Question

construct a table x and y which satisfies the equation x+2y=6​

Answer 1

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

Given equation is

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

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

$\rm :\longmapsto\:0 + 2y = 6$

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

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

Substituting 'y = 0' in the given equation,

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

$\rm :\longmapsto\:x + 2 \times 0 = 6$

$\rm :\longmapsto\:x = 6$

Substituting 'y = 1' in the given equation,

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

$\rm :\longmapsto\:x + 2 \times 1 = 6$

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

$\rm :\longmapsto\:x = 6 - 2$

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

Hᴇɴᴄᴇ,

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

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 0 & \sf 3 \\ \\ \sf 6 & \sf 0 \\ \\ \sf 4 & \sf 1 \end{array}} \\ \end{gathered}$

Answer 2

Question :-

construct a table x and y which satisfies the equation x+2y=6.

Solution :-

Given Equation :

• x + 2y = 6

From given equation,

$\sf : \longmapsto{x + 2y = 6} \: \: \\ \\ \sf : \longmapsto{x + y = \frac{6}{2} } \: \: \\ \\ \sf : \longmapsto{x + y = 3} \: \: \\ \\ \bf : \longmapsto \red{y = 3 - x}$

Now Substitute x = 0

$\sf : \longmapsto{y = 3 - 0} \\ \\\sf : \longmapsto{y = 3} \: \: \: \: \: \: \:$

$\sf : \longmapsto{(x,y) = (0,3)}$

Substitute x = 1

$\sf:\longmapsto{y = 3 - 1 } \: \: \: \: \: \: \: \: \: \: \\ \\ \sf:\longmapsto{y = 2 } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \sf:\longmapsto{(x, y) =( 1,2) } \:$

Substitute x = 2

$\sf:\longmapsto{y = 3 - 2} \: \: \: \: \: \: \: \: \: \\ \\\sf:\longmapsto{y = 1} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \sf:\longmapsto{(x, y) = (2, 1)}$

Table :-

$\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y\\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 0 & \sf 3 \\ \\ \sf 1 & \sf 2 \\ \\ \sf 2 & \sf 1\end{array}} \\ \end{gathered}\end{gathered}$

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