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Thread: a + b + c

  1. #1
    Super Member harpazo's Avatar
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    a + b + c

    If a+b = 11, b+c = 15, and a+c =10, what is the value of a+b+c?

    I know we have 3 equations in two unknowns.
    Can someone show how to find one of the variables?
    I can then proceed to find the other two.
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    Re: a + b + c

    Quote Originally Posted by harpazo View Post
    If a+b = 11, b+c = 15, and a+c =10, what is the value of a+b+c?
    TAKE NOTE: $(a+b)+(b+c)+(a+c)=2a+2b+2c=2(a+b+c)=36$ HOW?
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    Re: a + b + c

    Since I am not nearly as smart as Plato, I would have done this the hard way:

    Subtract b+ c= 15 from a+ b= 11 to get a- c= -4. Add that to a+ c= 10: 2a= 6 so a= 3. Then a+ c= 3+ c= 10 so c= 7. a+ b= 3+ b= 11 so b= 8. a+ b+ c= 3+ 8+ 7= 18.
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    Re: a + b + c

    Quote Originally Posted by harpazo View Post
    If a+b = 11, b+c = 15, and a+c =10, what is the value of a+b+c?
    Since I'm not nearly as smart as Halls, I'd do it this way (which is the best way!):

    a+b = 11 [1]
    b+c = 15 [2]
    a+c = 10 [3]
    Code:
     a + b =  11 [1]
    -a - c = -10 [3]*-1
    ============
     b - c =   1
     b + c =  15 [2]
    ============
    2b     =  16 
    b = 16/2 = 8
    Now go get a and c....

    3 cheers fer me!!
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    Forum Admin topsquark's Avatar
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    Re: a + b + c

    Quote Originally Posted by DenisB View Post
    3 cheers fer me!!
    With you it's more like "3 beers for me!!"

    -Dan
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    Re: a + b + c

    Quote Originally Posted by HallsofIvy View Post
    Subtract b+ c= 15 from a+ b= 11 to get a- c= -4. Add that to a+ c= 10: 2a= 6 so a= 3. Then a+ c= 3+ c= 10 so c= 7. a+ b= 3+ b= 11 so b= 8. a+ b+ c= 3+ 8+ 7= 18.
    @ harpazo, In the testing business this question is known as an "eater". There are two ways to do.
    One is to solve for $a,~b,~\&~c$ so to be able find the value of $a+b+c.$ That eats up time, so the name.
    The other as I used really requires only simple addition & division. From other posts I know you are preparing for tests.
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    Super Member harpazo's Avatar
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    Re: a + b + c

    Quote Originally Posted by Plato View Post
    TAKE NOTE: $(a+b)+(b+c)+(a+c)=2a+2b+2c=2(a+b+c)=36$ HOW?
    I added all 3 equations.

    a+b = 11
    b+c = 15
    a+c =10

    2a+2b+2c = 36

    I divided both sides by 2.

    a+b+c = 18
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  8. #8
    Super Member harpazo's Avatar
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    Re: a + b + c

    Quote Originally Posted by Plato View Post
    @ harpazo, In the testing business this question is known as an "eater". There are two ways to do.
    One is to solve for $a,~b,~\&~c$ so to be able find the value of $a+b+c.$ That eats up time, so the name.
    The other as I used really requires only simple addition & division. From other posts I know you are preparing for tests.
    Do not step beyond the question. I am not preparing for a test. I am 53 years old. Born in 1965. Do the math. Do not assume. I simply use test prep books to practice word problems. It's that simple, bro.
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    Re: a + b + c

    Plato, you need to kiss his feet...
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    Re: a + b + c

    Since I would do it Halls of Ivy's way, and Denis said that he is not as smart as Halls of Ivy, does it mean that I am smarter than Denis?
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    Re: a + b + c

    Quote Originally Posted by Monoxdifly View Post
    Since I would do it Halls of Ivy's way, and Denis said that he is not as smart as Halls of Ivy, does it mean that I am smarter than Denis?
    On set of all people the relation $\mathscr{S}$ defined as $\mathscr{S}(A,B)\iff A\text{ is smarter than }B$ is
    • irreflexive
    • asymmetric
    • transitive

    I can absolutely assure that $(plato,HallsofIvey)\notin\mathscr{S}$ that said the jury is out on if $(HallsofIvey,plato)\in\mathscr{S}~.$
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    Re: a + b + c

    Quote Originally Posted by Monoxdifly View Post
    Since I would do it Halls of Ivy's way, and Denis said that he is not as smart as Halls of Ivy, does it mean that I am smarter than Denis?
    Mr.Fly, I ain't smart enough to figure that out!
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    Re: a + b + c

    Quote Originally Posted by DenisB View Post
    Mr.Fly, I ain't smart enough to figure that out!
    Me, too. My brain short-circuited.
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    Re: a + b + c

    Quote Originally Posted by HallsofIvy View Post
    Since I am not nearly as smart as Plato, ...
    Why are people shortchanging themselves, even in jest!?
    In imparting lessons, having arrogance/being full of yourself is a debit in
    a ledger. You have a credit in the ledger if you are wise, and you
    do not discredit yourself (or others).
    Last edited by greg1313; Dec 12th 2018 at 05:57 AM.
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    Re: a + b + c

    a=11-b
    b+c=15
    c+a=10

    b+c=15
    11-b+c=10

    b+c=15
    -b+c= -1

    2c=14
    c=7
    b+7=15
    b=8
    a+8=11
    a=3

    Here are unknowns
    a=3
    b=8
    c=7
    And
    a+b+c=3+8+7=18
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