Question:

Let the cost of one pencil be $p$ rupees and the cost of one eraser be $e$ rupees.\
\
1. Form the equations\
\[\begin{aligned}
2p + 3e &= 9 \quad\text{(Equation 1)}\\
4p + 6e &= 18 \quad\text{(Equation 2)}
\end{aligned}\]\
2. Observe the relationship between the equations\
Equation 2 is exactly twice Equation 1: \[4p+6e = 2(2p+3e) = 2\times9 = 18.\]
Hence the two equations are not independent; they represent the same straight line.
3. Solve for one variable in terms of the other\
From Equation 1,\
\[2p = 9 - 3e \quad\Rightarrow\quad p = \frac{9-3e}{2}.\]
Alternatively,\
\[3e = 9 - 2p \quad\Rightarrow\quad e = \frac{9-2p}{3}.\]
4. State the general solution\
Since the system is dependent, there are infinitely many ordered pairs $(p,e)$ satisfying the relation.\
Let $e = t$ (where $t$ can be any real number such that $p$ remains non‑negative). Then\
\[p = \frac{9-3t}{2}.\]
Conversely, let $p = s$, then\
\[e = \frac{9-2s}{3}.\]
5. Interpretation\
Any pair of values that satisfy the above relation gives the required costs. For example, if we choose $e = 2\,\text{rupees}$, then $p = \frac{9-3\times2}{2}=\frac{3}{2}=1.5\,\text{rupees}$. Similarly, $e = 1\,\text{rupee}$ gives $p = 3\,\text{rupees}$, etc.

Thus the system has infinitely many solutions; the cost of a pencil can be expressed as $p = \frac{9-3e}{2}$ rupees (or the cost of an eraser as $e = \frac{9-2p}{3}$ rupees).