Equation of the sides are,
AB: Y=0
BC: X=-1
CD: Y=-4
DA; X=-4
Area=4 X 3
=12 sq. units

y=6x;y=60

\(y={5x+20\over4};\ (-4,0);\ (0,5)\)

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

-2x+3y=6
Comparing with ax+by+c=0, we get
a=2, b=3, c=-6

(0,-2),(2,-6/5),(5,0),(10,2)

x = 3 represents a line parallel to y-axis at a distance of 3 units to the right of the origin.
4x= 3y
\(\Rightarrow\ \ \ y={4x\over 3}\)
Table of solution

We plot the points (0,0) and (3, 4) on a graph paper and join the same by a ruler to get the line which is the graph of the equation 4x = 3y.

Area of the triangle GAB formed by the given two lines and the x-axis \(={3\times 4\over2}=6\) square units

2.x + y = 8
⇒ y=8-2x
Table of solution

We plot the points (4, 0) and (0, 8) on a graph paper and join the same by a ruler to get the line which is the graph of the equation 2.x + y = 8.

From graph, we see that this line intersects the x-axis at the point (4, 0) and the y-axis at the point (0,8).

If x=-2, y=6 is solution of equation
3ax+2by=6 then
3a(-2)+2b(6)=6
⇒ -6a+12b=6
⇒ -4+2b=1  ...(1)
Also,
2(a-1)+2(3b-4)=4
⇒ 2a-2+6b-8=4
⇒ 2a+6b=14
⇒ a+3b=7    ...(2)
Adding (1) and (2) we get
5b=8 ⇒ \(b={8\over 5}\)
Putting \(b={8\over 5}\) in (1), we get
-a+2\(\left(8\over 5\right)\)=1
\(\Rightarrow\ \ \ a={11\over5}-1={11\over 5}\)
Hence, \(a={11\over 5}, b={8\over 5}\)

7=2x
2x-7=0
2x+0y-7=0
Comparing with ax+by+c=0, we get
a=2
b=0
​​​​​​​c=-7

Putting x=2k-3, y=k+2 in 2x+3y=15=0, we get
2(2k-3)+3(k+2)+15=0
\(\Rightarrow 4k-6+3k+6+15=0\)
\(\Rightarrow k=\frac{-15}{7}\)

If point (3,4) lies on 
3y=ax+7
\(\therefore 3\times 4=3a+7\)
\(\Rightarrow 3a=12-7=5\Rightarrow a=\frac{5}{3}\)

Let father's present age= x years
Son's present age= y years
After 5 years father's age will be =(x+5) years
After 5 years son's age will be =(y+5) years
According to the question, x+5=2(y+5)
\(\Rightarrow\) x+5=2y+10
\(\Rightarrow\) x-2y=10-5
\(\Rightarrow\) x-2y=5

Let the abscissa of the point be x and the ordinate of the
point be y. According to the question,
y = 3x (1)
When x = 1, then y = 3 x 1 = 3
When x = 2, then y = 3 x 2 = 6
When x = 3, then y = 3 x 3 = 9

Here, we find three points A(1, 3), B(2, 6) and C(3, 9).Now, we can see that any point on the line joining these points has an ordinate 3 times its abscissa.

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).

(0,2) satisfies y=2.

\(ax+b(0)+c=0 \ \ \ \Rightarrow\ \ \ x=-{c\over a}\)
\(a(0)+by+c=0\ \ \ \Rightarrow\ \ \ y=-{c\over b}\)

a+(-a)=0

(a, 3a) satisfies y=3x

(2,2) and (-2,-1) both satisfy y=x

(1,1) satisfies y=x

(1, 1) satisfies 5x-3y=2

O(0,0) satisfies y=x

y\(\infty\)x ⇒ y=kx

x=2y+10 ⇒ x-2y-10=0

x+5=6 ⇒ x=1

-x+2y-7=0

definition

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) False. [Because ax + by + c = 0 is a linear equation in two variables if both 'a' and 'b' are non-zero.]
(ii) False. [Because a linear equation in two variables has infinitely many solutions.]
(iii) False. [Because the points (2, 0) and (-3, 0) lie on x-axis, (0, 5) lie on y-axis whereas the point (4, 2) lies in the first quadrant.]
(iv) True.
(v) False. [Because the point (0, 0) i.e., x = 0 + y = 0 does not satisfy the equation]

Given, total distance covered = x km = 1+ (x - 1) km
Fare for first kilometre = Rs 8
Fare for subsequent distance = Rs 5 per km
\(\therefore \)Fare for next (x - 1) km = (x -1) X 5 = 5 (x - 1)
According to the question,
Total fare = y
\(\therefore\) 8+5(x-1)=y
\(\Rightarrow\)8+5x-5=y\(\Rightarrow \)5x-y+3=0
which is the required linear equation.
It can also be written as y = 5x +3.
When x = 0, then y = 3
When x = 1, then y = 5 +3 = 8
When x = 2, then y = 5(2)+3 = 13
Then, we have the following table

Now, plot the points A (0,3), B (1, 8) and e(2, 13) on a graph paper and join them to form a line BC, which represents the required graph of linear equation.

(i) Given equation is x - 2y= 4.
On putting x = 0 and y = 2 in LHS, we get
LHS = x - 2y = 0 - 2 X 2
=0-4=-4≠4
⇒LHS≠RHS
Hence, (0,2) is not a solution of x - 2y = 4.
(ii) Given equation is x - 2y= 4.
On putting x = 2 and y = 0 in LHS, we get
LHS = x - 2y = 2 - 2 X 0
=2-0=2≠4
⇒LHS≠RHS
Hence, (2, 0) is not a solution of x - 2y = 4.
(iii) Given equation is x - 2y= 4.
On putting x = 4 and y = 0 in LHS, we get
LHS =x - 2y = 4 - 2 X 0
= 4 -0 =RHS
Hence, (4, 0) is a solution of x - 2Y = 4.