Question

solve the following pair of equations by graphical method as well as elimination method. 1. 2x+ y= 6 , 2x-y=2 ​

Answer 1

$\large\underline{\sf{Solution-}}$

Given pair of linear equations are

$\rm :\longmapsto\:2x + y = 6 - - - (1)$

and

$\rm :\longmapsto\:2x - y = 2 - - - (2)$

$\green{\large\underline{\sf{Solution-using \:graphical \: method }}}$

Consider first equation

$\rm :\longmapsto\:2x + y = 6$

➢ Pair of points of the given equation are shown in the below table.

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 0 & \sf 6 \\ \\ \sf 3 & \sf 0 \end{array}} \\ \end{gathered}$

Consider, Second equation

$\rm :\longmapsto\:2x - y = 2$

➢ Pair of points of the given equation are shown in the below table.

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 0 & \sf - 2 \\ \\ \sf 1 & \sf 0 \end{array}} \\ \end{gathered}$

➢ Now draw a graph using the points

➢ See the attachment graph.

So, Solution of pair of linear equations is

$\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\:-\begin{cases} &\sf{x = 2} \\ \\ &\sf{y = 2} \end{cases}\end{gathered}\end{gathered}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

$\red{\large\underline{\sf{Solution-elimination \: method}}}$

Given pair of linear equations are

$\rm :\longmapsto\:2x + y = 6 - - - (1)$

and

$\rm :\longmapsto\:2x - y = 2 - - - (2)$

On adding equation (1) and (2), we get

$\rm :\longmapsto\:4x = 8$

$\rm :\longmapsto\:x = 2$

On substituting x = 2, in equation (1) we get

$\rm :\longmapsto\:2(2) + y = 6$

$\rm :\longmapsto\:4 + y = 6$

$\rm :\longmapsto\:y = 6 - 4$

$\rm :\longmapsto\:y = 2$

So, Solution of pair of linear equations is

$\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\:-\begin{cases} &\sf{x = 2} \\ \\ &\sf{y = 2} \end{cases}\end{gathered}\end{gathered}$

Answer 2

$\large\underline{\sf{Solution-}}$

Given pair of linear equations are

$\rm :\longmapsto\:2x + y = 6 - - - (1)$

and

$\rm :\longmapsto\:2x - y = 2 - - - (2)$

$\green{\large\underline{\sf{Solution-using \:graphical \: method }}}$

Consider first equation

$\rm :\longmapsto\:2x + y = 6$

➢ Pair of points of the given equation are shown in the below table.

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 0 & \sf 6 \\ \\ \sf 3 & \sf 0 \end{array}} \\ \end{gathered}$

Consider, Second equation

$\rm :\longmapsto\:2x - y = 2$

➢ Pair of points of the given equation are shown in the below table.

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 0 & \sf - 2 \\ \\ \sf 1 & \sf 0 \end{array}} \\ \end{gathered}$

➢ Now draw a graph using the points

➢ See the attachment graph.

So, Solution of pair of linear equations is

$\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\:-\begin{cases} &\sf{x = 2} \\ \\ &\sf{y = 2} \end{cases}\end{gathered}\end{gathered}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

$\red{\large\underline{\sf{Solution-elimination \: method}}}$

Given pair of linear equations are

$\rm :\longmapsto\:2x + y = 6 - - - (1)$

and

$\rm :\longmapsto\:2x - y = 2 - - - (2)$

On adding equation (1) and (2), we get

$\rm :\longmapsto\:4x = 8$

$\rm :\longmapsto\:x = 2$

On substituting x = 2, in equation (1) we get

$\rm :\longmapsto\:2(2) + y = 6$

$\rm :\longmapsto\:4 + y = 6$

$\rm :\longmapsto\:y = 6 - 4$

$\rm :\longmapsto\:y = 2$

So, Solution of pair of linear equations is

$\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\:-\begin{cases} &\sf{x = 2} \\ \\ &\sf{y = 2} \end{cases}\end{gathered}\end{gathered}$