(CONSISTENT) Given pair of equations is:
$x+y=3 .....(i)$
$3x+3y=9 .....(ii)$
On comparing with $ax+by+c=0$,we get:
$a_1=1, b_1=1, c_1=-3$ [from Eq. $(i)$]
$a_2=3, b_2=3, c_2=-9$ [from Eq. $(ii)$]
$\frac{a_1}{a_2} = \frac{1}{3}, \frac{b_1}{b_2} = \frac{1}{3}, \frac{c_1}{c_2} = \frac{-3}{-9} = \frac{1}{3}$
Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$,the given pair of lines is coincident. Therefore,these lines have infinitely many solutions. Hence,the given pair of linear equations is consistent.
Now,for $x+y=3 \Rightarrow y=3-x$:
| $x$ | $0$ | $3$ |
| $y$ | $3$ | $0$ |
| Points | $A$ | $B$ |
For $3x+3y=9 \Rightarrow y = \frac{9-3x}{3} = 3-x$:
| $x$ | $0$ | $1$ | $3$ |
| $y$ | $3$ | $2$ | $0$ |
| Points | $C$ | $D$ | $E$ |
Plotting the points $A(0,3)$ and $B(3,0)$,we get the line $AB$. Similarly,plotting the points $C(0,3), D(1,2)$,and $E(3,0)$,we get the same line. We observe that the lines represented by Eqs. $(i)$ and $(ii)$ are coincident.