⦁ Determine if the equation represents a direct variation. .8x = 1.6y ⦁ Suppose y varies directly with x. Write a direction variation equation that relates x and y. Then find the value of y when x = 12. y=5 when x=2. ⦁ For the data in the table, does y vary directly with x? If it does, write an equation for the direct variation. X Y -6 9 1 -1.5 8 -12 ⦁ For the data in the table, does y vary directly with x? If it does, write an equation for the direct variation. X Y -2 1 3 6 8 11

Part A:

Given that 

[tex]0.8x=1.6y\Rightarrow x= \frac{1.6}{0.8} y=2y[/tex]

Thus, x = ky where k = 2.

Therefore, the given equation is a direct variation.


Part B:

If y varies directly as x, then y = kx.

Given that y = 5 when x = 2, then

[tex]5=2k \\ \\ \Rightarrow k=\frac{5}{2} =2.5 \\ \\ \Rightarrow y=2.5x[/tex]

When x = 12, y = 2.5(12) = 30.



Part C:

If y varies directly as x, then y = kx.

Given that y = 9 when x = -6, then

[tex]9=-6k \\ \\ \Rightarrow k=-\frac{9}{6} =-\frac{3}{2} \\ \\ \Rightarrow y=-\frac{3}{2}x[/tex]

It can be seen that the equation above satisfies all the rows of the given table, therefore, y varies with x for the data in the question and the equation for the direct variation is [tex]y=-\frac{3}{2}x[/tex]



Part 4:

If y varies directly as x, then y = kx.

Given that y = 1 when x = -2, then

[tex]1=-2k \\ \\ \Rightarrow k=-\frac{1}{2} \\ \\ \Rightarrow y=-\frac{1}{2}x[/tex]

It can be seen that the equation above does not satisfies the other rows of the given table, therefore, y does not vary with x for the data in the question.