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March 30, 2017

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I also do not get these inverses questions

y = 5(x-3)^3
x = 5(y-33)^3
I divided 5 on both sides and got

1/5x = (y-3)^3

Then do I cube root the 1/5x or something to get rid of of the cubed on (y-3)?

Also

y = x-2/5x-3

  • Pre Calculus 12 - ,

    From 1/5x = (y-3)^3 , which is the inverse and correct, you now have to solve for y
    So, take the cube root of both sides

    (x/5)^(1/3) = y-3

    y = (x/5)^(1/3) + 3

    check:
    in original let x = 4
    y = 5(1)^3 = 5

    now use x=5 in your inverse

    y = (5/5)^(1/3) + 3 = 4 , which is what we started with.

    now another, x = 8
    y = 5(5^3) = 625

    now use x=625 in inverse:
    y = (625/5)^(1/3) + 3
    = (125)^(1/3) + 3
    = 5 +3 = 8
    all looks good!!!

  • Pre Calculus 12 - ,

    Where did you get 4 from? For the first one? Where you wrote in original let x = 4?

  • Pre Calculus 12 - ,

    And I do not get this one

    y = x-2/5x-3
    x = y-2/5y-3

    ?

  • Pre Calculus 12 - ,

    I could have picked any x I wanted, since it must be true for all values of x
    I simply picked x = 4, because it gave me a "nice" answer.
    Same reason I picked x=8
    I could have picked x = 45.678

    As to your second post
    is that a new equation??

    did you mean:
    y = (x-2)/(5x-3)
    those brackets are critically important.

  • Pre Calculus 12 - ,

    Oh okay that makes sense, Im just sort of still confused with how the answer to the first one needs to be layed out

    I got y = ³√(1/5x) +3

    And yes that is a new equation on my second post, and yeah I meant that, sorry will remember to put brackets next time.

  • Pre Calculus 12 - ,

    y = ³√(1/5x) +3 is good

    Just make sure when you type the bracket part (1/5x) , it is only 5 that is divided and not the 5 as well
    that is why I wrote it as (x/5)

    Your new problem:

    y = (x-2)/(5x-3) -----> x = (y-2)/(5y-3) as the inverse

    cross-multiply
    5xy - 3x = y-2
    5xy - y = 3x - 2
    y(5x - 1) = 3x - 2
    y = (3x-2)/(5x - 1)

    test by taking any suitable value for x in the original
    I suggest x = 2, (worked for me)

  • Pre Calculus 12 - ,

    Could you explain why it wouldnt be 1/5x again? Because isnt there an invisible 1 beside the x, and so when we divide the 5 doesnt it become 1/5x?

    For the second equation to get rid of 5y-3 on the bottom you multiply it on the other side? Could you explain step by step the whole cross multiplication?

  • Pre Calculus 12 - ,

    I didn't say it shouldn't be 1/5x
    I just wanted you to be aware of the difference between
    1/5x and 1/(5x)

    for your second part of your question ....
    look at
    x = (y-2)/(5y-3) , so far ok ?
    or
    x/1 = (y-2)/(5y-3)
    the common denomator is 5y-3 , so multiply each side by that

    x(5y-3) = (y-2)/(5y-3) * (5y-3) , the * means multiply
    so after you cancel the 5y-3 on the right, you get

    x(5y-3) = 1(y-2)
    now look at that pattern.
    Did we not multiply the top of the left fraction by the bottom of the second fraction, and the bottom of the first fraction by the top of the 2nd fraction?
    That is why we call it cross-multiplying
    (use it only if you have a single fraction on the left, and a single fraction on the right of the equal sign)

  • Pre Calculus 12 - ,

    How is 5y-3 is a common denominator?

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