maths, variations checking!
posted by hayley .
question: given that y is inversely proportional to (x1),
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(x1)
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
Your expression for y is incorrect, try this instead...
A) y is inversely proportional to (x1) if there exists a nonzero constant k such that y = k/(x1).
B) Given that when y=5 x=3, we can solve for k in y = k/(x1) as follows:
5 = k/(31)
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/(x1)
5 = 10/(31)
5 = 10/2
5 = 5 is true, therefore k=10 is correct
Now, substituting k=10, find x when y =25:
y = k/(x1)
25 = 10/(x1)
25(x1) = 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/(x1)
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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