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Math Help - Systems of Equations :: Elimination using addition and subtraction

  1. #1
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    Systems of Equations :: Elimination using addition and subtraction

    I'm extremely confused with these two problems:

     7f + 3g = -6
    7f - 2g = -31

    What would I do first... would I subtract or add? I don't know when I should subtract or when I should add.

    5m - p = 7
    7m - p = 11
    would I add for this one?
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    Quote Originally Posted by AbstractHero View Post
    I'm extremely confused with these two problems:

     7f + 3g = -6
    7f - 2g = -31

    What would I do first... would I subtract or add? I don't know when I should subtract or when I should add.

    5m - p = 7
    7m - p = 11
    would I add for this one?
    \displaystyle\begin{bmatrix}7&3&-6\\7&-2&-31\end{bmatrix}\Rightarrow\text{Row1 - Row2}\begin{bmatrix}7&3&-6\\0&5&25\end{bmatrix}\Rightarrow\text{1/5Row2}\begin{bmatrix}7&3&-6\\0&1&5\end{bmatrix}

    g=5

    Now solve

    7f+3(5)=-6

    Do the same for the second system.
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    I don't understand it the way you set it up.
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    Quote Originally Posted by AbstractHero View Post
    I don't understand it the way you set it up.
    I set it up as a coefficient matrix. I take all the numbers associated with the variables and plugged them into a matrix and then used elementary row operations.
    Last edited by dwsmith; February 7th 2011 at 06:50 PM. Reason: used the wrong word
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    Quote Originally Posted by AbstractHero View Post
     7f + 3g = -6
    7f - 2g = -31

    5m - p = 7
    7m - p = 11
    Idea is to ELIMINATE a variable.
    If you subtract in 1st set of equation, f's are eliminated.
    If you subtract in 2nd set of equation, p's are eliminated.
    See that?
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    Use Gaussian elimination with row operations to find g. Then back substituting the answer for g into equation one to solve for f.

    \begin{cases}7f+3g=-6\\7f-2g=-31\end{cases}

    first multiply equation 2 by -1.

    \begin{cases}7f+3g=-6\\-7f+2g=31\end{cases}

    then add equation 1 and equation 2. the sum is the new equation 2

    \begin{cases}7f+3g=-6\\5g=25\end{cases}

    multiply equation 2 by 1/5.

    \begin{cases}7f+3g=-6\\g=5\end{cases}

    now back substitute g into equation 1 and solve for f.
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    Quote Originally Posted by AbstractHero View Post
    I'm extremely confused with these two problems:

     7f + 3g = -6 .... (1)
    7f - 2g = -31 .... (2)

    What would I do first... would I subtract or add? I don't know when I should subtract or when I should add.

    [snip]
    (1) - (2): 5g = 25 => g = 5 etc.

    Quote Originally Posted by AbstractHero View Post
    [snip]
    5m - p = 7 .... (1)
    7m - p = 11 .... (2)
    would I add for this one?
    (1) - (2): ...........
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  8. #8
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    Quote Originally Posted by AbstractHero View Post
    I'm extremely confused with these two problems:

     7f + 3g = -6
    7f - 2g = -31

    What would I do first... would I subtract or add? I don't know when I should subtract or when I should add.
    The first step in know what to do is understanding why you would want to do it! You object is to eliminate one of the unknown letters. You should see that the coefficient of "f" in both equations is "7". Subtracting will give 7f- 7f= 0 which eliminates "f". Adding them would not.

    5m - p = 7
    7m - p = 11
    would I add for this one?
    For 5m- p= 7, 7m- p= 11, p has coefficient -11 in both equations. Subtracting will give -11p- (-11p)= 0.

    Notice that skoker multiplied an equation by -1 and then added. Of course, that is the same as subtracting the equations.
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    Quote Originally Posted by HallsofIvy View Post
    The first step in know what to do is understanding why you would want to do it! You object is to eliminate one of the unknown letters. You should see that the coefficient of "f" in both equations is "7". Subtracting will give 7f- 7f= 0 which eliminates "f". Adding them would not.


    For 5m- p= 7, 7m- p= 11, p has coefficient -11 in both equations. Subtracting will give -11p- (-11p)= 0.

    Notice that skoker multiplied an equation by -1 and then added. Of course, that is the same as subtracting the equations.
    Thank you, this is what I was looking for.

    Everybody else was helpful but you explained it the best way for me to understand.

    EDIT: How should I determine which variable to eliminate?
    Last edited by AbstractHero; February 8th 2011 at 01:26 PM.
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    Quote Originally Posted by AbstractHero View Post
    Thank you, this is what I was looking for.

    Everybody else was helpful but you explained it the best way for me to understand.

    EDIT: How should I determine which variable to eliminate?
    Whichever happens to be simplest to eliminate.
    For the first problem,
    5m- p= 7
    7m- p= 11

    You can see that each p has a "coefficient" of -1. If we simply subtract the second equation from the first equation, those will cancel: (5m- 7m)- (p- p)= 7- 11 or -2m= 4. Or subtract the first equation from the first, if you don't like negative numbers: (7m- 5m)- (p- p)= 11- 7 or 2m= 4. Once you have found that m= 2, put that back into either equation: 5(2)- p= 7 so that 10- p= 7 and p= 3.

    But I could have multiplied each part of the first equation by 7 to get 35m- 7p= 49 and multiplied each part of the second equation by 5 to get 35m- 5p= 55 so that now the each m has the same coefficient, 35. If I subtract the second equation from the first, I get (35m- 35m)- (7p- 5p)= 49- 55 or 2p= 6 so that p= 3. Put that back into either of the first equations: 5m- (3)= 7, 5m= 7+ 3= 10, m= 2.
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