How do you solve the system using the elimination method for $3x + y = 4$ and $6x + 2y = 8$?

Hint: First, multiply the equations so as to make the coefficients of the variables to be eliminated equal. Next, add or subtract the equation the equations obtained according as the terms having the same coefficients are of opposite or of the same sign. Solve the equation in one variable. Substitute the value found in any of the equations after substitution and find the value of another variable. The values of the variables constitute the solution of the given system of equations.

Complete step by step solution:
In this method, we eliminate one of the two variables to obtain an equation in one variable which can be easily solved. Putting the value of this variable in any one of the given equations, the value of another variable can be obtained.
The given system of equations is
$3x + y = 4$…(i)
$6x + 2y = 8$…(ii)
Now, multiply the equations so as to make the coefficients of the variables to be eliminated equal.
So, multiply equation (i) by 2.
$6x + 2y = 8$…(iii)
Next, add or subtract the equation the equations obtained according as the terms having the same coefficients are of opposite or of the same sign.
So, subtract equations (iii) from (i).
$0 = 0$

Hence, the given system of equations has infinitely many solutions.

Note: We can directly check whether the system of equations is consistent with infinitely many solutions or not using below property:
The system of equations ${a_1}x + {b_1}y + {c_1} = 0$ and ${a_2}x + {b_2}y + {c_2} = 0$ is consistent with infinitely many solutions, if
$\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} = \dfrac{{{c_1}}}{{{c_2}}}$……(i)
Step by step solution:
First, we have to compare $3x + y = 4$ and $6x + 2y = 8$ with ${a_1}x + {b_1}y + {c_1} = 0$ and ${a_2}x + {b_2}y + {c_2} = 0$.
${a_1} = 3,{b_1} = 1,{c_1} = - 4$ and ${a_2} = 6,{b_2} = 2,{c_2} = - 8$
Now we have to find $\dfrac{{{a_1}}}{{{a_2}}},\dfrac{{{b_1}}}{{{b_2}}},\dfrac{{{c_1}}}{{{c_2}}}$ and check whether it satisfy (i) or not.
$\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{3}{6} = \dfrac{1}{2}$, $\dfrac{{{b_1}}}{{{b_2}}} = \dfrac{1}{2}$ and $\dfrac{{{c_1}}}{{{c_2}}} = \dfrac{{ - 4}}{{ - 8}} = \dfrac{1}{2}$
Therefore, $\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} = \dfrac{{{c_1}}}{{{c_2}}}$.