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  1. #1
    Junior Member Rheanna's Avatar
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    Help on this question

    This isn't in our books. How would you solve this system of equation when it's set up like this?

    7x+7y+z=1
    x+8y+8z=8
    9x+y+9z=9


    ??
    17x+16y+18z=18?? or this a graphing linear equation? Thank you.
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Rheanna View Post
    This isn't in our books. How would you solve this system of equation when it's set up like this?

    7x+7y+z=1
    x+8y+8z=8
    9x+y+9z=9


    ??
    17x+16y+18z=18?? or this a graphing linear equation? Thank you.
    You can rewrite it using matrices:

    \begin{bmatrix}7 & 7 & 1 & 1\\ 1 & 8 & 8 & 8\\ 9 & 1 & 9 & 9\end{bmatrix}

    Then reduce it to the form \begin{bmatrix}1 & 0 & 0 & x_0\\ 0 & 1 & 0 & y_0\\ 0 & 0 & 1 & z_0\end{bmatrix}

    Where \left(x_0,y_0,z_0\right) is the solution to the system.

    Does this help?
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  3. #3
    Junior Member Rheanna's Avatar
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    I don't get it. I got the matrix part, but what you did after that you lost me.
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  4. #4
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    If you don't want to do the matrix method (which is fairly easy), you can do it just like any other system of three equations:

    (A) 7x+7y+z=1
    (B) x+8y+8z=8
    (C) 9x+y+9z=9

    Let's solve for either y or z first, so we need to eliminate x. Multiple equation (B) by 7 and then do A-B. Then do a similar subtraction process with either (A) and (C) or (B) and (C).

    After that, you'll have a system of equations with just y,z. So you can treat it like a two variable system of equations. Solve for y and z, then plug back in to get x.
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  5. #5
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Rheanna View Post
    I don't get it.
    The other way then would be to use substitutions.

    Note that in the first equation 7x+7y+z=1\implies z=1-7x-7y

    If we now substitute this into the second equation, we have x+8y+8(1-7x-7y)=8\implies x+8y+8-56x-56y=8\implies -55x-48y=0 \implies y=-\frac{55x}{48}.

    Now if we substitute this into the first equation, we get z=1-7x-7\left(-\frac{55x}{48}\right)\implies z=1-7x+\frac{7\cdot 55x}{48}\implies z=1-\frac{49x}{48}

    Finally, we substitute both values into the final equation to get

    9x+\left(-\frac{55x}{48}\right)+9\left(1-\frac{49x}{48}\right)=9

    Simplifying, we get

    9x-\frac{55x}{48}+9-\frac{9\cdot49x}{48}=9\implies \frac{9(48-49)x}{48}-\frac{55x}{48}=0\implies -\frac{64x}{48}=0\implies x=0

    Since x=0, then y=-\frac{55(0)}{48}=0 and z=1-\frac{49(0)}{48}=1.

    Thus, the solution to this system of equations is \left(0,0,1\right)

    Does this make sense? Its a lot of messy work, but it shouldn't be too bad to follow.
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  6. #6
    Junior Member Rheanna's Avatar
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    okie doke, thank you. I'll keep re writing this over and over until I understand it but I think I pretty much get what your saying.
    On the matrix I was trying to follow you but lost me when you did this
    Then reduce it to the form \begin{bmatrix}1 & 0 & 0 & x_0\\ 0 & 1 & 0 & y_0\\ 0 & 0 & 1 & z_0\end{bmatrix}
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  7. #7
    Junior Member Rheanna's Avatar
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    Quote Originally Posted by Chris L T521 View Post
    You can rewrite it using matrices:



    Then reduce it to the form \begin{bmatrix}1 & 0 & 0 & x_0\\ 0 & 1 & 0 & y_0\\ 0 & 0 & 1 & z_0\end{bmatrix}

    Where \left(x_0,y_0,z_0\right) is the solution to the system.

    Does this help?
    This is where you lost me. How you managed to get the 0, and the 1 , Did you subtract it? Duh, nevermind.. lol reduce=subtract haha lol
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  8. #8
    Junior Member Rheanna's Avatar
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    The Matrice?

    Does it matter what order it comes in?

    Like the Answer is 0,0,1

    but if you put 1,0,0 would that be wrong?
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