x=4y
\(\Rightarrow\ \ y={x\over 4}\)
Put x=0, we get \(y={0\over4}=0\)
Put x=4, we get \(y={4\over4}=1\)
Put x=-4, we get \(y={-4\over 4}=-1\)
Put x=2, we get \(y={2\over4}={1\over 2}\)
Four solutions are (0,0), (4, 1), (-4, -1) and \(\left(2,{1\over 2}\right)\)

x=3y
x-3y+0=0
Comparing with ax+by+c=0, we get
a=1, b=-3, c=0.

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

x=1

m=7

x+y=3; infinitely many; x+y-3=0

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).

2x+3y-4.37=0;
a=2, b=3 and c=-4.37

x+y=10\(\Rightarrow\)y=10-x

2x-y=5
y=2x-5

Plot these point on the graph paper

\(3x-12+{3\over7}x=2(x-1)\)
\(\Rightarrow\ \ {24\over7}x-12=2x-2\)
\(\Rightarrow\ \ \ {24\over 7}x-2x=12-2\)
\(\Rightarrow\ \ \ {10x\over 7}=10\)
\(x=7\)
The graph of this equation is a line parallel to y-axis at a distance of 7 units to the right of origin O.

2x+5y=20
\(x={5\over2}y\)
\(\therefore 2\left({5\over2},y\right)+5y=20\)
⇒ 10y=20
⇒ y=2
\(\therefore x={5\over 2}(2)=5\)
Hence the required point is (5, 2).

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, \(a={11\over 5}, b={8\over 5}\)

Evident

2x+0y-5=0

definition