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

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 a 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:
$(\frac{3}{2},\frac{4}{5}),(\frac{1}{2},\frac{6}{5})$ …
Hence, in natural numbers, there exists only one pair; therefore, the solution is natural numbers.

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:
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) $(-\frac{9}{2},m)$
(B) $(n,-\frac{9}{2})$
(C) $(0,-\frac{9}{2})$
(D) (–9, 0)

Answer:
(A) $(-\frac{9}{2},m)$
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 infinitely many values of y.
Hence the solution of this question is $(-\frac{9}{2},m)$
Here, m stands for an 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 the 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 the 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 on the x-axis, the y coordinate is zero
So, the equation of the 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 the 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 on the x-axis, the y coordinate is zero
i.e., y = 0
Hence, the equation of the x-axis is y = 0
Therefore, option (B) is correct.

Question:10

The graph of y = 6 is a line
(A) parallel to the x-axis at a distance of 6 units from the origin
(B) parallel to the y-axis at a distance of 6 units from the origin
(C) making an intercept 6 on the x-axis.
(D) making an intercept 6 on both axes.

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

Hence in the above graph shows a line parallel to the 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
$=9\ne 7(RHS)$
(B) Putting x = 5, y = 2 in 5x + 2y = 7 we get
LHS = 5(5) + 2(2)
= 25 + 4
$=29\ne 7(RHS)$
(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
$=27\ne 7(RHS)$
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
$=4\ne 0(RHS)$
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 satisfies 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 the 1st quadrant, the values of x and y are positive
In the 2nd quadrant, the values of x is negative and y is positive
In the 3rd quadrant, the values of x are negative and y are also negative
In the 4th quadrant, the values of x are positive and y are 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

Question:15

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

Answer:
(C) (1, 1)
Solution:
The given linear equation is y = x
i.e., y – x = 0
First point is $(\frac{3}{2},-\frac{3}{2})$
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 $(0,\frac{3}{2})$
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 $(\frac{1}{2},-\frac{1}{2})$
Putting the value of $x=\frac{1}{2}$ in the given equation we get
$y-(-\frac{1}{2})=0$
$y=-\frac{1}{2}$
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 the 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 the same.
Now divide both sides of equation (1) by 2
$\begin{array}{r}\frac{x}{2}+\frac{y}{2}=1,\text{Putting}\mathrm{x}=1,\mathrm{y}=1\text{in L.H.S. we get}\\ \frac{1}{2}+\frac{1}{2}=\frac{1+1}{2}=\frac{2}{2}=1=\text{R.H.S}\end{array}$
Also, on dividing an equation by a non-zero number, the solution remains the 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 are 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 the x-axis
On the x-axis, the value of y is always zero
Therefore, point (a, a) cannot lie on the x-axis as the value of the y-coordinate is given as a.
Now we will discuss the y-axis
On the y-axis, the value of x is always zero.
Therefore point of the form (a, a) cannot lie on the y-axis as the value of the 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.

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 the 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,

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

Question:3

Write whether the following statement is True or False. Justify your answer.
The graph given below represents the linear equation x + y = 0.

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

Write whether the following statement is True or False. Justify your answer.
The graph given below represents the linear equation x = 3
Figure:

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:

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:

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 entries 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 values 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.

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 lines is a straight line graph and meets 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
$y=\frac{3}{2}x$ ... (1)
Putting the value of y in the above equation (i), we get
$2x+5\left(\frac{3}{2}x\right)=19\Rightarrow 2x+\frac{15x}{2}=19\Rightarrow \frac{4x+15x}{2}=19\Rightarrow 19x=19\times 2$
⇒ $x=\frac{19\times 2}{19}=2$
⇒ 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 the y-intercept at a distance of 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 the sum of the ordinates is 10 units
x + y = 10 …(1)
We can find such points as:

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:

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 the number line:

On the 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 the Cartesian plane, so there are infinitely many solutions.

Question:8

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

Answer:
(8, 0)
Solution:
The given equation is
x + 2y = 8
At the 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 the 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:
$c=\frac{8-2y}{y}(y\ne 0)$
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
$c=\frac{8-2y}{y}(y\ne 0)$
Hence the correct answer is $c=\frac{8-2y}{y}(y\ne 0)$

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.

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 they must 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

The following observed values of x and y are thought to satisfy a linear equation. Write the linear equation :

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 cuts the y-axis at (0, 2).

Question:3

Draw the graph of the linear equation 3x + 4y = 6. At what points, the graph cut 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 the same in both 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:5

If the temperature of a liquid can be measured in Kelvin units as $x$K or in Fahrenheit units as $y$K, the relation between the two systems of measurement of temperature is given by the linear equation $y=\frac{9}{5}(x-273)+32$.(i) Find the temperature of the liquid in Fahrenheit if the temperature of the liquid is $313$K. (ii) If the temperature is $158°$F, then find the temperature in Kelvin.

Answer:
$104°$F

Given that,

$y=\frac{9}{5}(x-273)+32$

Putting the value of x = $313$K in equation (1), we get

$\begin{array}{r}y=\frac{9}{5}(313-273)+32\\ y=\frac{9}{5}(40)+32\\ y=\frac{360}{5}+32\\ y={104}^{\circ }F\end{array}$

Given that,

$\begin{array}{r}y=\frac{9}{5}(x-273)+32\\ \text{Put}y={158}^{\circ }\text{in equation (1)}\\ \text{we get}\\ {158}^{\circ }=\frac{9}{5}(x-273)+32\\ {158}^{\circ }=\frac{9(x-273)+160}{5}\end{array}$

790 – 160 = 9x – 2457630 + 2457 = 9x
⇒ 3087 = 9x
⇒ 343 = x
Hence, answer is 343K

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 a proportionality constant and also mass of body}
It is also given that m = 6
$\therefore F=6a$

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 a proportionality constant and also the mass of the body}
It is also given that m = 6
$\therefore F=6a$

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.