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GCSE simultaneous equation help please Watch

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    3x+5y=19
    4x-2y=-18

    so basiccally i am stuck and got this far:

    12x+20y=76
    12x-6y=54
    20y+6y=76--54?
    26y=?

    Any help could be good.
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    it is going well...

    if 26y = 130

    then y = ...
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    (Original post by Green22)
    3x+5y=19
    4x-2y=-18

    so basiccally i am stuck and got this far:

    12x+20y=76
    12x-6y=54
    20y+6y=76--54?
    26y=?

    Any help could be good.
    3x+5y=19
    4x-2y=-18

    (x4) 12x+20y=76
    (x3) 12x-6y=-54

    -12x-20y=-76
    12x-6y=-54

    -26y=-130
    -y=-5
    y=5

    Then substitute y into the equation or do the same for x

    (Hope I did this right)
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    (Original post by the bear)
    it is going well...

    if 26y = 130

    then y = ...
    I thought too it was something to do with 130. Anyway 130/26=?
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    (Original post by Kangie)
    3x+5y=19
    4x-2y=-18

    (x4) 12x+20y=76
    (x3) 12x-6y=-54

    -12x-20y=-76
    12x-6y=-54

    -26y=-130
    -y=-5
    y=5

    Then substitute y into the equation or do the same for x

    (Hope I did this right)
    May i ask where did you get the -26y from:confused:
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    (Original post by Green22)
    I thought too it was something to do with 130. Anyway 130/26=?
    Y = 130/26 yah
    What you aiming to get in your maths gcse? xd
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    (Original post by Green22)
    May i ask where did you get the -26y from:confused:
    When you change the top equation so that it's negative the y changes to -20, then when you minus -6 from -20 it goes to -26
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    (Original post by Kangie)
    When you change the top equation so that it's negative the y changes to -20, then when you minus -6 from -20 it goes to -26
    ohh ok.
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    (Original post by Placeboo123)
    Y = 130/26 yah
    What you aiming to get in your maths gcse? xd
    God knows tbh. I have tried a question before posting this one and i got the answers rights but this one is just messed up.
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    (Original post by Kangie)
    3x+5y=19
    4x-2y=-18

    (x4) 12x+20y=76
    (x3) 12x-6y=-54

    -12x-20y=-76
    12x-6y=-54

    -26y=-130
    -y=-5
    y=5

    Then substitute y into the equation or do the same for x

    (Hope I did this right)
    I am doubting this for some reason.
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    (Original post by the bear)
    it is going well...

    if 26y = 130

    then y = ...
    I am starting to think one of the answers will not be a whole number? I will keep trying.
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    (Original post by Green22)
    I am doubting this for some reason.
    Why? I even did the whole simultaneous equation and the answer is correct.
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    (Original post by Kangie)
    3x+5y=19
    4x-2y=-18

    (x4) 12x+20y=76
    (x3) 12x-6y=-54

    -12x-20y=-76
    12x-6y=-54

    -26y=-130
    -y=-5
    y=5

    Then substitute y into the equation or do the same for x

    (Hope I did this right)
    I thought it would be 20y+6y=26y? thats what i got.?
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    (Original post by Green22)
    God knows tbh. I have tried a question before posting this one and i got the answers rights but this one is just messed up.
    Haha private message me or quote if you need any help, should be able to give you some tips at least
    year 10 or 11?
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    (Original post by Green22)
    I thought it would be 20y+6y=26y? thats what i got.?
    No the bottom equation has a negative y number unless you wrote the question wrong in your original post
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    (Original post by Kangie)
    Why? I even did the whole simultaneous equation and the answer is correct.
    Thanks for that but it is just that may working out is a bit different. Maybe your answer is right though.
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    (Original post by Placeboo123)
    Haha private message me or quote if you need any help, should be able to give you some tips at least
    year 10 or 11?
    Cool. i have done tons of these over the past and have got them right but this question just messed me up. I am 16 btw.
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    (Original post by the bear)
    it is going well...

    if 26y = 130

    then y = ...
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    (Original post by Green22)
    ohh ok.
    Why did you make the top equation negative?
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    (Original post by Kangie)
    No the bottom equation has a negative y number unless you wrote the question wrong in your original post
    I am sure the question i wrote in my op is right.
    Here is the link:
    http://keshgcsemaths.files.wordpress...-equations.pdf
    It is the second sim equation question on that paper.

    Thanks for trying to help me.
 
 
 
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