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Math Help - Solving for one of the variables help

  1. #1
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    Solving for one of the variables help

    These are the problems i have been trying to solve.

    Solve for n.

    1.) na + n= cd



    2.) n/x + 1= x - d

    My teacher gave me no explanation for these types of problems. Can someone help me figure them out?
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  2. #2
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    The first problem, factor out the n. The second problem, first multiply both sides by x. Does this help?
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  3. #3
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    1.) n= cda/2 ?

    2.) n= x^2 -dx -1 ?

    Are these right? or am i completely off.
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  4. #4
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    To answer your question, take your answer and substitute in the original problems to see if the left-hand side equals the right-hand side. If they do, then your answer is correct.
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  5. #5
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    My work says that they are both wrong. I need a math book! I'm on summer break.
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  6. #6
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    Can you help me factor the first one for n?
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  7. #7
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    Quote Originally Posted by Youngice11 View Post
    Can you help me factor the first one for n?
    Have you studied factoring? Please show all your work you've done so far.
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  8. #8
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    We did not use factoring in Geometry. Obviously. This is my summer assignment for physics. I don't remember how to factor.

    na + n= cd
    n= cd -na
    n/a = cd- n
    n + n/a = cd
    n+ n = cda
    2n=cda
    n= cda/2
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  9. #9
    Senior Member eumyang's Avatar
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    I see a number of mistakes here.
    Quote Originally Posted by Youngice11 View Post
    na + n= cd
    n= cd -na
    n/a = cd- n
    This isn't right. If you want to divide both sides by a here, you have to divide the cd term by a too, like this:
    n/a = cd/a - n

    n + n/a = cd
    n+ n = cda
    This isn't right either, for the same reason. If you want to multiply both sides by a, you have to multiply ALL terms by a, like this:
    na + n = cda

    Here's what you should do: one way of factoring is to take out the greatest common monomial factor -- in a sense, you are reversing the distributive property. Since on the left side there is a common factor of n, you "take it out":
    \begin{aligned}<br />
na + n &= cd \\<br />
na + n(1) &= cd \\<br />
n(a + 1) &= cd \\<br />
\end{aligned}

    You can see that this is true by going backwards and distributing the n. Now that you have n times something, you can divide both sides by that other factor:
    \begin{aligned}<br />
n(a + 1) &= cd \\<br />
\frac{n(a + 1)}{(a + 1)} &= \frac{cd}{(a + 1)} \\<br />
n &= \frac{cd}{a + 1}<br />
\end{aligned}
    Last edited by eumyang; July 27th 2010 at 10:21 AM. Reason: Added some steps for clarity
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  10. #10
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    OK. Thank you i understand the distributing and how you got the answer. The second problem....

    n/x + 1 =x -d
    first you want to multiply times x.

    n + x = x^2 -dx or n + 1 = x^2 -dx?

    What is next?
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  11. #11
    Senior Member eumyang's Avatar
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    Quote Originally Posted by Youngice11 View Post
    OK. Thank you i understand the distributing and how you got the answer. The second problem....

    n/x + 1 =x -d
    first you want to multiply times x.

    n + x = x^2 -dx or n + 1 = x^2 -dx?

    What is next?
    It's the first one. Here's how to get it, step by step:
    \begin{aligned}<br />
\frac{n}{x} + 1 &= x - d \\<br />
x \left( \frac{n}{x} + 1 \right) &= x(x - d) \\<br />
x \left( \frac{n}{x} \right) + x(1) &= x(x) - x(d) \\<br />
n + x &= x^2 - dx<br />
\end{aligned}

    Then subtract x from both sides, and you can clean up along the way:
    \begin{aligned}<br />
n + x &= x^2 - dx \\<br />
n &= x^2 - dx - x \\<br />
n &= x^2 - x(d + 1) \\<br />
n &= x^2 - (d + 1)x \\<br />
 \end{aligned}
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  12. #12
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    OK. On the third line once you subtracted x, you factored the x. correct? Also.. Can you tell me if this answer is correct to this problem?


    solve for n.


    n+2/c = n+c/b

    my answer was..

    n= c^2 -2b/ b-c
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  13. #13
    Senior Member eumyang's Avatar
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    You should learn some LaTeX in typing these equations, because I'm not sure what your equation is. I could read it in one of two ways.

    (A) n + \frac{2}{c} = n + \frac{c}{b}

    (B) \frac{n + 2}{c} = \frac{n + c}{b}

    So which one is it? (A) or (B)? Because if it is (A), the n's would cancel from both sides.
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  14. #14
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    B)
    How do i use LaTeX?
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  15. #15
    MHF Contributor Unknown008's Avatar
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    LaTeX is the use of math tabs. You put the square brackets around 'math' like this [tex] and you close by inserting a similar tab but with '/' in front of 'math'. This allow the ease of display of some functions and fractions

    If it's B, then

    \frac{n+2}{c} = \frac{n+c}{b}

    Multiply throughout by bc. Some b and c will cancel out, giving;

    b(n+2) = c(n+c)

    Expand;

    bn + 2b = nc + c^2

    Remove nc from both sides, and remove 2 b too from both sides;

    bn + 2b  -2b - nc = nc + c^2 - 2b - nc

    bn  - nc = c^2 - 2b

    Factor n,

    n(b  - c) = c^2 - 2b

    Divide by (b-c);

    n= \frac{c^2 - 2b}{b-c}

    Looks like you did well!
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