Question

Is the relationship shown by the data linear? If so, model the data with an equation. x y 1 –4 7 –7 13 –10 19 –13

Answer 1

You can see that there is a proportional relationship between X and Y because every 6 of X  decreases 3 in Y. In that way you can say is a linear relationship and you can construct a linear equation: Now, to calculate the equation you use: (Y - y1) = y2 - y1  (X - x1)                x2 - x1 Let's peek two pairs (1,-4) and (7,-7) (Y - (-4)) = (-7 - (-4))  (X - 1)                    (7 - 1) Y + 4 =  -3 (X - 1)               6 If you want to express the equation as  aX + bY + C = 0 Y + 1/2 X + 7/2 = 0 If you want to express the equation as Y= ax + b Y = -1/2X - 7/2

Answer 2

Answer:

Yes, the relationship shown by the data linear.

The equation is given by:

             $y=\dfrac{-1}{2}x-\dfrac{7}{2}$

Step-by-step explanation:

The table of values is given by:

           x            y

           1            -4

           7            -7

           13          -10

           19          -13

We know that the difference in each of the x-value is: 6

( Since, 7-1=6

13-7=6

19-13=6)

and if the difference in the each of the y-value is same then the relationship is linear.

( Since, a table of values represent a linear relationship if the rate of change is constant.

i.e. the ratio of change in y-values to the change in x-values)

Hence, we find the difference in y-value:

-7-(-4)= -3

-10-(-7)= -3

-13-(-10)= -3

Since, the difference in y-value is constant.

             Hence, the relationship is linear.

Also, we know that the equation of the line will pass through (1,-4) and (7,-7)

Hence, the equation of line is calculated by:

$y-(-4)=\dfrac{-7-(-4)}{7-1}\times (x-1)\\\\i.e.\\\\y+4=\dfrac{-3}{6}\times (x-1)\\\\i.e.\\\\y+4=\dfrac{-1}{2}\times (x-1)\\\\i.e.\\\\y+4=\dfrac{-1}{2}x+\dfrac{1}{2}\\\\i.e.\\\\y=\dfrac{-1}{2}x+\dfrac{1}{2}-4\\\\i.e.\\\\y=\dfrac{-1}{2}x-\dfrac{7}{2}$