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Math Help - Changing the subject of a formula

  1. #1
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    Changing the subject of a formula

    How do I make 'x' the subject for these? Can you please show all the steps and working out:

    1) h = d - b/x
    2) r - m/x = e^2
    3) t^2 = b - n/x


    4) 3M = M + N/P+x
    5) m^2/x - n = -p
    6) t = w - q/x


    Thanks if you can help!
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  2. #2
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    Quote Originally Posted by Lalis View Post
    How do I make 'x' the subject for these? Can you please show all the steps and working out:

    1) h = d - b/x
    h= d-\frac{b}{x}

    h-d= -\frac{b}{x}

    d-h= \frac{b}{x}

    x(d-h)= b

    x= \frac{b}{d-h}

    Now you have a go.
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  3. #3
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    Thanks. But the others are hard as well

    I'll give it a go and if I can't I'll come back.
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  4. #4
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    Quote Originally Posted by pickslides View Post

    h-d= -\frac{b}{x}

    d-h= \frac{b}{x}
    I didn't get this bit. Could you explain please?
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  5. #5
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    I multiplied -1 through both sides.

    Here's a bit more detail

    h-d= -\frac{b}{x}

    -1 \times (h-d)= -1 \times-\frac{b}{x}

    -h+d=\frac{b}{x}

    Then swapping the order on the LHS

    d-h=\frac{b}{x}
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  6. #6
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    Hi sorry to bother you again, but I used the other one as an example and did #2. Can you check it?

    r - m/x = e^2
    -m/x = e^2 - r
    m/x = r - e^2
    m = x(r - e^2)
    m/r - e^2 = x

    And also I have two questions. Why did you multiply both sides by -1, and what is the purpose of swapping th order on the LHS? Thanks so much!
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  7. #7
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    Quote Originally Posted by Lalis View Post
    Hi sorry to bother you again, but I used the other one as an example and did #2. Can you check it?

    r - m/x = e^2
    -m/x = e^2 - r
    m/x = r - e^2
    m = x(r - e^2)
    m/r - e^2 = x
    This is correct, be careful though, use brackets.


    m/(r - e^2) = x


    Quote Originally Posted by Lalis View Post
    And also I have two questions. Why did you multiply both sides by -1, and what is the purpose of swapping th order on the LHS? Thanks so much!
    -1 on both sides gets rid of any negative values.

    Swapping the order makes it all neater.
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  8. #8
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    You wrote m/r- e^2= x and Pickslides wrote "be careful, though, use brackets m/(r- e^2)= x".

    Which makes me ask- is your original problem r- m/x= e^2 (which is r- \frac{m}{x}= e^2) or (r- m)/x= e^2 (which is \frac{r- m}{x}= e^2)? Those are very different equations with very different solutions.

    Similarly, was your first problem h= d- b/x ( h= d- \frac{b}{x}) or h= (d- b)/x ( h= \frac{d- b}{x}?

    pickslides assumed it was the first in both of those.
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