RHS = 4
Clearly, LHS $\ne$ RHS
$\therefore$ (0, 2) is not the solution of equation
x – 2y = 4.

Answer:

Put x = 2 and y = 0 in equation
x – 2y = 4
LHS, x – 2y = 2 – 2 × 0
= 2 – 0 = 2
RHS = 4
Clearly LHS $\ne$ RHS
$\therefore$ (2, 0) is not the solution of x – 2y = 4.

Answer:

Put x = 4 and y = 0 in the equation
x – 2y = 4
LHS, x – 2y = 4 – 2 × 0 = 4 – 0 = 4
RHS = 4
Clearly, LHS = RHS
$\therefore$ (4, 0) is the solution of equation x – 2y = 4.

Answer:

Answer:

Put x = 1, y = 1 in the equation x – 2y = 4
LHS, x – 2y = 1 – 2 × 1 = 1 – 2 = – 1
RHS = 4
Clearly, LHS $\ne$ RHS
$\therefore$ (1, 1) is not the solution of equation
x – 2y = 4.

Answer:

If x = 2, y = 1 is the solution of equation 2x + 3y = k, then it definitely satisfy the equation.
$\therefore$ 2 × 2 + 3 × 1 = k
(Substituting x = 2, y = 1 in the equation)
$\Rightarrow$ 4 + 3 = k
$\Rightarrow$ 7 = k
or k = 7
Hence, required value of k is 7.

Answer:

x + y = 4 $\Rightarrow$ y = 4 – x
Table showing the values of y corresponding to the values of x is as follows:

Plot the points A(0, 4), B(2, 2) and C(4, 0) on a
graph paper. Join these points and obtain a line.
Scale chosen:
On X-axis: 5 divisions = 1 unit
On Y-axis: 5 divisions = 1 unit.

Answer:

x – y = 2 $\Rightarrow$ – y = 2 – x
or y = x – 2
Table showing the values of y corresponding
to the values of x is as follows:

Plot the points P(0, –2), Q(1, –1) and R(2, 0) on a graph paper. Join these points and obtain a line.
Scale chosen:
On X-axis: 5 divisions = 1 unit
On Y-axis: 5 divisions = 1 unit

Answer:

y = 3x
Table showing the values of y corresponding to the values of x is as follows:

Plot the points P(–1, –3), Q(0, 0) and R(1, 3) on a graph paper. Join these points and obtain a line.
Scale chosen:
On X-axis: 5 divisions = 1 unit
On Y-axis: 5 divisions = 1 unit

Answer:

3= 2x + y
$\Rightarrow$ 3 – 2x = y
or y = 3 – 2x
Table showing the values of y corresponding to the values of x is as follows:

Plot the points L(0, 3), M(1, 1) and N(2, –1) on a graph paper. Join these points and obtain a line.
Scale chosen:
On X-axis: 5 divisions = 1 unit
On Y-axis: 5 divisions = 1 unit

Answer:

As we know that through a point, infinitely many lines can be drawn. So out of these; any two linear equations which are satisfied by the point (2, 14) are as follows:
Let ax + by = c be linear equation in x and y which is satisfied by the point (2, 14).
It means (2, 14) lie on it:
$\therefore$ 2a + 14b = c

Answer:

Given that point (3, 4) lie on the graph corresponding to:
3y = ax + 7
So, it satisfy this linear equation.
There by
3 × 4 = a × 3 + 7
$\Rightarrow$ 12 = 3a + 7
$\Rightarrow$ 12 – 7 = 3a
$\Rightarrow$ 5 = 3a
or 3a = 5

Answer:

Taking total distance travelled as x km.
For first kilometer, fare = `8
For subsequent distance i.e., for remaining distance;
fare per kilometer = `5
$\therefore$ Fare for remaining (x – 1) km
= `5 (x – 1)
Total fare = `y
$\therefore$ `8 + `5(x – 1) = `y
$\Rightarrow$ 8 + 5(x – 1) = y
$\Rightarrow$ 8 + 5x – 5 = y
$\Rightarrow$ 5x + 3 = y
$\Rightarrow$ 5x – y + 3 = 0 ...(1)
Which is a linear equation for the given information. To plot the graph of equation (1) the table showing the value of y corresponding to the values
of x is as follows:

Answer:

Every point on the graph (a) is satisfied in the equation; (ii) x + y = 0
Therefore, for graph (a) the correct equation out of four equations is:
x + y = 0
$\Rightarrow$ y = –x
For x = 0, y = 0
For x = 1, y = –1
For x = –1, y = 1
Now, every point on the graph (b) is satisfied in the equation (iii) y = – x + 2
Therefore, for graph (b) the correct equation out of four equations is:
y = –x + 2
For x = 2,
y = – 2 + 2
$\Rightarrow$ y = 0
For x = 0,
y = – 0 + 2
$\Rightarrow$ y = 2
For x = –1,
y = – (–1) + 2
$\Rightarrow$ y = 1 +
$\Rightarrow$ y = 3

Answer:

Let work done by constant force (say k units) is y units and distance travelled by the body is x units. As we know that work done by a force is directly proportional to the distance travelled by the body. From ratio and proportion; we can express this fact as:
y = kx
Taking k = 5 units
$\therefore$ y = 5x ...(1)
Which is a linear equation in two variables.
To plot the graph of equation (1); the table showing the value of y corresponding to the values of x is as follows:

From the graph, we observe that when
(i) x = 2 units
then y = 10 units
Hence, when distance moved by the body is 2 units then work done is 10 units.
(ii) When x = 0 units
then y = 0 units
Hence, when no distance is moved by the body then no work is done on body.

Answer:

Let Yamini contributed `x and Fatima contributed `y towards Prime Minister’s Relief Fund. Contribution of both = `100
$\Rightarrow$ x + y = 100 ...(1)
which is a linear equation in two variables.
To plot the graph of equation (1); we draw the table showing the values of y corresponding to the values of x.

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Answer:

From the graph; if temperature is 30$°$C, then temperature in Fahrenheit is 86$°$F.

Answer:

From the graph; if temperature is 95$°$F then temperature in celsius is 35$°$C.

Answer:

From the graph; if temperature is 0$°$C then temperature in Fahrenheit is 32$°$F. If temperature is 0$°$F, then temperature in Celsius is –17.8$°$C (approximately).

Answer:

Yes; there is a temperature which is numerically same in both Fahrenheit and Celsius. From the graph we observe that point (–40, –40) lie on line.
Hence, –40$°$ (both in F and C) is numerically the same temperature.
(Note: – 40$°$F $\ne$ – 40$°$C).

Answer:

y = 3 in terms of one variable is represented on a number line.

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y = 3 in terms of two variables is represented in cartesian coordinate system.

Answer:

2x + 9 = 0 in terms of one variable is represented on number line:
2x + 9 = 0
$\Rightarrow$ 2x = –9

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