Expert: Sherman D. - 4/18/2006

Question
Hi it's me again,
1. Suppose y varies jointly as x and z. If y=10 when x=2 and z=4, find y when x=4 and z=3.

So far i got :

y1= x1z1
  -------  and y=kxz
y2= x2z2

10=  (2)(4)
--   ------   
y2   (4)(3)

2. How can you tell what type of variation is ?
a1 = a2
---  ---     or  a1b1 = b2a2
b1  b2                                          

3. Simplify x^2-1x-6
        ------------
       x^2+2x-15

3. 5xy* 4x^2y
 ----- ------
  2yz   10y^2

Can you break it down into steps.
Thanks

Answer
1.) Suppose y varies jointly as x and z. If y=10 when x=2 and z=4, find y when x=4 and z=3.

z = kxy
4 = k(2)(10)
4 = 20k
k = (1/5)

3 = (1/5)(4)y
3 = (4/5)y
y = 3/(4/5)
y = (3/1)/(4/5)
y = (3/1)*(5/4)
y = 15/4

ANS : y = 3.75

Info found at
www.okc.cc.ok.us/maustin/Variation/Variation-ma.htm

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2.) How can you tell what type of variation is ?
a1b1 = b2a2                               

Its inverse variation

y = k/x
k = xy

When trying to find the second set, its the same as saying

k = x1y1
k = x2y2

or in this case x1y1 = x2y2

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3.)
(x^2 - x - 6)/(x^2 + 2x - 15)
((x - 3)(x + 2))/((x + 5)(x - 3))
(x + 2)/(x + 5)

ANS : (x + 2)/(x + 5)

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4.)
((5xy)/(2yz)) * ((4x^2y)/(10y^2))

To make this easier, lets simplify each fraction

(5xy)/(2yz) = (5x)/(2z)
(4x^2y)/(10y^2) = (2x^2)/(5y)
So now you have
((5x)/(2z)) * ((2x^2)/(5y))
Multiply across
(10x^3)/(10yz)
Simplifies to
(x^3)/(yz)

On this one, you can also go to www.quickmath.com, Click on Simplify, and type in your problem like i typed it in, and it will give you the same value.

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Direct Variation :
y = kx or (x1/y1) = (x2/y2) or x1y2 = x2y1
Inverse Variation :
y = k/x or x1y1 = x2y2 or (x1/y2) = (x2/y1)
Jointly Variation :
z = kxy or k = z/(xy) or (z1/(x1y1)) = (z2/(x2y2)) or x2y2z1 = x1y1z2

The x1,y1,z1 just stands for the 1st set of values and the x2,y2,z2 just stands for the 2nd set of values.