Variations

Received PM asking for clarification. Frankenstein is right on but let me see if a slight variation in the notation might clear it up for anyone that isn't 100% with us!

If Y moves directly with x, that means that there exists a constant (let's call it j) where y=jx. If Y were to move inversely with x, then that means there exists a constant (let's call this one k) where y=k/x.

This problem says that the value of Y is the SUM of two values - one that varies directly with x and one that varies inversely with x, so we can use what we learned above.

Y = jX + k/X.

Now, we are given 2 cases {y=6,x=4} and {y=10/3, x=3}. We can plug these each into our equation above to get the following 2 equations:

(a) 6 = j(4) + k/(4)
(b) 10/3 = j(3) + k/(3)

Let's multiply both equations through with their denominators to simplify:

(a) 24 = 16j + k
(b) 10 = 9j + k

Now we have 2 equations in 2 unknowns, the constants j and k. We can subtract equation (b) from equation (a):

24 = 16j + k
-10 = -9j - k
---------------
14 = 7j
2 = j

Now plug back in to either to solve for k:

10 = 9(2) + k
-8 = k

We can then plug these into our original equation:

Y = jX + k/X
Y = 2X - 8/X

So the answer is C[/spoiler]


Whit