MATH Topic - Coordinate system and Linear graphs Q.1) Complete the following tables and plot the points on the graph paper to represent the equations given below 1 2 3 X y=x+1 y=-3x (x, y) (x, y)​

Step-by-step explanation:

Given Question

Complete the following tables and plot the points on the graph paper yo represents the equations given below :-

[tex]\begin{gathered}\boxed{\begin{array}{c|c|c|c} \bf x & \bf 1 & \bf 2& \bf 3\\ \frac{\qquad}{} & \frac{\qquad}{}\frac{\qquad}{} &\frac{\qquad}{} & \frac{\qquad}{} &\\ \sf y = x + 1 & \sf & \sf & \sf \\ \\ \sf (x,y)& \sf & \sf & \sf \\ \end{array}} \\ \end{gathered} \\ [/tex]

and

[tex]\begin{gathered}\boxed{\begin{array}{c|c|c|c} \bf x & \bf 1 & \bf 2& \bf 3\\ \frac{\qquad}{} & \frac{\qquad}{}\frac{\qquad}{} &\frac{\qquad}{} & \frac{\qquad}{} &\\ \sf y = - 3x & \sf & \sf & \sf \\ \\ \sf (x,y)& \sf & \sf & \sf \\ \end{array}} \\ \end{gathered} \\ [/tex]

[tex]\large\underline{\sf{Solution-}}[/tex]

Given that,

[tex]\rm \longmapsto\:y = x + 1[/tex]

On substituting x = 1, we get

[tex]\rm \longmapsto\:y = 1 + 1[/tex]

[tex]\rm \longmapsto\:y = 2[/tex]

On substituting x = 2, we get

[tex]\rm \longmapsto\:y = 2 + 1[/tex]

[tex]\rm \longmapsto\:y = 3[/tex]

On substituting x = 3, we get

[tex]\rm \longmapsto\:y = 3 + 1[/tex]

[tex]\rm \longmapsto\:y = 4[/tex]

Hence,

[tex]\begin{gathered}\boxed{\begin{array}{c|c|c|c} \bf x & \bf 1 & \bf 2& \bf 3\\ \frac{\qquad}{} & \frac{\qquad}{}\frac{\qquad}{} &\frac{\qquad}{} & \frac{\qquad}{} &\\ \sf y = x + 1 & \sf 2 & \sf 3 & \sf 4\\ \\ \sf (x,y)& \sf (1,2) & \sf (2,3) & \sf (3,4)\\ \end{array}} \\ \end{gathered}[/tex]

Now, draw a graph using the points (1 , 2), (2 , 3) & (3 , 4)

---------------------------------------------

Given equation is

[tex]\rm \longmapsto\:y = - 3x[/tex]

On substituting x = 1, we get

[tex]\rm \longmapsto\:y = - 3 \times 1[/tex]

[tex]\rm \longmapsto\:y = - 3[/tex]

On substituting x = 2, we get

[tex]\rm \longmapsto\:y = - 3 \times 2[/tex]

[tex]\rm \longmapsto\:y = - 6[/tex]

On substituting x = 3, we get

[tex]\rm \longmapsto\:y = - 3 \times 3[/tex]

[tex]\rm \longmapsto\:y = - 9[/tex]

Hence,

[tex]\begin{gathered}\boxed{\begin{array}{c|c|c|c} \bf x & \bf 1 & \bf 2& \bf 3\\ \frac{\qquad}{} & \frac{\qquad}{}\frac{\qquad}{} &\frac{\qquad}{} & \frac{\qquad}{} &\\ \sf y = - 3x & \sf - 3 & \sf - 6 & \sf - 9\\ \\ \sf (x,y)& \sf (1, - 3) & \sf (2, - 6) & \sf (3, - 9)\\ \end{array}} \\ \end{gathered}[/tex]

Now, draw a graph using the points (1 , - 3), (2 , - 6) & (3 , - 9)

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