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Thread: Solving Systems

  1. #1
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    Solving Systems

    How do I solve this using any algebraic method?

    3x-2y=10
    5x+3y=15

    2x-4y=-6
    -x+2y=3
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  2. #2
    Member integral's Avatar
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    solve for x on one equation and substitute it in the other
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    Quote Originally Posted by Vuong View Post
    How do I solve this using any algebraic method?

    3(3x-2y=10)
    2(5x+3y=15)

    2x-4y=-6
    2(-x+2y=3)
    solve using elimination ... distribute, add the equations and solve for the remaining variable. back substitute to find the other variable.
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  4. #4
    Super Member Bacterius's Avatar
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    $\displaystyle 3x-2y=10$
    $\displaystyle 5x+3y=15$

    From the second equation you know that :

    $\displaystyle 5x = 15 - 3y$

    Thus $\displaystyle 15x = 45 - 9y$

    And so, $\displaystyle 3x = 9 - 1.8y$.

    Substituting back this result into the first equation gives :

    $\displaystyle 9 - 1.8y - 2y = 10$

    $\displaystyle 9 - 3.8y = 10$

    $\displaystyle 3.8y = -1$

    $\displaystyle y = \frac{-1}{3.8}$

    Now substitute the value of $\displaystyle y$ found into any of the equations to retrieve $\displaystyle x$.

    Can you do it for the second system now ?
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  5. #5
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    Quote Originally Posted by Vuong View Post

    3x-2y=10
    5x+3y=15

    You can also use the elimination method

    make

    $\displaystyle 3x-2y=10$ ...(1)
    $\displaystyle 5x+3y=15$ ...(2)

    To eliminate x take (1) $\displaystyle \times 5 $ and (2) $\displaystyle \times 3
    $

    gives

    $\displaystyle 15x-10y=50$ ...(3)
    $\displaystyle 15x+9y=45$ ...(4)

    Now taking (4) form (3) gives

    $\displaystyle -19y = 5$

    Can you take it from here?
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  6. #6
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    Quote Originally Posted by Bacterius View Post
    $\displaystyle 3x-2y=10$
    $\displaystyle 5x+3y=15$

    From the second equation you know that :

    $\displaystyle 5x = 15 - 3y$

    Thus $\displaystyle 15x = 45 - 9y$

    And so, $\displaystyle 3x = 9 - 1.8y$.

    Substituting back this result into the first equation gives :

    $\displaystyle 9 - 1.8y - 2y = 10$

    $\displaystyle 9 - 3.8y = 10$

    $\displaystyle 3.8y = -1$

    $\displaystyle y = \frac{-1}{3.8}$

    Now substitute the value of $\displaystyle y$ found into any of the equations to retrieve $\displaystyle x$.

    Can you do it for the second system now ?
    How did you get the 9 for $\displaystyle 9 - 1.8y - 2y = 10$?
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  7. #7
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    Quote Originally Posted by pickslides View Post
    You can also use the elimination method

    make

    $\displaystyle 3x-2y=10$ ...(1)
    $\displaystyle 5x+3y=15$ ...(2)

    To eliminate x take (1) $\displaystyle \times 5 $ and (2) $\displaystyle \times 3
    $

    gives

    $\displaystyle 15x-10y=50$ ...(3)
    $\displaystyle 15x+9y=45$ ...(4)

    Now taking (4) form (3) gives

    $\displaystyle -19y = 5$

    Can you take it from here?
    y = -3.8 then plug it in?
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  8. #8
    Super Member Bacterius's Avatar
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    Quote Originally Posted by Vuong View Post
    How did you get the 9 for $\displaystyle 9 - 1.8y - 2y = 10$?
    Comes from the substitution of $\displaystyle 3x = 9 - 1.8y$.
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  9. #9
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    Quote Originally Posted by Vuong View Post
    y = -3.8 then plug it in?

    To either (1),(2),(3) or (4) to find x.
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    Quote Originally Posted by pickslides View Post

    To either (1),(2),(3) or (4) to find x.
    Thanks, I got 5.86
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  11. #11
    Super Member Bacterius's Avatar
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    That's not right, you should find something around $\displaystyle x \approx 3.15$
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  12. #12
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    Quote Originally Posted by Bacterius View Post
    That's not right, you should find something around $\displaystyle x \approx 3.15$
    I plugged -3.8 into 3x-2y=10 x_x
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  13. #13
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    Quote Originally Posted by Vuong View Post
    y = -3.8 then plug it in?
    Are you sure this is correct?

    $\displaystyle -19y = 5 \implies y = -\frac{5}{19} \approx -0.263$
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  14. #14
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    Quote Originally Posted by pickslides View Post
    Are you sure this is correct?

    $\displaystyle -19y = 5 \implies y = -\frac{5}{19} \approx -0.263$
    So I plug -0.263 into one of them?
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  15. #15
    Super Member Bacterius's Avatar
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    Quote Originally Posted by Vuong View Post
    So I plug -0.263 into one of them?
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