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GCSE simultaneous equation help please Watch

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    (Original post by Blaze3211)
    Why did you make the top equation negative?
    Huh? please enlighten me
    :confused:
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    (Original post by Green22)
    I thought too it was something to do with 130. Anyway 130/26=?
    Wait, so is your problem just with working out the value of 130/26? Do you understand how to get to y=130/26?
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    Hello


    I'm an A2 Mathematician and Year 11 Maths mentor and I've solved your Maths problem using a method I came up with in year 11.

    Follow the letters through a-e.

    Steps a and b are rearranging the two equations so that there is an equal element to both.

    Step c puts the two equal equations you found in steps a and b together. You then rearrage this to get y on its own.

    Step d puts the newfound y value into the equation 'x=' which you found in step a.

    Step e puts your already-found x and y values into the equation you found in part b. If your answer to this is right (in this case, you get -2=-2, which is true), you know you have the right answer.

    I know it's quite complex to understand at first, but I've found that it's a really reliable method which I still use for A-level Maths.

    I hope this helps

    P.S. Sorry the picture's sideways - I'll try and fix it
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    (Original post by Green22)
    Thanks for that but it is just that may working out is a bit different. Maybe your answer is right though.
    Oh I see what you're doing now, I've never seen anyone do a simultaneous equation that way

    But if you do it your way:

    12x+20y=76
    12x-6y=54
    20y+6y=76--54?
    26y=130 (because two negatives is a positive)
    y=5

    Then substitute it in the original equations...

    3x+5(5)=19
    3x+25=19
    (-25) 3x=-6
    x=-2

    3(-2)+5(5) -> -6+25 = 19
    4(-2)-2(5) -> -8-10 = -18





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    (Original post by brittanna)
    Wait, so is your problem just with working out the value of 130/26? Do you understand how to get to y=130/26?
    i believe so because 76--54=130 right? the question isn non calculator btw
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    (Original post by Kangie)
    Why? I even did the whole simultaneous equation and the answer is correct.
    I got x to be -2
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    (Original post by Green22)
    i believe so because 76--54=130 right? the question isn non calculator btw
    Yeah, 76--54=76+54=130. (You could always write it out with the columns lined up if you needed to).

    You could always do the following to help you work out the division:

    \frac{130}{26} = \frac{13 * 10}{13 * 2} = \frac{10}{2} = 5
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    Hope this is better
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    You guys sure do know how to to turn an easy question into a lot of working... x = -2, y = 5

    /thread
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    (Original post by Green22)
    Huh? please enlighten me
    :confused:
    Oh, i meant Kangie sorry
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    (Original post by Kangie)
    Oh I see what you're doing now, I've never seen anyone do a simultaneous equation that way
    There are many ways for this but my way is the way i was thought.

    But if you do it your way:

    12x+20y=76
    12x-6y=54
    20y+6y=76--54?
    26y=130 (because two negatives is a positive)
    y=5

    Then substitute it in the original equations...

    3x+5(5)=19
    3x+25=19
    (-25) 3x=-6
    x=-2

    3(-2)+5(5) -> -6+25 = 19
    4(-2)-2(5) -> -8-10 = -18





    I understand now. Thanks.
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    (Original post by danniegee)


    Hope this is better
    Thanks for this. I understand now.
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    (Original post by Green22)
    Thanks for this. I understand now.
    You're welcome! Good luck with your GCSE
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    (Original post by danniegee)


    Hello


    I'm an A2 Mathematician and Year 11 Maths mentor and I've solved your Maths problem using a method I came up with in year 11.

    Follow the letters through a-e.

    Steps a and b are rearranging the two equations so that there is an equal element to both.

    Step c puts the two equal equations you found in steps a and b together. You then rearrage this to get y on its own.

    Step d puts the newfound y value into the equation 'x=' which you found in step a.

    Step e puts your already-found x and y values into the equation you found in part b. If your answer to this is right (in this case, you get -2=-2, which is true), you know you have the right answer.

    I know it's quite complex to understand at first, but I've found that it's a really reliable method which I still use for A-level Maths.

    I hope this helps

    P.S. Sorry the picture's sideways - I'll try and fix it
    Name:  ImageUploadedByStudent Room1419887997.762882.jpg
Views: 52
Size:  123.3 KB


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    (Original post by Blaze3211)
    Oh, i meant Kangie sorry
    That were my thoughts exactly.
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    (Original post by danniegee)
    You're welcome! Good luck with your GCSE
    Thanks but just wondering if there is even a thing as "luck"? I am not sure tbh as in maths you either know it or you dont.
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    (Original post by brittanna)
    Yeah, 76--54=76+54=130. (You could always write it out with the columns lined up if you needed to).

    You could always do the following to help you work out the division:

    \frac{130}{26} = \frac{13 * 10}{13 * 2} = \frac{10}{2} = 5
    Cool.
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    I hope you understand this:

    Make x the same in both equations by finding a number that has 3 and 4 as a factor, in this case 12.
    Subtract both equations, so 12x - 12x becomes 0x, because of the two minus signs, 20y - -6y becomes 20y + 6y = 26y.
    Also, 76 - -54 becomes 76 + 54 = 130. Therefore, 26y = 130.
    130 / 26 is 5 = y.

    Substitute the value of y into the equation to find x.
    3x + 5y(25) = 19.
    19 - 25 = -6 = 3x
    Therefore, x = -6 / 3 = -2.

    Finally, x = -2 and y = 5

    Name:  ImageUploadedByStudent Room1419895725.730572.jpg
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Size:  176.8 KB

    Took me ages to type this, but hope it helps, typing on the iPad sucks :/


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    (Original post by Green22)
    Thanks but just wondering if there is even a thing as "luck"? I am not sure tbh as in maths you either know it or you dont.
    Well, basically. for maths anyways.
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    (Original post by Blaze3211)
    Well, basically. for maths anyways.
    Cool.

    Thanks everyone for your your help.
 
 
 
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