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Math Help - solve by elimination

  1. #1
    Newbie annabananzaa's Avatar
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    solve by elimination

    5x-y=9
    x-1.5y=-3

    i dont even kno what it means by elimination. i know how to solve by substitution, but not elimination....
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  2. #2
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    Substitution

    Hello annabananzaa
    Quote Originally Posted by annabananzaa View Post
    5x-y=9
    x-1.5y=-3

    i dont even kno what it means by elimination. i know how to solve by substitution, but not elimination....
    It's the same thing! When you substitute for x, you eliminate (= get rid of) it!

    Grandad

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  3. #3
    Newbie annabananzaa's Avatar
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    so i take the first problem, turn it into y=5x-9 and then just plug that into the first equation so it looks like x-1.5(5x-9)=9

    would that be right?
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  4. #4
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    Substitution

    Hello annabananzaa
    Quote Originally Posted by annabananzaa View Post
    so i take the first problem, turn it into y=5x-9 and then just plug that into the first equation so it looks like x-1.5(5x-9)=9

    would that be right?
    Yes. Just be careful with the minus signs when you multiply out the brackets, won't you?

    Grandad

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  5. #5
    Newbie annabananzaa's Avatar
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    of course i will!

    so this is what im coming up with step by step, and it doesnt seem right still....

    x-1.5(5x-9)=9
    x-7.5x+13.5=9
    -6.5x+13.5=9
    -6.5x=-4.5

    -4.5/-6.5

    x=.69?
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  6. #6
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    Quote Originally Posted by Grandad View Post
    Hello annabananzaaIt's the same thing! When you substitute for x, you eliminate (= get rid of) it!

    Grandad
    Crazily enough, it's not ..... Elimination and substitution are taught as two seperate methods.

    One way of applying the elimination method:

    5x - y = 9 .... (1)

    x - 1.5y = -3 .... (2)

    Multiply equation (2) by -5:

    -5x + 7.5y = 15 .... (3)

    Now add equation (1) and equation (3) to eliminate x:

    6.5 y = 24 \Rightarrow y = \, ....

    Now substitute the value of y into any equation and solve for x.

    This is what is taught in schools (so don't shoot me ....)
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  7. #7
    Newbie annabananzaa's Avatar
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    so i was right when saying my answer looked a little crazy?

    im sorry, im still confused... i dont see where you got the 3 from in the begining and then i dont get why i would add the equations....
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  8. #8
    Flow Master
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    Quote Originally Posted by annabananzaa View Post
    so i was right when saying my answer looked a little crazy?

    im sorry, im still confused... i dont see where you got the 3 from in the begining and then i dont get why i would add the equations....
    What 3? I have labelled the equations (1), (2) and (3).

    You add so that x gets eliminated.

    I have (reasonably) assumed that you have some familiarity with the method ....

    Read this: Systems of Linear Equations: Solving by Addition / Elimination
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  9. #9
    Newbie annabananzaa's Avatar
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    so would my first answer that i posted be correct or this new answer that i got after following the steps on the link??
    x=3.69

    iim tryingg really hard!
    just not very good at math
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  10. #10
    Flow Master
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    Quote Originally Posted by annabananzaa View Post
    so would my first answer that i posted be correct or this new answer that i got after following the steps on the link??
    x=3.69

    iim tryingg really hard!
    just not very good at math
    If you do it how I showed you get y = 24/6.5 = 48/13 which is approximately equal to 3.69.

    Now substitute and solve for x. You should get x \approx 2.538 (Note:an exact answer might be required).
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  11. #11
    Newbie annabananzaa's Avatar
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    ok. now i get it, i was forgetting the last step. thats what was really confusing me.
    thank you so much for your patience!
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