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Thread: Help with fairly simple, quick problem.

  1. #1
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    Help with fairly simple, quick problem.

    Can someone solve this and show the work for me, please?

    3(x-y)=2(y-x) + 8
    -(2x - y) = 3(y + 3x - 1)



    Thanks very much. i\I just need to know how to work it.
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  2. #2
    MHF Contributor Quick's Avatar
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    Quote Originally Posted by alkali112 View Post
    3(x-y)=2(y-x) + 8
    -(2x - y) = 3(y + 3x - 1)
    hmm...

    first let's distribute out the equations to get:

    $\displaystyle 3(x-y)=2(y-x) + 8\quad\rightarrow\quad 3x-3y=2y-2x + 8$
    $\displaystyle -(2x - y) = 3(y + 3x - 1)\quad\rightarrow\quad -2x+y=3y+9x-3$

    Alright, now let's solve for x (or y, it's your choice) for the first problem:

    $\displaystyle 3x-3y=2y-2x + 8\quad\rightarrow\quad 5x=5y+8\quad\rightarrow\quad x=\frac{5y+8}{5}$

    now take the second equation:

    $\displaystyle -2x+y=3y+9x-3$

    and also solve for x:

    $\displaystyle -2x+y=3y+9x-3\quad\rightarrow\quad -11x=2y-3\quad\rightarrow\quad x=\frac{2y-3}{-11}$

    and now we can start solving stuff. First things first, you have to write down that:

    $\displaystyle x=x$

    So now we substitute:

    $\displaystyle \overbrace{\frac{5y+8}{5}}^{\text{from the first equation}}=\underbrace{\frac{2y-3}{-11}}_{\text{from the second equation}}$

    so now multiply both sides by -11 and 5 to get:

    $\displaystyle -11(5y+8)=5(2y-3)$

    now use the distributive property:

    $\displaystyle -55y-88=10y-15$

    now move things around:

    $\displaystyle -65y=73$

    then divide:

    $\displaystyle y=\frac{-73}{65}$

    Now you can solve for x.


    On a side note, $\displaystyle \frac{-73}{65}$ is not a nice number, you should double check my work and see if my arithmetic is right.
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    a "quick" problem indeed
    Quote Originally Posted by Quick View Post
    hmm...

    first let's distribute out the equations to get:

    $\displaystyle 3(x-y)=2(y-x) + 8\quad\rightarrow\quad 3x-3y=2y-2x + 8$
    $\displaystyle -(2x - y) = 3(y + 3x - 1)\quad\rightarrow\quad -2x+y=3y+9x-3$

    Alright, now let's solve for x (or y, it's your choice) for the first problem:

    $\displaystyle 3x-3y=2y-2x + 8\quad\rightarrow\quad 5x=5y+8\quad\rightarrow\quad x=\frac{5y+8}{5}$

    now take the second equation:

    $\displaystyle -2x+y=3y+9x-3$

    and also solve for x:

    $\displaystyle -2x+y=3y+9x-3\quad\rightarrow\quad -11x=2y-3\quad\rightarrow\quad x=\frac{2y-3}{-11}$

    and now we can start solving stuff. First things first, you have to write down that:

    $\displaystyle x=x$

    So now we substitute:

    $\displaystyle \overbrace{\frac{5y+8}{5}}^{\text{from the first equation}}=\underbrace{\frac{2y-3}{-11}}_{\text{from the second equation}}$

    so now multiply both sides by -11 and 5 to get:

    $\displaystyle -11(5y+8)=5(2y-3)$

    now use the distributive property:

    $\displaystyle -55y-88=10y-15$

    now move things around:

    $\displaystyle -65y=73$

    then divide:

    $\displaystyle y=\frac{-73}{65}$

    Now you can solve for x.


    On a side note, $\displaystyle \frac{-73}{65}$ is not a nice number, you should double check my work and see if my arithmetic is right.
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