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**Hint:**We find the general equations and their coefficients to find ratios. We define the relations between those ratios to find the number of solutions possible. Then we put the values from $x+2y=3$ and $2x+4y=6$ to find a number of solutions.

**Complete step-by-step solution:**

The coefficients of the given equation define the number of solutions for the given equations.

Let’s take two equations and two unknowns. $ax+by=m$ and $cx+dy=n$

In the equations the terms a, b, c, d, m, n all are constants.

We take the ratios of the coefficients of the same variables. The ratios are $\dfrac{a}{c},\dfrac{b}{d},\dfrac{m}{n}$.

The relation among these ratios determines the equations.

If $\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{m}{n}$, then we have an infinite number of solutions for those equations.

If $\dfrac{a}{c}=\dfrac{b}{d}\ne \dfrac{m}{n}$, then we have no solution for those equations.

If $\dfrac{a}{c}\ne \dfrac{b}{d}$, then we have only one solution for those equations.

In case of $\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{m}{n}$, both equations become a single equation. For a particular value of x, we can determine the value of y.

In case of $\dfrac{a}{c}=\dfrac{b}{d}\ne \dfrac{m}{n}$, for particular values of x and y, we have different values of same function $ax+by=cx+dy$.

Now, for $x+2y=3$ and $2x+4y=6$. We take the coefficients and place them in the formula. The ratios are $\dfrac{1}{2},\dfrac{2}{4},\dfrac{3}{6}$.

**The relation is $\dfrac{1}{2}=\dfrac{2}{4}=\dfrac{3}{6}=\dfrac{1}{2}$. So, we have an infinite number of solutions for those equations.**

**Note:**The solution of $x+2y=3$ and $2x+4y=6$. Both the equations are the same. We take any of the equations and then proceed with an assumed value of $x$. Let $x=-1$. We put that value in the equation of $x+2y=3$.

$\begin{align}

& x+2y=3 \\

& \Rightarrow -1+2y=3 \\

& \Rightarrow 2y=3+1=4 \\

& \Rightarrow y=\dfrac{4}{2}=2 \\

\end{align}$

One of those solutions is $x=-1,y=2$.

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