Posted by TyTy on Saturday, January 5, 2008 at 5:34pm.
This is kind of like the first question I asked, and I'm still having a little trouble on it.
The Ordered pairs in each exercise are for the same direct variation.
Find the missing value.
(2,8) and (x,12)
I'm not sure how to work out the problem.

Stuck on Algebra 1  TyTy, Saturday, January 5, 2008 at 6:18pm
I'm not sure but if i cross multiply the two coordinates, i think the answer is...
x=3?

Stuck on Algebra 1  Writeacher, Saturday, January 5, 2008 at 6:33pm
Crossmultiplying seems to be the process:
http://www.jiskha.com/display.cgi?id=1199568727
So ...
2/8 = x/12
8x = 24
x = 3
Right?

Stuck on Algebra 1  TyTy, Saturday, January 5, 2008 at 6:46pm
yes, x=3

Stuck on Algebra 1  Damon, Saturday, January 5, 2008 at 6:54pm
Although I am sure I will be called on the carpet for saying why, I will try anyway.
If it is a direct variation it is a straight line of form:
y = k x
put in your first point
8 = 2 x
so k = 4
NOW any other point on this line satisfies
y = 4 x
so put in 12 for y
12 = 4 x
x = 3 as you know from "cross multiplying"

Stuck on Algebra 1  Damon, Saturday, January 5, 2008 at 6:59pm
By the way, these problems are a special case of direct variation, linear.
The relation could be quadratic, cubic quartic ....
but is always of form y = k x^n
Here so far we are assuming that n = 1

Stuck on Algebra 1  Damon, Saturday, January 5, 2008 at 7:00pm
Do you get it now?

Stuck on Algebra 1  Damon, Saturday, January 5, 2008 at 7:03pm
Fixing typo and expanding
y = k x
put in your first point
8 = 2 k
solve that for k
so k = 4
NOW any other point on this line satisfies
y = 4 x
so put in 12 for y
12 = 4 x
x = 3 as you know from "cross multiplying"

Stuck on Algebra 1  Damon, Saturday, January 5, 2008 at 7:41pm
I do not think you will get direct variation that is quadratic etc in this course or they would have mentioned it. For now assume direct variation means y = k x
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