Math, asked by dumeshverma288, 1 month ago

A situation is represented by the equations. x+2y-4=0 and 2x+4y-12=0. represent this situation graphically. write your comment

Answers

Answered by mathdude500
12

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

Given pair of lines is

\rm :\longmapsto\:x + 2y - 4 = 0

and

\rm :\longmapsto\:2x + 4y - 12 = 0

Consider,

 \red{\rm :\longmapsto\:x + 2y - 4 = 0}

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

 \red{\rm :\longmapsto\:0 + 2y - 4 = 0}

 \red{\rm :\longmapsto\:2y = 4}

 \red{\rm :\longmapsto\:y = 2}

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

 \red{\rm :\longmapsto\:x + 2(0) - 4 = 0}

 \red{\rm :\longmapsto\:x - 4 = 0}

 \red{\rm :\longmapsto\:x = 4}

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 2 \\ \\ \sf 4 & \sf 0 \end{array}} \\ \end{gathered}}

➢ Now draw a graph using the points.

➢ See the attachment graph.

Now, Consider

 \purple{\rm :\longmapsto\:2x + 4y - 12 = 0}

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

 \purple{\rm :\longmapsto\:2(0) + 4y - 12 = 0}

 \purple{\rm :\longmapsto\:4y - 12 = 0}

 \purple{\rm :\longmapsto\:4y = 12}

 \purple{\rm :\longmapsto\:y = 3}

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

 \purple{\rm :\longmapsto\:2x + 4(0) - 12 = 0}

 \purple{\rm :\longmapsto\:2x - 12 = 0}

 \purple{\rm :\longmapsto\:2x = 12}

 \purple{\rm :\longmapsto\:x = 6}

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 6 & \sf 0 \end{array}} \\ \end{gathered}}

➢ Now draw a graph using the points.

➢ See the attachment graph.

From graph we concluded that,

Lines are parallel, and hence the given system of equations has no solution.

Attachments:
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