Given,
$a–2b+11=0$……. (i)
$3a–6b+33=0$……. (ii)
From equation (i),
⇒ $b=(a+11)/2$
When $a=-1$, we get $b=(-1+11)/2=5$.
When $a=-3$, we get $b=(-3+11)/2=4$.
Thus, we have the following table giving points on the line $a–2b+11=0$.
| a | $-1$ | $-3$ |
| b | $5$ | $4$ |
From equation (ii),
Solve for b:
⇒ $b=(3a+33)/6$
So, when $a=1$
$b=(3(1)+33)/6=6$
And, when $a=-1$
⇒ $b=(3(-1)+33)/6=5$
Thus, we have the following table giving points on the line $3a–6b+33=0$
| a | $1$ | $-1$ |
| b | $6$ | $5$ |
Graph of the equations (i) and (ii) is given below:
Thus, the given graphs of the two equations are coincident. Hence, the system of equations has infinitely many solutions.