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

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! -

    Your 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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