# maths, variations checking!

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question: given that y is inversely proportional to (x-1),

A) express y in terms of x and a constant k
b) give that y=5 when x=3, find x when y=25

y=-kx+k

B)y=5, x=3, k(constant)=3x5
k=15
25=-15x+15
10=-15x
x=-10/15

• maths, variations checking! -

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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