Solve the pair of linear equations graphically.
$
  2x + y - 6 = 0 \\
  4x - 2y - 4 = 0 \\
 $

Hint: To find a solution to a given linear equation graphically. We first take different values of either of one variable and then solve the other variable by the help of the given equation and then plot a graph of both equations on the same graph to find where both lines meet or we can say the solution of the given equations.

Complete step-by-step answer:
Given, linear equations are
$
  2x + y - 6 = 0 \\
  4x - 2y - 4 = 0 \;
 $
To draw a graph of a given linear equation.
We take different values of either or of one variable and solve another variable using one variable.
We first consider an equation:
$2x + y - 6 = 0$
Taking $x = 1$ in above equation. We have
$
  2(1) + y - 6 = 0 \\
   \Rightarrow 2 + y - 6 = 0 \\
   \Rightarrow y - 4 = 0 \\
   \Rightarrow y = 4 \;
 $
We have $\left( {1,4} \right)$
Taking $x = 2\,$we have,
$
  2(2) + y - 6 = 0 \\
   \Rightarrow 4 + y - 6 = 0 \\
   \Rightarrow y - 2 = 0 \\
   \Rightarrow y = 2 \;
 $
We have $\left( {2,2} \right)$
Taking $x = 3$we have
$
  2\left( 3 \right) + y - 6 = 0 \\
  6 + y - 6 = 0 \\
  y = 0 \;
 $
We have $\left( {3,0} \right)$
Hence, table of values given as



Now, considering other equation:
$
  4x - 2y - 4 = 0 \\
  or\,\,we\,\,can\,\,write\,\,it\,\,as: \\
  2x - y - 2 = 0 \;
 $
Now take the different values of x in the above equation and solve for y.
Taking $x = 1$
$
  2(1) - y - 2 = 0 \\
   \Rightarrow 2 - y - 2 = 0 \\
   \Rightarrow - y = 0 \\
   \Rightarrow y = 0 \;
 $
We have $\left( {1,0} \right)$
Taking $x = 2$
$
  2(2) - y - 2 = 0 \\
   \Rightarrow 4 - y - 2 = 0 \\
   \Rightarrow 2 - y = 0 \\
   \Rightarrow y = 2 \;
 $
We have $\left( {2,2} \right)$
Taking $y = 3$
$
  2(3) - y - 2 = 0 \\
   \Rightarrow 6 - y - 2 = 0 \\
   \Rightarrow 4 - y = 0 \\
   \Rightarrow y = 4 \;
 $
We have $\left( {3,4} \right)$
Hence, table of values given as:


Now plotting all points obtained in the x y plane to draw a graph of both equations.

From the above graph we see that graphs of both lines intersect or meet at a point$\left( {2,2} \right)$.
Hence the solution of the given equation is$\left( {2,2} \right)$.
So, the correct answer is “{2,2} ”.

Note: While plotting deterrent points on x y plane if graph is not coming straight line then we can say that there is a mistake in calculations as we know that graph of linear equation is a straight line. Also, we can cross check graphs by putting points of intersection in given equations as points of intersection will satisfy both linear equations.