NCERT Exemplar Class 9 Maths Solutions Chapter 4 Linear Equations in Two Variables

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# NCERT Exemplar Class 9 Maths Solutions Chapter 4 Linear Equations in Two Variables

Edited By Ravindra Pindel | Updated on Aug 31, 2022 11:41 AM IST

NCERT exemplar Class 9 Maths solutions chapter 4 provide students with detailed answers for the chapter 4 Linear Equation in two variables. These NCERT Solutions are important as the concepts used here will be applied in mathematics courses of higher Classes. NCERT exemplar solutions for Class 9 Maths chapter 4 gives a good number of practice questions.

These NCERT exemplar Class 9 Maths solutions chapter 4 are also useful for competitive exams like JEE Main, advanced and NEET and other state and class 9 CBSE syllabus. The Class 9 Maths NCERT exemplar chapter 4 solutions are useful to solve Class 9 Physics problems.

## NCERT Exemplar Class 9 Maths Solutions Chapter 4: Exercise 4.1

Question:1

The linear equation $2x -5y = 7$ has
(A) A unique solution
(B) Two solutions
(C) Infinitely many sol

Answer:
(C) Infinitely many solutions
Solution:
Linear equation: The equation of two variables which gives a straight line graph is called linear equation.
Here the linear equation is 2x – 5y = 7
Let y = 0 then the value of x is:
$2x - 5(0) = 7$
$2x = 7$
$x=\frac{7}{2}$
Now, let y = 1 then the value of x is:
$2x - 5(1) = 7$
$2x = 12$
$x = \frac{12}{2}=6$
$x =6$
Here for different values of y, we are getting different values of x
Therefore the equation has infinitely many solutions

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

Answer:
(A) Natural numbers
Solution:
Unique solution: for Unique solution, there should exist only one pair of x and y.
In natural numbers, there exists only one pair which is (1, 1)
2x + 5y = 7
(2) + (5) = 7
7 = 7
In positive real numbers there exists more than one pair.
For example: (0.5, 1.2), (1.5, 0.8)…
In real numbers there exists more than one pair.
For example: (–0.5, 1.6), (–1.5, 2)…
Also in rational numbers, there exists more than one pair.
For example:
$\left (\frac{3}{2},\frac{4}{5} \right ),\left ( \frac{1}{2},\frac{6}{5} \right )$
Hence in natural numbers there exists only one pair therefore the solution is natural numbers.

Question:3

Solution:
In this question we have been given a pair (2, 0)
Here the value of x is 2 and the value of y is 0.
The linear equation is: 2x+ 3y = k
Now put the value of x and y in this equation we get
2x + 3y = k
2(2) + 3(0) = k
4 + 0 = k
4 = k
Hence the value of k is equal to 4

Question:4

Any solution of the linear equation 2x + 0y + 9 = 0 in two variables is of the form
(A) $\left (-\frac{9}{2},m \right )$
(B) $\left (n,-\frac{9}{2} \right )$
(C) $\left (0,-\frac{9}{2} \right )$
(D) (–9, 0)

Answer:
(A) $\left (-\frac{9}{2},m \right )$
Solution:
Here the linear equation is 2x + 0y + 9 = 0
Let,
2x = – 9
$x = -\frac{9}{2}$
In equation (1), y is multiplied by zero so we can put any value of y. Therefore there will be infinite many values of y.
Hence the solution of this question is $\left (-\frac{9}{2},m \right )$
Here m stands for infinite many values.

Question:5

The graph of the linear equation 2x + 3y = 6 cuts the y-axis at the point
(A) (2, 0)
(B) (0, 3)
(C) (3, 0)
(D) (0, 2)

Answer:
(D) (0, 2)
Solution:
Here the given equation is 2x + 3y = 6
We know that at y-axis x = 0
Now put x = 0 in the given equation we get
2x + 3y = 6
2(0) + 3y = 6
3y = 6
$y= \frac{6}{3}=2$

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

Answer:
(B) 1.x + 0.y = 7
Solution:
The given equation is x = 7
In the given equation there is no second variable.
If we want to write the given equation in the form of two variables then the value of second variable must be zero.
So, the solution of the given equation is 1.x + 0.y = 7
Therefore option (B) is correct.

Question:7

Any point on the x-axis is of the form
(A) (x, y)
(B) (0, y)
(C) (x, 0)
(D) (x, x)

Answer:
(C) (x, 0)
Solution:
We know that at x-axis, y coordinate is zero
So, equation of x-axis is y = 0
Therefore any point on the x-axis is of the form (x, 0).
Hence the solution of the given question is (x, 0)

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)

Answer:
(A) (a, a)
Solution:
The given line is y = x
This can be written as:
y – x = 0
Putting y = 1,
1 – x = 0
x = 1
Point is (1, 1)
Putting y = 2,
2 – x = 0
x = 2
Point is (2, 2)
Hence every point is in the form of (a, a)
Therefore option (A) is correct.

Question:9

The equation of x-axis is of the form
(A) x = 0
(B) y = 0
(C) x + y = 0
(D) x = y

Answer:
(B) y = 0
Solution:
We know that at x-axis, y coordinate is zero
i.e., y = 0
Hence the equation of x-axis is y = 0
Therefore option (B) is correct.

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.

Answer:
(A) parallel to x-axis at a distance 6 units from the origin
Solution:
The graph of y = 6 is a line as shown below:

Hence in the above graph is a line parallel to x-axis at a distance 6 units from the origin.
Therefore option (A) is correct.

Question:11

x = 5, y = 2 is a solution of the linear equation
(A) x + 2 y = 7
(B) 5x + 2y = 7
(C) x + y = 7
(D) 5 x + y = 7

Answer:
(C) x + y = 7
Solution:
(A) Putting x = 5, y = 2 in x + 2y = 7 we get
LHS = 5 + 2 (2)
= 5 + 4
$= 9 \neq 7 (RHS)$
(B) Putting x = 5, y = 2 in 5x + 2y = 7 we get
LHS = 5(5) + 2(2)
= 25 + 4
$= 29 \neq 7 (RHS)$
(C) Putting x = 5, y = 2 in x + y = 7 we get
LHS = 5 + 2
= 7 = 7 (RHS)
(D) Putting x = 5, y = 2 in 5x + y = 7 we get
LHS = 5(5) + 2
= 25 + 2
$= 27 \neq 7 (RHS)$
Hence only x + y = 7 satisfies the given values of x and y
Therefore option (C) is correct.

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

Answer:
(B) x + y = 0
Solution:
(A) Put the given points in y – x = 0
Putting (–2, 2)
LHS = 2 – (–2)
= 2 + 2
$= 4\neq 0 (RHS)$
The given equation does not satisfy this point, so there is no need to check for other points.
(B) Now put the given points in x + y = 0
Putting (–2, 2)
LHS = –2 + 2 = 0 (RHS)
Putting (0, 0)
LHS = 0 + 0 = 0 (RHS)
Putting (2, –2)
LHS = 2 – 2 = 0 (RHS)
The given equation is satisfying all the points
(C) Put the given points in –2x + y = 0
Putting (–2, 2)
LHS = (–2) (–2) + (2)=6, which is not equal to RHS

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

Answer:
(A) 1st quadrant
Solution:
In 1st quadrant the value of x and y are positive
In 2nd quadrant the value of x is negative and y is positive
In 3rd quadrant the value of x is negative and y is also negative
In 4th quadrant the value of x is positive and y is negative.

Hence for the positive solution of the given equation, both the values of x and y must be positive. Therefore option (A) is correct.

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)

Answer:
(C) (3, 0)
Solution:
At x-axis, y = 0
Put y = 0 in 2x + 3y = 6
2x + 3(0) = 6
2x = 6
$x =\frac{6}{2} = 3$
x = 3
So the required point is (3, 0)
The graph of equation 2x + 3y = 6 is
 x 0 3 y 2 0

Question:15

The graph of the linear equation y = x passes through the point
(A) $\left (\frac{3}{2}, -\frac{3}{2} \right )$
(B) $\left (0, \frac{3}{2} \right )$
(C) (1, 1)
(D) $\left (\frac{1}{2}, -\frac{1}{2} \right )$

Answer:
(C) (1, 1)
Solution:
The given linear equation is y = x
i.e., y – x = 0
First point is $\left (\frac{3}{2}, -\frac{3}{2} \right )$
Putting the value of $x=\frac{3}{2}$ in the above equation we get
$y-\frac{3}{}2=0$
$y=\frac{3}{2}$
We get $y=\frac{3}{2}$ which does not satisfy our y-coordinate
Second point is $\left (0, \frac{3}{2} \right )$
Putting the value of x = 0 in the given equation, we get
y = 0
We get y = 0 which does not satisfy our y-coordinate
Third point is (1, 1)
Putting the value of x = 1 in the given equation we get
y – 1 = 0
y = 1
We get y = 1 which satisfies our y-coordinate
Fourth point is $\left (\frac{1}{2}, -\frac{1}{2} \right )$
Putting the value of $x=\frac{1}{2}$ in the given equation we get
$y-\left (-\frac{1}{2} \right )=0$
$y=-\frac{1}{2}$
We get $y=-\frac{1}{2}$ which does not satisfy our y-coordinate
Hence only one point which is (1, 1) satisfies the given linear equation.
Therefore option (C) is correct.

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) Changes
(B) Remains the same
(C) Changes in case of multiplication only
(D) Changes in case of division only

Answer:
(B) Remains the same
Solution:
Let the linear equation is x + y = 2 ……… (1)
Now put x = 1 in equation (1) we get
y = 1
Here x = 1, y = 1 is the solution of the equation
Now multiply both sides of equation (1) by 2
2x + 2y = 4
Put x = 1, y = 1 in left hand side we get
1(1) + 2(1) = 2 + 2 = 4 = RHS
Hence on multiply an equation by a non-zero number the solution remains same
Now divide both sides of equation (1) by 2
Put x = 1, y = 1 in L.H.S. we get
Also on divide an equation by non-zero number the solution remains same.

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

Answer:
(C) Infinitely many
Solution:
Let the line’s equation be ax + by + c = 0
Now put x = 1 and y = 2 in above equation
a + 2 b + c = 0
Here only one equation is not possible because a, b and c are real numbers. Hence different values of a, b and c can satisfy the equation a + 2b + c = 0
For example, (a = 1, b = –1, c = 1)
(a = 2, b = –2, c = 2) and many more.
Similarly two and three equations is not the correct answer.
Therefore infinitely many linear equations in x and y can be satisfied by x = 1 and y = 2.
Therefore option (C) is correct.

Question:18

The point of the form (a, a) always lies on:
(A) x-axis
(B) y-axis
(C) On the line y = x
(D) On the line x + y = 0

Answer:
(C) On the line y = x
Solution:
First of all we will discuss about x-axis
On x-axis, the value of y is always zero
Therefore point (a, a) cannot lie on x-axis as value of y-coordinate is given as ‘a’.
Now we will discuss about y-axis
On y-axis, the value of x is always zero.
Therefore point of the form (a, a) cannot lie on y-axis as value of x-coordinate is given as ‘a’.
On the line y = x, for each and every value of x there is an equal value of y
For example for x = 1, we get y = 1
Therefore point of the form (a, a) lies on the line y = x.
On the line x + y = 0
Let the value of x is 1
We get y = –1
Hence point of the form (a, a) does not lies on the line x + y = 0.
Therefore, the point of the form (a, a) always lies on the line y = x
Therefore option (C) is correct.

Question:19

The point of the form (a, – a) always lies on the line
(A) x = a
(B) y = – a
(C) y = x
(D) x + y = 0

Answer:
(D) x + y = 0
Solution:
The given point of the form (a, –a) shows that for every value of x there must be equal value of y with opposite sign.
In x = a, only the value of x is given
Here we do not know the value of y, therefore it is not possible to determine (a, –a) point.
In y = –a, only the value of y is given
Here we do not know the value of x, therefore it is not possible to determine (a, –a) point.
On the line y = x, for each and every value of x there is an equal value of y
For example for x = 1, we get y = 1
Therefore point of the form (a, a) lies on the line y = x.
On the line x + y = 0, i.e., x = -y
for each and every value of x there is an equal value of y (which is negative)
For example for x = 1, we get y = -1
Therefore point of the form (a, -a) lies on the line x + y = 0.
Therefore option (D) is correct.

## NCERT Exemplar Class 9 Maths Solutions Chapter 4: Exercise 4.2

Question:1

Write whether the following statement is True or False. Justify your answer.
The point (0, 3) lies on the graph of the linear equation 3x + 4y = 12

Answer:
True
Solution:
The given linear equation is 3x + 4y = 12 and the point is (0, 3)
Here the value of x is 0 and y is 3
Now put the value of x and y in above equation. We get,
LHS = 3(0) + 4 (3)
= 0 + 12
= 12 = RHS
Hence the given point (0, 3) satisfies the linear equation 3x + 4y = 12.
Therefore the given statement is True.

Question:2

Write whether the following statement is True or False. Justify your answer.
The graph of the linear equation x + 2y = 7 passes through the point (0, 7).

Answer:
False
Solution:
First of all, let us plot the graph of the linear equation x + 2y = 7
For x = 0,
0 + 2y = 7
$y=\frac{7}{2}$
For y = 0,
x + 2 (0) = 7
x = 7
For x = 1,
1 + 2y = 7
2y = 6
y = 3
So we have,
 x 0 7 1 y $\frac{7}{2}$ 0 3

The above graph shows that the linear equation does not passes through the point (0, 7)
Hence the given statement is false.

Question:3

Answer:
True
Solution:
Given graph is:

The points of the graph are (–1, 1) and (–3, 3)
Given linear equation is x + y = 0
Putting x = –1 in above equation
–1 + y = 0
y = 1
Putting x = –3 in above equation
-3 + y = 0
y = 3

Question:4

Answer:
True
Solution:
Since x = 3 is a vertical line. That is there is no y-intercept, also the slope is not defined.
The given graph is:

Therefore the above graph represents the linear equation x = 3 as it passes through x = 3 and there is no y-intercept, also the slope is not defined.
Hence the given statement is True.

Question:5

Write whether the following statement is True or False. Justify your answer.
The coordinates of points in the table:

 x 0 1 2 3 4 y 2 3 4 -5 6
represent some of the solutions of the equation x – y + 2 = 0.
Answer:
False
Solution:
The given equation is x – y + 2 = 0
Put x = 0 in the given equation
0 – y + 2 = 0
y = 2, i.e., the point is (0, 2)
Put x = 1 in the given equation
1 – y + 2 = 0
y = 3, i.e., the point is (1, 3)
Put x = 2 in the given equation
2 – y + 2 = 0
4 = y, i.e., the point is (2, 4)
Put x = 3 in the given equation
3 – y + 2 = 0
5 = y, i.e., the point is (3, 5)
Put x = 4 in the given equation
4 – y + 2 = 0
6 = y, i.e., the point is (4, 6)
The coordinates of points in the table:
 x 0 1 2 3 4 y 2 3 4 -5 6
It is given that the above represents some of the solutions of the equation x – y + 2 = 0.
Here all the points satisfy the given equation except (3, -5).
Therefore the given statement is false because one of the table entry is an incorrect solution for the given equation.

Question:6

Write whether the following statement is True or False. Justify your answer.
Every point on the graph of a linear equation in two variables does not represent a solution of the linear equation.

Answer:
False
Solution:
The graph of a linear equation in two variables is a line which can be plotted by finding two solutions (a1, b1) and (a2, b2).
Hence the graph of a linear equation is constructed by joining these points. So any point on the graph must be a solution of this linear equation.
Hence the given statement is False.

Question:7

Write whether the following statement is True or False. Justify your answer.
The graph of every linear equation in two variables need not be a line.

Answer:
False
Solution:
We know that the standard form of any linear equation is ax + by + c = 0
If we put different value of x we get different values of y corresponding to the values of x.
So we have x proportional to y.
Thus we can say $\frac{x}{y}$ = constant.
Now, the general equation of a line is y=mx+c, which also gives a direct proportionality between x and y.
So if we plot a graph with the help of such points, it will always be a line
Therefore the given statement is false.

## NCERT Exemplar Class 9 Maths Solutions Chapter 4: 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?

Answer:

First of all the us plot the graph of linear equation y = x and y = –x
The corresponding x and y-coordinates can be written as:

Plotting the above in the same Cartesian plane, we get:

Here we observe that the graph of both the lines is straight line graph and meet only at the origin.

Question:2

Determine the point on the graph of the linear equation 2x + 5y = 19, whose ordinate is times its abscissa.

Answer:
(2, 3)
Solution:
The given linear equation is 2x + 5y = 19 … (i)
Here it is given that the ordinate is $\frac{3}2$ times its abscissa
$y=\frac{3}2x$ ... (1)
Put the value of y in above equation (i) we get

$x = \frac{19 \times 2}{19}=2$
x = 2
Put the value of x in equation (i)
$y=\frac{3}{2}\times 2$
y = 3
Hence the point 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.

Answer:

We know that the straight line which is parallel to the x-axis is on the y-intercept, i.e. there is no x-intercept on it.
Now let us plot a straight line on y-intercept at a distance 3 units below the x-axis

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.

Answer:

It is given that sum of the ordinates is 10 units
x + y = 10 …(1)
We can find such points as:
 x 1 2 3 y 9 8 7
The points are (1, 9), (2, 8) , (3, 7) and so on.
Now plotting the graph of the linear equation

Question:5

Write the linear equation such that each point on its graph has an ordinate 3 times its abscissa.

Answer:

It is given that the ordinate is 3 times its abscissa.
That is y = 3x … (i)
Such points are:

 x 0 1 2 y 0 3 6
The points are (0, 0), (1, 3) and (2, 6)
Now plot the graph of equation (1)

Hence y = 3x is the required solution such that each point on its graph has 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.

Answer:
$a=\frac{5}{3}$
Solution:
The given linear equation is 3y = ax + 7 and the point is (3, 4)
If the point (3, 4) lies on the graph of equation 3y = ax + 7 then it will surely satisfy the equation
Put value of x = 3 and y = 4 in equation 3y = ax + 7
3(4) = a(3) + 7
12 = 3a + 7
12 – 7 = 3a
5 = 3a
$\frac{5}{3}=a$

Question:7

How many solution(s) of the equation 2x + 1 = x – 3 are there on the number line?

How many solution(s) of the equation 2x + 1 = x – 3 are there on the Cartesian plane?

Answer:
Only one solution
Solution:
The given equation is
2x + 1 = x – 3
2x – x = – 3 – 1
x = –4
Representation on number line:

On number line, there is only one solution

Answer:
Infinite solutions
Solution:
The given equation is
2x + 1 = x – 3
2x – x = – 3 – 1
x = –4
Cartesian plane representation:

There are infinite many points on line x = –4 in Cartesian plane so there are infinite solutions.

Question:8

Find the solution of the linear equation x + 2y = 8 which represents a point on x-axis.

Find the solution of the linear equation x + 2y = 8 which represents a point on y-axis.

Answer:
(8, 0)
Solution:
The given equation is
x + 2y = 8
at x-axis, we have y = 0
Now put the value of y = 0 in the given equation
x + 2(0) = 8
x = 8
So the required point is (8, 0)

Answer:
(0, 4)
Solution:
The given equation is
x + 2y = 8
At y-axis, we have x = 0
Now put the value of x = 0 in the given equation
0 + 2y = 8
$y=\frac{8}{2}=4$
y = 4
So the required point 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.

Answer:

Answer:
$c=\frac{8-2y}{y}(y\neq 0)$
Solution:
The given linear equation is
2x + cy = 8 … (i)
and it is also given that the value of x and y is equal
i.e., x = y
Putting x = y in the equation (i) we get
2y + cy = 8
cy = 8– 2y
$c=\frac{8-2y}{y}(y\neq 0)$
Hence the correct answer is $c=\frac{8-2y}{y}(y\neq 0)$

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?

Answer:
y = 3x, y = 15 when x = 5.
Solution:
It is given that y varies directly as x
$y\propto x$
y = kx (Here k is constant)
For finding k, we can use the given condition y = 12 when x = 4
12 = k(4)
$\frac{12}{4}=k$
k = 3
Hence the equation is y = 3x
When x = 5 then the value of y is:
y = 3(5)
y = 15
Hence y = 15 when x = 5.

## NCERT Exemplar Class 9 Maths Solutions Chapter 4: 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.

Answer:

The given linear equation is y = 9x – 7 and the points are A(1, 2), B(–1, –16) and C(0, –7)
If the points lie on the graph then it must be satisfy the equation y = 9x – 7
Put x = 1, y = 9 – 7
y = 2
Put x = –1 y = –9 – 7
y = –16
Put x = 0 y = 0 – 7
y = -7
Here all the points satisfy the equation y = 9x – 7.
Therefore all the points lie on the graph of the linear equation y = 9x – 7
Hence proved

Question:2

 x 6 -6 y -2 6
Draw the graph using the values of x, 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

Answer:

Let the linear equation is y = mx + c … (i), and it satisfies the points (6, –2), (–6, 6)
So, we have:
–2 = 6m + c … (ii)
6 = – 6m + c …(iii)
Adding equation (ii) & (iii) we get
4 = 2c
$\frac{4}{2}=c$
c = 2
On putting the value of c = 2 in equation (ii) we get
–2 = 6 m + 2
– 2 – 2 = 6 m
–4 = 6 m
$-\frac{4}{6}=m$
$m=-\frac{2}{3}$
Now, equation (i) becomes y=2x/3 + 2
which gives
3y + 2x = 6 … (iv)
Which is required equation and its graph is

Putting y = 0 in equation (iv), we get
2x = 6
x = 3
Hence the graph cut x-axis at (3, 0)
Putting x = 0 in equation (iv), we get
3y = 6
y = 2
Hence the graph cut 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.

Answer:

The given equation is 3x + 4y = 6 …(i)
For x =0,
0 + 4y = 6
$y = \frac{3}{2}=1.5$
For y = 0,
3x = 6
x = 2
For x = -2,
3 (-2) + 4y = 6
- 6 - 6 = -4y
y = 3
Plotting the points, we get:

Hence the graph cut x-axis at (2, 0) and the graph cut y-axis at (0, 1.5)

Question:4

The linear equation that converts Fahrenheit (F) to Celsius (C) is given by the relation $\frac{5F-160}{9}$ If the temperature is 86°F, what is the temperature in Celsius?

The linear equation that converts Fahrenheit (F) to Celsius (C) is given by the relation $\frac{5F-160}{9}$.If the temperature is $35^{\circ}C$, what is the temperature in Fahrenheit (F)?

The linear equation that converts Fahrenheit (F) to Celsius (C) is given by the relation $\frac{5F-160}{9}$.If the temperature is $0^{\circ}C$ what is the temperature in Fahrenheit and if the temperature is $0^{\circ}F$, what is the temperature in Celsius?

The linear equation that converts Fahrenheit (F) to Celsius (C) is given by the relation $C=\frac{5F-160}{9}$ .What is the numerical value of the temperature which is same in both the scales?

Answer:
$C =30^{\circ}$
Solution:
Given, $\frac{5F-160}{9}$
Put F = 86° in equation (1)
$C =\frac{5(86)-160}{9}$
$C =\frac{430-160}{9}=\frac{270}{9}=30^{\circ}$
Hence, $C =30^{\circ}$

Answer:
$95^{\circ}F$
Solution:
Given, $\frac{5F-160}{9}$
Put C = $35^{\circ}C$ in above equation
$35^{\circ}=\frac{5F-160}{9}$
315 = 5F – 160
315 + 160 = 5F
475 = 5F
$F=\frac{475}{5}=95^{\circ}$

Answer:
$32^{\circ}F,-17.78^{\circ}C$
Solution:
Given $\frac{5F-160}{9}$ …(1)
Put $C=0^{\circ}$ in equation (1)
$0=\frac{5F-160}{9}$
5F = 160
$F=\frac{160}{5}=32^{\circ}$
Now put F = 0 in (1), we get
$C=\frac{160}{9}=-17.78^{\circ}C$

Answer:
– 400
Solution:
Given that $C=\frac{5F-160}{9}$
It is asked that the numerical value of the temperature which is same in both the scales.
It means that the value of $^{\circ}C$ and $^{\circ}F$ are same
$\therefore F =\frac{5F-160}{9}$
9F – 5F = –160
4F = –160
F = –400

Question:6

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 5 m/sec2.

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 6 m/sec2.

Answer:

It is given that force is directly proportional to acceleration
i.e.,$F \propto a$
F = ma {Here m is proportionality constant and also mass of body}
It is also given that m = 6
$\therefore F=6a$
 F 6 12 a 1 2

The given equation is F = 6a
When acceleration is 5 m/sec2
Put a = 5
$F = 6 \times 5 = 30$ kgm/sec2
Hence the required answer is 30 kgm/sec2

It is given that force is directly proportional to acceleration
i.e.,$F \propto a$
F = ma {Here m is proportionality constant and also mass of body}
It is also given that m = 6
$\therefore F=6a$
 F 6 12 a 1 2

The given equation is
F = 6a
When acceleration is 6 m/sec2
Put a = 6
$F = 6 \times 6 = 36$ kgm/sec2
Hence the required answer is 36 kgm/sec2

## NCERT Exemplar Solutions Class 9 Maths Chapter 4 Important Topics:

The major pointers of this chapter covered in the NCERT exemplar Class 9 Maths solutions chapter 4 are mentioned below:

◊ Making two-variable linear equations for any given situation or statement.

◊ We will learn to solve these linear equations to find out the values of variables.

◊ NCERT exemplar Class 9 Maths solutions chapter 4 include methods to solve these linear equations to determine the values of variables.

◊ These linear equations can be drawn on a Cartesian plane as straight lines.

◊ We can find out the values of variables by geometric representations of these equations.

◊ We learn about linear equations representing lines parallel to X-axis or Y-axis and the line passing through the origin.

## NCERT Class 9 Maths Exemplar Solutions for Other Chapters:

 Chapter 1 Number System Chapter 2 Polynomials Chapter 3 Coordinate geometry Chapter 5 Introduction to Euclid’s Geometry Chapter 6 Lines and Angles Chapter 7 Triangles Chapter 8 Quadrilaterals Chapter 9 Area of Parallelograms and Triangles Chapter 10 Circles Chapter 11 Constructions Chapter 12 Heron’s Formula Chapter 13 Surface Areas and Volumes Chapter 14 Statistics and Probability

## Features of NCERT Exemplar Class 9 Maths Solutions Chapter 4:

These Class 9 Maths NCERT exemplar chapter 4 solutions will help the students grasp the basics of linear equations in two variables. It gives two relations between Binomial of two variables of a single degree. With the help of those two equations, we have to determine the values of two variables individually. The chapter on Linear Equations in Two Variables can be studied and practiced using these NCERT exemplar Class 9 Maths chapter 4 solutions Linear Equations in Two Variables and will be enough to solve other books such as A textbook of Mathematics by Monica Kapoor, NCERT Class 9 Maths, RD Sharma Class 10 Maths, RS Aggarwal Class 9 Maths et cetera. NCERT exemplar Class 9 Maths solutions chapter 4 pdf download is a free-to-use feature. The solutions can be downloaded using any tools that are available to convert the page to pdf.

### Check the solutions of questions given in the book

 Chapter No. Chapter Name Chapter 1 Number Systems Chapter 2 Polynomials Chapter 3 Coordinate Geometry Chapter 4 Linear Equations In Two Variables Chapter 5 Introduction to Euclid's Geometry Chapter 6 Lines And Angles Chapter 7 Triangles Chapter 8 Quadrilaterals Chapter 9 Areas of Parallelograms and Triangles Chapter 10 Circles Chapter 11 Constructions Chapter 12 Heron’s Formula Chapter 13 Surface Area and Volumes Chapter 14 Statistics Chapter 15 Probability

### Frequently Asked Question (FAQs)

1. Can we solve any two equations of two variables?

No, we cannot solve any two equations of two variables all the time, however, yes, if these two equations are linear, we can always solve them provided they have a solution.

2. Is it necessary for a linear equation of two variables to have a solution?

No, two equations written in the linear form of two variables can have zero solution, infinite solution and precisely one unique solution.

3. Can we solve the linear equations of three variables?

Yes, we can solve linear equations of three variables. However, the same method we use in linear equations of two variables will not be used as later we will study solving these equations using a matrix method.

4. What is the mark distribution of Linear equation in two variables for the final examination?

Generally 8-10% of the marks of the final paper account for Linear equation in two variables. The class 9 maths NCERT exemplar solutions of chapter 4 are equipped with the detailed solutions to ace the final examinations.

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Get answers from students and experts

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

 Option 1) Option 2) Option 3) Option 4)

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

 Option 1) 2.45×10−3 kg Option 2)  6.45×10−3 kg Option 3)  9.89×10−3 kg Option 4) 12.89×10−3 kg

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 Option 1) Option 2) Option 3) Option 4)

A particle is projected at 600   to the horizontal with a kinetic energy . The kinetic energy at the highest point

 Option 1) Option 2) Option 3) Option 4)

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 Option 1)   at STP  is produced for every mole   consumed Option 2)   is consumed for ever      produced Option 3) is produced regardless of temperature and pressure for every mole Al that reacts Option 4) at STP is produced for every mole Al that reacts .

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 Option 1) 0.02 Option 2) 3.125 × 10-2 Option 3) 1.25 × 10-2 Option 4) 2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

 Option 1) decrease twice Option 2) increase two fold Option 3) remain unchanged Option 4) be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

 Option 1) Molality Option 2) Weight fraction of solute Option 3) Fraction of solute present in water Option 4) Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

 Option 1) twice that in 60 g carbon Option 2) 6.023 × 1022 Option 3) half that in 8 g He Option 4) 558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

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##### Corporate Executive

Are you searching for a Corporate Executive job description? A Corporate Executive role comes with administrative duties. He or she provides support to the leadership of the organisation. A Corporate Executive fulfils the business purpose and ensures its financial stability. In this article, we are going to discuss how to become corporate executive.

2 Jobs Available
##### Product Manager

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.

3 Jobs Available
##### Quality Controller

A quality controller plays a crucial role in an organisation. He or she is responsible for performing quality checks on manufactured products. He or she identifies the defects in a product and rejects the product.

A quality controller records detailed information about products with defects and sends it to the supervisor or plant manager to take necessary actions to improve the production process.

3 Jobs Available
##### Production Manager

Production Manager Job Description: A Production Manager is responsible for ensuring smooth running of manufacturing processes in an efficient manner. He or she plans and organises production schedules. The role of Production Manager involves estimation, negotiation on budget and timescales with the clients and managers.

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##### QA Manager

Quality Assurance Manager Job Description: A QA Manager is an administrative professional responsible for overseeing the activity of the QA department and staff. It involves developing, implementing and maintaining a system that is qualified and reliable for testing to meet specifications of products of organisations as well as development processes.

2 Jobs Available
##### QA Lead

A QA Lead is in charge of the QA Team. The role of QA Lead comes with the responsibility of assessing services and products in order to determine that he or she meets the quality standards. He or she develops, implements and manages test plans.

2 Jobs Available
##### Reliability Engineer

Are you searching for a Reliability Engineer job description? A Reliability Engineer is responsible for ensuring long lasting and high quality products. He or she ensures that materials, manufacturing equipment, components and processes are error free. A Reliability Engineer role comes with the responsibility of minimising risks and effectiveness of processes and equipment.

2 Jobs Available
##### Safety Manager

A Safety Manager is a professional responsible for employee’s safety at work. He or she plans, implements and oversees the company’s employee safety. A Safety Manager ensures compliance and adherence to Occupational Health and Safety (OHS) guidelines.

2 Jobs Available
##### Corporate Executive

Are you searching for a Corporate Executive job description? A Corporate Executive role comes with administrative duties. He or she provides support to the leadership of the organisation. A Corporate Executive fulfils the business purpose and ensures its financial stability. In this article, we are going to discuss how to become corporate executive.

2 Jobs Available
##### ITSM Manager

ITSM Manager is a professional responsible for heading the ITSM (Information Technology Service Management) or (Information Technology Infrastructure Library) processes. He or she ensures that operation management provides appropriate resource levels for problem resolutions. The ITSM Manager oversees the level of prioritisation for the problems, critical incidents, planned as well as proactive tasks.

3 Jobs Available
##### Information Security Manager

Individuals in the information security manager career path involves in overseeing and controlling all aspects of computer security. The IT security manager job description includes planning and carrying out security measures to protect the business data and information from corruption, theft, unauthorised access, and deliberate attack

3 Jobs Available
##### Computer Programmer

Careers in computer programming primarily refer to the systematic act of writing code and moreover include wider computer science areas. The word 'programmer' or 'coder' has entered into practice with the growing number of newly self-taught tech enthusiasts. Computer programming careers involve the use of designs created by software developers and engineers and transforming them into commands that can be implemented by computers. These commands result in regular usage of social media sites, word-processing applications and browsers.

3 Jobs Available
##### Product Manager

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.

3 Jobs Available
##### IT Consultant

An IT Consultant is a professional who is also known as a technology consultant. He or she is required to provide consultation to industrial and commercial clients to resolve business and IT problems and acquire optimum growth. An IT consultant can find work by signing up with an IT consultancy firm, or he or she can work on their own as independent contractors and select the projects he or she wants to work on.

2 Jobs Available
##### Data Architect

A Data Architect role involves formulating the organisational data strategy. It involves data quality, flow of data within the organisation and security of data. The vision of Data Architect provides support to convert business requirements into technical requirements.

2 Jobs Available
##### Security Engineer

The Security Engineer is responsible for overseeing and controlling the various aspects of a company's computer security. Individuals in the security engineer jobs include developing and implementing policies and procedures to protect the data and information of the organisation. In this article, we will discuss the security engineer job description, security engineer skills, security engineer course, and security engineer career path.

2 Jobs Available
##### UX Architect

A UX Architect is someone who influences the design processes and its outcomes. He or she possesses a solid understanding of user research, information architecture, interaction design and content strategy.

2 Jobs Available
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