How do you solve the following linear system \[2x-4y=3,2x-5y=3\]?

Hint: If we solve the linear system, the equation must be solved simultaneously. First of all we should add or subtract the equations so that the common term is eliminated. Since the coefficient of the variable \[x\] is the same and both the equations have the same sign, we should eliminate \[x\]. Here in this question eliminating \[x\] will be easier than multiplying the equations with any other number.

Complete step by step solution:
We have the equations with us, which are \[2x-4y=3,2x-5y=3\].
\[\begin{align}
  & 2x-4y=3.....\left( 1 \right) \\
 & 2x-5y=3....\left( 2 \right) \\
\end{align}\]
We should subtract the equations. If we subtract the equations then the variable \[y\] will be added and the variable \[x\] will be eliminated. Variables \[x\] will be eliminated because they are the same in both the equations. The constants will also get subtracted. Doing so, we get:
\[\begin{align}
  & \underline{\begin{align}
  & 2x-4y=3 \\
 & 2x-5y=3
\end{align}} \\
 & \,\,\,\,\,\,\,\,\, y\,=0 \\
\end{align}\]
Now we have to solve to find the value of \[x\]. To do so we should put \[y=0\] in \[2x-5y=3\], such that we get:
\[\begin{align}
  & \Rightarrow 2x-5y=3 \\
 & \Rightarrow 2x-5\cdot 0=3 \\
 & \Rightarrow 2x=3 \\
\end{align}\]
Now we have to divide by 2 on both the sides of \[2x=3\]. Such that the coefficient of \[x\]becomes 1.
\[\begin{align}
  & \Rightarrow 2x=3 \\
 & \Rightarrow \dfrac{2x}{2}=\dfrac{3}{2} \\
 & \Rightarrow x=\dfrac{3}{2} \\
\end{align}\]
Hence we have obtained the results of the equation. The values of variable \[x\] and $y$ is \[x=0\] and \[y=\dfrac{3}{2}\].

Note: We have obtained the results. We can also solve the equation by using the substitution method but the substitution method is very lengthy and requires more time as compared to this method. In the substitution method we take any variable either x or y and find their value and then substitute it in the other equation.