Question

solve the following simultaneous equations by graphically :- x+y=5 ; x-y=3​

Answer 1

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

Given pair of linear equations are

$\rm :\longmapsto\:x + y = 5 - - - (1)$

and

$\rm :\longmapsto\:x - y = 3 - - - (2)$

Consider Line (1)

$\rm :\longmapsto\:x + y = 5$

Substituting 'x = 0' in the given equation, we get

$\rm :\longmapsto\:0+ y = 5$

$\bf\implies \:y = 5$

Substituting 'y = 0' in the given equation, we get

$\rm :\longmapsto\:x + 0 = 5$

$\bf\implies \:x = 5$

Hᴇɴᴄᴇ,

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

$\red{\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 0 & \sf 5 \\ \\ \sf 5 & \sf 0 \end{array}} \\ \end{gathered}}$

➢ Now draw a graph using the points (0 , 5) & (5 , 0)

➢ See the attachment graph.

Consider Line (2)

$\rm :\longmapsto\:x - y = 3$

Substituting 'x = 0' in the given equation, we get

$\rm :\longmapsto\:0 - y = 3$

$\rm :\longmapsto\: - y = 3$

$\bf\implies \:y = - 3$

Substituting 'y = 0' in the given equation, we get

$\rm :\longmapsto\:x - 0 = 3$

$\bf\implies \:x = 3$

Hᴇɴᴄᴇ,

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

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

➢ Now draw a graph using the points (0 , - 3) & (3 , 0)

➢ See the attachment graph.

From graph we concluded that

$\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\:\begin{cases} &\sf{x = 4} \\ \\ &\sf{y = 1} \end{cases}\end{gathered}\end{gathered}$