How do you solve the system \[5x+6y=4\] and \[10x+12y=8\]?

Hint: Assume the given equations as equation (1) and (2) respectively. Write the given equations in their standard form given as: - \[{{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0\] and \[{{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}=0\]. Find the respective coefficients of x, y and the constant term. Now, consider the ratios \[\dfrac{{{a}_{1}}}{{{a}_{2}}},\dfrac{{{b}_{1}}}{{{b}_{2}}}\] and \[\dfrac{{{c}_{1}}}{{{c}_{2}}}\]. If \[\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}\ne \dfrac{{{c}_{1}}}{{{c}_{2}}}\] then there will not be any solution. If \[\dfrac{{{a}_{1}}}{{{a}_{2}}}\ne \dfrac{{{b}_{1}}}{{{b}_{2}}}\] then there will be a unique solution, so solve the equations algebraically using the elimination method. If \[\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}=\dfrac{{{c}_{1}}}{{{c}_{2}}}\] then there will be infinite number of solutions.

Complete step-by-step solution:
Here, we have been provided with the system of equations \[5x+6y=4\] and \[10x+12y=8\] and we are asked to solve it, that means we have to find the values of the variables x and y. But first we need to check if the system has no solution, unique solution or infinite solution because we can solve the equations algebraically only when there will be a unique solution.
Now, let us assume the given equations as equation (1) and (2) respectively, so we have,
\[\Rightarrow 5x+6y=4\] - (1)
\[\Rightarrow 10x+12y=8\] - (2)
Converting the above equations in their standard forms given as: - \[ax+by+c=0\], we get,
\[\Rightarrow 5x+6y-4=0\]
\[\Rightarrow 10x+12y-8=0\]
Here, we can consider the above equations of the form \[{{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0\] and \[{{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}=0\], so on comparing we get the respective coefficients of x, y and the constant terms as: -
\[\begin{align}
  & \Rightarrow {{a}_{1}}=5,{{b}_{1}}=6,{{c}_{1}}=4 \\
 & \Rightarrow {{a}_{2}}=10,{{b}_{2}}=12,{{c}_{2}}=-8 \\
\end{align}\]
Considering the ratios \[\dfrac{{{a}_{1}}}{{{a}_{2}}},\dfrac{{{b}_{1}}}{{{b}_{2}}}\] and \[\dfrac{{{c}_{1}}}{{{c}_{2}}}\], we have,
\[\Rightarrow \dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{5}{10}=\dfrac{1}{2}\]
\[\Rightarrow \dfrac{{{b}_{1}}}{{{b}_{2}}}=\dfrac{6}{12}=\dfrac{1}{2}\]
\[\Rightarrow \dfrac{{{c}_{1}}}{{{c}_{2}}}=\dfrac{-4}{-8}=\dfrac{1}{2}\]
Clearly, we can see that we have the condition \[\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}=\dfrac{{{c}_{1}}}{{{c}_{2}}}\], so we can say that the above two equations have infinitely many solutions.

Note: One may note that if you will multiply equation (1) with 2 then you will get the same equation as given in equation (2). So, from this we can directly say that both are the same, it means we are actually having one equation with two variables, hence there will be an infinite number of solutions. Graphically we can also say that there will be infinite solutions because when we will draw their graph we will see that they will overlap with each other.