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    simplify the following expression

    (a) 2(3r + 1) + 2r
    - for this i multiplied out of the brackets to give 6r + 1 + 2r, then collect like terms to give 8r + 1 - that is my final answer. my book says the correct answer is 8r + 2. can someone explain to me what i have done wrong here.

    can someone check my answer for another question please as it is not in my book! :mad:

    the question is:

    solve these simultaneous equations by the method of elimination

    4d – 3c = 10
    2d + 3c = 23

    i know that the two lhs added will equal the two rhs added, so i have done this:

    4d - 3c + 2d + 3c = 10 + 23

    the c is cancelled out because of the status sign and this leaves

    4d + 2d = 33
    (6d = 33)

    6d/6 = 33/6

    .:. d = 5½

    then i use this and plug it into the second simultaneous equation

    11 + 3c = 23

    11 + 3c = 23
    -11 -11

    3c = 12

    3c/3 = 12/3

    .:. c = 4

    hence the solution is d = 5½, c = 4. can someone tell me if this is correct? i think it might be, because it works when i plug them into the other equation. my book which i am learning from is crap and does not tell me some answers. i am starting gcses next year and i wish to have a head start!
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    a)

    2(3r + 1) + 2r
    6r + 2 + 2r
    8r + 2

    you didn't multiply the 1 in brackets by 2.

    b)

    4d – 3c = 10
    2d + 3c = 23

    so:

    3c = 4d - 10
    3c = 23 - 2d

    4d - 10 = 23 - 2d
    6d = 33
    d = 5.5

    plugging in...

    2(5.5) + 3c = 23
    3c = 23 - 11
    3c = 12
    c = 4

    so yeah, you're right. well done.
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    (Original post by einsci)
    (a) 2(3r + 1) + 2r
    - for this i multiplied out of the brackets to give 6r + 1 + 2r, then collect like terms to give 8r + 1 - that is my final answer. my book says the correct answer is 8r + 2. can someone explain to me what i have done wrong here.
    2(3r + 1) + 2r = 6r + 2 + 2r = 8r + 2
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    (Original post by Solomon)
    2(3r + 1) + 2r = 6r + 2 + 2r = 8r + 2
    ahh yews, careless mistake, thankyou.
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    (Original post by chewwy)
    a)

    2(3r + 1) + 2r
    6r + 2 + 2r
    8r + 2

    you didn't multiply the 1 in brackets by 2.

    b)

    4d – 3c = 10
    2d + 3c = 23

    so:

    3c = 4d - 10
    3c = 23 - 2d

    4d - 10 = 23 - 2d
    6d = 33
    d = 5.5

    plugging in...

    2(5.5) + 3c = 23
    3c = 23 - 11
    3c = 12
    c = 4

    so yeah, you're right. well done.
    thanks mates, i like the helpfullness of some members here
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    (Original post by einsci)
    simplify the following expression

    (a) 2(3r + 1) + 2r
    - for this i multiplied out of the brackets to give 6r + 1 + 2r, then collect like terms to give 8r + 1 - that is my final answer. my book says the correct answer is 8r + 2. can someone explain to me what i have done wrong here.

    can someone check my answer for another question please as it is not in my book! :mad:

    the question is:

    solve these simultaneous equations by the method of elimination

    4d – 3c = 10
    2d + 3c = 23

    i know that the two lhs added will equal the two rhs added, so i have done this:

    4d - 3c + 2d + 3c = 10 + 23

    the c is cancelled out because of the status sign and this leaves

    4d + 2d = 33
    (6d = 33)

    6d/6 = 33/6

    .:. d = 5½

    then i use this and plug it into the second simultaneous equation

    11 + 3c = 23

    11 + 3c = 23
    -11 -11

    3c = 12

    3c/3 = 12/3

    .:. c = 4

    hence the solution is d = 5½, c = 4. can someone tell me if this is correct? i think it might be, because it works when i plug them into the other equation. my book which i am learning from is crap and does not tell me some answers. i am starting gcses next year and i wish to have a head start!
    need help with gcse maths? add [email protected] to your msn, hopefully i can provide yet another year of help
 
 
 
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