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November 24, 2014

November 24, 2014

Posted by **hayley** on Tuesday, December 18, 2007 at 5:57am.

A) express y in terms of x and a constant k

b) give that y=5 when x=3, find x when y=25

answer: A)y=-k(x-1)

y=-kx+k

B)y=5, x=3, k(constant)=3x5

k=15

25=-15x+15

10=-15x

x=-10/15

is this correct? i'm not too sure about this variation thing, i would appreciate your help!

- maths, variations checking! -
**Cathy**, Tuesday, December 18, 2007 at 6:12pmYour expression for y is incorrect, try this instead...

A) y is inversely proportional to (x-1) if there exists a non-zero constant k such that y = k/(x-1).

B) Given that when y=5 x=3, we can solve for k in y = k/(x-1) as follows:

5 = k/(3-1)

5 = k/2

5(2) = k [cross multiply]

10 = k or k=10

Check the result: Substitute k=10 when y=5 and x=3:

y = k/(x-1)

5 = 10/(3-1)

5 = 10/2

5 = 5 is true, therefore k=10 is correct

Now, substituting k=10, find x when y =25:

y = k/(x-1)

25 = 10/(x-1)

25(x-1) = 10 [cross multiply]

25x - 25 = 10

25x - 25 + 25 = 10 + 25

25x = 35

x = 35/25 or x=7/5

Check the result: Substitute x=7/5 when y=25 and k=10:

y = k/(x-1)

25 = 10/((7/5) - 1)

25((7/5) - 1) = 10 [cross multiply

25 ((7/5) - (5/5)) = 10

25 (2/5) = 10

25(2)/5 = 10

50/5 = 10

10 = 10 is true, therefore x=7/5 is correct

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