The given equation is
y=3
(i)In one variable
The representation of y = 3 on the number line is as shown below:

(ii)In two variables

It is a linear equation in two variables x and y .This is represented by a line. All the values of x are permissible because O.x is always O.However, y must satisfy the relation y = 3. Hence, two solutions of the given equation are x = 0, y = 3 and x = 2, y = 3.Thus the graph AB is a line parallel to the x-axis at a distance of 3 units above it

Let the contributions ofYarnini and Fatima be Rs.x and Rs.y respectively.
Then according to the question
x + y = 100
This is the linear equation which the given data satisfies.
Now, x + y = 100
⇒ y = 100-x
Table of solution

We plot the points (0, 100) and (50, 50) on the graph paper and join the same by a ruler to get the line which is the graph of the equation x + y = 100.

The given equation is x-2y = 4
Put x = 1 and y1 in (1), we get
x - 2y = 1-2(1) = 1-2 = -1, which is not 4.
(1, 1) is not solution of (1).

The given equation is x - 2y = 4
Put \(x=\sqrt{2}\), y=\(4\sqrt{2}\) in (1), we get
x-2y =\(\sqrt{2}-2(4\sqrt{2})\)
\(=\sqrt{2}-8\sqrt{2}=-7\sqrt{2}\) which is not 4.
\((\sqrt{2},\ 4\sqrt{2})\) which is not 4.
\((\sqrt{2}, 4\sqrt{2})\) is not a soluton of (1)

The given equation is x-2y = 4
(4,0)
 Put x = 4 and y = 0 in (1) we get
x - 2y = 4 - 2(0) = 4
(4,0) is a solution of (1).

The given equation is x-2y = 4
(2,0)
Put x = 2 and y = 0 in (1), we get
x - 2y = 2 - 2(0) = 2 - 0 = 2, which is not 4.
(2,0) is not a solution of (1)

The true option is (iii) y = 3x + 5 has infinitely many solution
Reason: For every value of x, there is a corresponding value of y and vice-versa.

(a) In Fig, the points on the line are (–1, –2), (0, 0), (1, 2). By inspection, y = 2x is the equation corresponding to this graph. You can find that the y-coordinate in each case is double that of the x-coordinate.
(b) In Fig., the points on the line are (–2, 0), (0, 4), (1, 6). You know that the coordinates of the points of the graph (line) satisfy the equation y = 2x + 4. So, y = 2x + 4 is the equation corresponding to the graph in Fig. 4.5 (ii).
(c) In Fig., the points on the line are (–1, –6), (0, –4), (1, –2), (2, 0). By inspection, you can see that y = 2x – 4 is the equation corresponding to the given graph (line).

Here the variables involved are force and acceleration. Let the force applied be y units and the acceleration produced be x units. From ratio and proportion, you can express this fact as y = kx, where k is a constant. (From your study of science, you know that k is actually the mass of the body.)
Now, since we do not know what k is, we cannot draw the precise graph of y = kx. However, if we give a certain value to k, then we can draw the graph. Let us take k = 3, i.e., we draw the line representing y = 3x.
For this we find two of its solutions, say (0, 0) and (2, 6)
From the graph, you can see that when the force applied is 3 units, the acceleration produced is 1 unit. Also, note that (0, 0) lies on the graph which means the acceleration produced is 0 units, when the force applied is 0 units.

(i) (0, 4) and (3, 0)
(ii) (0, 0) and (1, \(-\frac{2}{5}\))
(iii) (0, \(\frac{-4}{3}\)) and (1, \(\frac{-5}{3}\))

(i) 2x + 3y - 3.47 = 0; a = 2, b = 3 and c = -3.47
(ii) x - \(\sqrt{3}\)y - 9 = 0; a = I, b = -\(\sqrt{3}\) and c = -9
(iii) -5x + 8y + 4 = 0; a = -5, b = 8 and c = 4
(iv) -2x + y + 0 = 0; a = -2, b = 1 and c = 0

(i) x = –5 can be written as 1.x + 0.y = –5, or 1.x + 0.y + 5 = 0.
(ii) y = 2 can be written as 0.x + 1.y = 2, or 0.x + 1.y – 2 = 0.
(iii) 2x = 3 can be written as 2x + 0.y – 3 = 0.
(iv) 5y = 2 can be written as 0.x + 5y – 2 = 0.

(i)

(ii)

Here (1, 2) is a solution of a linear equation you are looking for. So, you are looking for any line passing through the point (1, 2). One example of such a linear equation is x + y = 3. Others are y – x = 1, y = 2x, since they are also satisfied by the coordinates of the point (1, 2). In fact, there are infinitely many linear equations which are satisfied by the coordinates of the point (1, 2).

By inspection, x = 2, y = 2 is a solution because for x = 2, y = 2
x + 2y = 2 + 4 = 6
Now, let us choose x = 0. With this value of x, the given equation reduces to 2y = 6 which has the unique solution y = 3. So x = 0, y = 3 is also a solution of x + 2y = 6. Similarly, taking y = 0, the given equation reduces to x = 6. So, x = 6, y = 0 is a solution of x + 2y = 6 as well. Finally, let us take y = 1. The given equation now reduces to x + 2 = 6, whose solution is given by x = 4. Therefore, (4, 1) is also a solution of the given equation. So four of the infinitely many solutions of the given equation are:
(2, 2), (0, 3), (6, 0) and (4, 1).

(i) We have F = (\(\frac{9}{5}\)) C + 32
When C = 0, F (\(\frac{9}{5}\)) x 0 + 32 = 32
When C = -15, F = \(\frac{9}{5}\) (-15) + 32 = -27 + 32 = 5
When C = -10, F = \(\frac{9}{5}\) (-10) + 32 = 9(-2) + 32 = 14
We have the following table:

Plot the ordered pairs (0, 32), (-15, 5) and (-10, 14) on a graph paper. Joining these points we get a straight line AB.

(ii) From the graph, we have
86°F corresponds to 30°C
(iii) From the graph, we have
95°F = 35°C
(iv) From the graph, we have
0°C = 32°F
and 0°F = 17.8°C
(v) Yes, from the graph, we have
40°F = -40°C

Given equation is x + y = 4.
To draw the graph, we need at least two solutions of the equation.
Given equation can be written as y = 4 - x.
When x = 0, then y = 4
When x = 4, then y = 4 - 4 = 0
Thus, we have the following table
So, plot the points A(0,4) and B(4,0) on thegraph paper and join them by a line.

Hence, line AB represents the required graph of given linear equation.

(ii) x - y = 2
⇒ y = x-2
If we have x = 0, then y = 0 -2 = -2
x = 1, then y = 1 - 2 = - 1
x = 2, then y = 2 - 2 = 0
∴ We have the following table:

Plot the ordered pairs (0, -2), (1, -1) and (2, 0) on the graph paper. Joining these points, we get a straight line PQ as shown below:
Thus, the line PQ is required graph of x - y = 2.

(iii) y = 3x
If x = 0, then y = 3(0) ⇒ y = 0
x = 1, then y = 3(1) ⇒ y = 3
x = -1, then y = 3(-1) ⇒ y = -3
We get the following table:

Plot the ordered pairs (0, 0), (1, 3) and (-1, -3) on the graph paper. Joining these points, we get the straight line LM.
Thus, LM is the required graph of y = 3x.

Note: The graph of the equation of the form y = kx is a straight line which always passes through the origin.
(iv) 3 = 2x + y ⇒ y = 3 - 2x
∴ If x = 0, then y = 3 - 2(0) ⇒ y = 3
If x = 1, then y = 3 - 2(1) ⇒ y = 1
If x = 2, then y = 3 - (2) ⇒ y = -1

Plot the ordered pairs (0, 3), (1, 1) and (2, -1) on the graph paper. Joining these points, we get a line CD.

Thus, the line CD is the required graph of 3 = 2x + y

(i) 2x + y = 7
When x = 0, 2(0) + y = 7
⇒ 0+ y = 7
⇒ y = 7
∴ Solution is (0, 7).
When x = 1, 2(1) + y = 7
⇒ y = 7 - 2
⇒ y = 5
∴ Solution is (1, 5).
When x = 2, 2(2) + y = 7
⇒ y = 7 - 4
⇒ y = 3
∴ Solution is (2, 3).
When x = 3, 2(3) + y = 7
⇒ y = 7 - 6
⇒ y = 1
∴ Solution is (3, 1).
(ii) πx + y = 9
When x = 0, π(0) + y = 9
⇒ y = 9 - 0 
⇒ y = 9
∴ Solution is (0, 9).
When x = 1, π(1) + y = 9
⇒ y =9-π
∴ Solution is {1, (9 - π)}.
When x = 2, π(2) + y = 9
⇒ y = 9 - 2π
∴ Solution is {2, (9 - 2π)}.
When x = -1, π(-1) + y = 9
⇒ -π+ y = 9
⇒ y=9+π
∴ Solution is {-1, (9 + π)}.
(iii) x = 4y
When x = 0, 4y = 0
⇒ y = 0
∴ Solution is (0, 0).
When x = 1, 4y = 1
⇒ y= \(\frac{1}{4}\)
∴ Solution is (1,\(\frac{1}{4}\)).
When x = 4, 4y = 4
⇒ y = \(\frac{4}{4}\) = 1
∴ Solution is (4, 1).
When x =-4, 4y =-4
y = \(\frac{-4}{4}\) =-1
∴ Solution is (-4, -1).

(i) We have 2x + 3y = 9.3\(\overline{5}\)
∴ (2)x + (3)y + (-9.3\(\overline{5}\)) = 0
Comparing it with ax + bx + c = 0, we have a = 2, b = 3 and c = -9.3\(\overline{5}\).
(ii) We have x - \(\frac{y}{5}\)- 10 = 0
or x + \(\left(-\frac{1}{5}\right)\) y + (-10) = 0
Comparing with ax + bx + c = 0, we get
a = 1, b = \(-\frac{1}{5}\) and c = -10
Note: Above equation can also be compared by:
Multiplying throughout by 5,
5(x) - (5x\(\frac{y}{5}\)) - (5 x 10) = 0
or 5x - y - 50 = 0
or 5(x) + (-1)y + (-50) = 0
Comparing with ax + by + c = 0, we get a = 5, b = -1 and c = -50.
(iii) We have -2x + 3y = 6
⇒ -2x + 3y - 6 = 0
⇒ (-2)x + (3)y + (-6) = 0
Comparing with ax + bx + c = 0, we get a = -2, b = 3 and c = -6.
(iv) We have x = 3y
⇒ x - 3y = 0
⇒ (1)x + (-3)y + 0 = 0
Comparing with ax + bx + c = 0, we get a = 1, b = -3 and c = 0.
(v) We have 2x = -5y
⇒ 2x + 5y = 0
⇒ (2)x + (5)y + 0 = 0
Comparing with ax + by + c = 0, we get a = 2, b = 5 and c = 0.
(vi) We have 3x + 2 = 0
⇒ 3x + 2 + 0y = 0
⇒ (3)x + (0)y + (2) = 0
Comparing with ax + by + c = 0, we get a = 3, b = 0 and c = 2.
(vii) We have y - 2 = 0
⇒ (0)x + (1)y + (-2) = 0
Comparing with ax + by + c = 0, we have a = 0, b = 1 and c = -2.
(viii) We have 5 = 2x
⇒ 5 - 2x = 0
⇒ -2x + 0y + 5 = 0
⇒ (-2)x + (0)y + (5) = 0
Comparing with ax + by + c = 0, we get a = -2, b = 0 and c = 5.

Given II'near equatIo..n is 2x + 9 = 0 \(\Rightarrow x=-\frac{9}{2}\)  ... (i)
(i) If \(x=-\frac{9}{2}\)  is treated as an equation in one variable, then it has a unique solution \(x=-\frac{9}{2}\) So, it is a point on the number line as shown below

(ii) Given equation 2x + 9 = 0 can be written as 2x + 0·y + 9 = 0, which is a linear equation in two variables x and y.This is represented by a line. Here, all the values of y are permissible, because 0·y is always 0. However, x must satisfy the equation 2x + 9 = 0.

Thus, the graph AB is a line parallel to the Y-axis at a distance of  \(\frac{9}{2}=4.5\) units in the direction of negative X-axis.

Given, work done by a body on application of a constant force is directly proportional to the distance travelled by the body,
\(i.e., Work done (W)\propto Distance(s)\)
\(\Rightarrow W=F.s\)
where, P is an arbitrary constant.
Now, let x be the distance and y be the work done.
Then, the equation will e y = Fx.
Which is the required linear equation in two variables x and y.
If the constant force is 5 units, i.e. P = 5, then equation is
y = 5x ... (i)
When x = 0, then y = 0.When x = 1, then y = 5
Thus,we have the following table

Now, plot the points O (0, 0) and A (1, 5) on graph paper and join them to get a line OA which represents the required graph.

(i) From point D(2, 0), draw a line parallel to OY to cut y = 4x at B and then from B, draw a line parallel to X-axis to intersect the Y-axis at C(O, 10).
Clearly, B represents (2,10). So, work done, when the distance travelled by the body is 2 units, is 10 units.
(ii) Clearly, y = 0, when x = 0, so the work done when the distance travelled by the body is 0 unit, is zero unit.

In Fig. (a), given points are (0,0), (-1, 1) and (1,-1).
So, these points will satisfy the equation of line.
At (-1, 1), x + y = - 1 + 1 = 0
At(0, 0), x + y = 0 +0 = 0
At (1,-1), x + y = 1- 1= 0
i.e. these points satisfy the equation x + y = 0.
Hence, given graph represents the equation x + y = 0.
In Fig. (b), given points are (-1, 3), (0,2) ,and (2, 0).
So, these points will satisfy the equation of line.
At( -1, 3), x + y = - 1 + 3 = 2
At (0, 2), x + y = 0 + 2 = 2
At(2, 0), x,. y = 2 + 0 = 2
i.e. these points satisfy the equation x +y = 2.
Hence, given graph represents the equation x +y = 2,
i.e.y = -x+2.

Let the cost of a notebook = Rs x
and the cost of a pen = Rs y
According to the question,
Cost of a notebook = 2 (Cost of a pen)
\(\Rightarrow\)x = 2y
\(\Rightarrow\) x-2y = 0
which is the required linear equation in two variables.