Question

solve graphically 2x+y=6, x+y=4 ​

Answer 1

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

Consider first equation

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

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

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

$\rm \: 0 + y = 6$

$\rm\implies \:y = 6$

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

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

$\rm \: 2x = 6$

$\rm\implies \:x = 3$

Hᴇɴᴄᴇ,

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

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

Consider second equation

$\rm \: x + y = 4$

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

$\rm \: 0 + y = 4$

$\rm\implies \:y = 4$

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

$\rm \: x + 0 = 4$

$\rm\implies \:x = 4$

Hᴇɴᴄᴇ,

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

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

➢ Now draw a graph using the points (0 , 6), (3 , 0), (0, 4) & (4 , 0)

➢ See the attachment graph.

Now, from graph we concluded that system of equations is consistent having unique solution and solution is

$\rm\implies \:\boxed{ \rm{ \:x = 2 \: \: \: and \: \: \: y = 2 }}$

Answer 2

$\huge\underline{\sf{Solution = > }}$

$\large\underline{\sf{consider \: first \: equation..}}$

$\implies \large\underline{\sf{2x + y = 6}}$

$\large{\sf{substituting \: x = 0 \: in \: the \: }}$

$\large\underline{\sf{equation \: we \: get}}$

$\large\underline{\sf{2 \times 0 + y = 6}}$

$\large\underline{\sf{0 + y = 6}}$

$\large\underline{\sf{y = 6}}$

$\large\underline{\sf{substituting \: y = 0 \: in \: the \: }}$

$\large\underline{\sf{equation \: we \: get}}$

$\large\underline{\sf{2x + 0 = 6}}$

$\large\underline{\sf{2x = 6}}$

$\large\underline{\sf{x = 6 \div 2}}$

$\large\underline{\sf{x = 3}}$

$\large\underline{\sf{ hence}}$

$\large\underline{\sf{ \rightarrow pair \: of \: points \: of \: the \: given}}$

$\large\underline{\sf{equation \: of \: points \: are}}$

$\large\underline{\sf{x \: \: \: \: \: \: \: \: \: y}} \\ \large\underline{\sf{0 \: \: \: \: \: \: \: \: 6}} \\ \large\underline{\sf{3 \: \: \: \: \: \: \: \: 0}}$

$\large\underline{\sf{consider \: {2}^{nd} \: equation}}$

$\large\underline{\sf{x + y = 4}}$

$\large\underline{\sf{substituting \: x = 0 \: in \: the}}$

$\large\underline{\sf{given \: equation \: we \: get}}$

$\large\underline{\sf{0 + y = 4}}$

$\large\underline{\sf{y = 4}}$

$\large\underline{\sf{Substituting \: y = 0 \: in \: the \: }}$

$\large\underline{\sf{given \: equation \: we \: get}}$

$\large\underline{\sf{x + 0 = 4}}$

$\large\underline{\sf{x = 4}}$

$\large\underline{\sf{hence}}$

$\large\underline{\sf{pair \: of \: points \: of \: the \: given}}$

$\large\underline{\sf{equation \: are:}}$

$\large\underline{\sf{x \: \: \: \: \: \: y}} \\ \large\underline{\sf{0 \: \: \: \: \: \: 4}} \\ \large\underline{\sf{4 \: \: \: \: \: \: 0}}$

$\large\underline{\sf{now \: draw \: a \: graph \: using }}$

$\large\underline{\sf{the \: point \: (0.6)(3.0)(0.4)}}$

$\large\underline{\sf{and \: (4.0)}}$

$\large\underline{\sf{see \: the \: attached \: graph}}$

$\large\underline{\sf{now \: from \: graph \: we \: concluded}}$

$\large\underline{\sf{that \: system \: of \: equation \: is}}$

$\large\underline{\sf{constitent \: having \: unique}}$

$\large\underline{\sf{solution \: ans \: solution \: is}}$

$\implies \large\underline{\sf{x = 2 \: \: \:. \: \: \: y = 2}}$