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Thread: Helping Daughter w/ math, this one is stumping me: 2=(ab+c)/a Solve for a.

  1. #11
    Senior Member
    Join Date
    Nov 2017
    Rochester, NY
    Quote Originally Posted by Narcosys View Post
    Solving for b leaves only a and c variables on one side. You then throw those in for b, leaving only a and c variables in the equation and you can solve for c, leaving only a variables. Tjen you attempt to solve for a.
    It sounds as if you are confusing this with a system of equations, where you can use one equation to eliminate a variable in another. Here you have only one equation, and the answer will be an expression containing b and c.

    Have you tried doing what was suggested? You have 2a = ab + c, which is better than the original because a is not in the denominator. Now you have to gather terms with a on one side, and then, as I hinted, factor a out of two terms, so that there is only one a in the equation.

  2. #12
    Full Member
    Join Date
    Feb 2004
    Ottawa, Ontario
    Quote Originally Posted by Narcosys View Post
    2a = ab + c

    c = 2a - ab + 2 ****

    2= (ab + 2a - ab)/a

    2= 2a/a

    **** WHERE does that "2" come from?!!
    I'm a man of few words...but I use 'em often!!

  3. #13
    Elite Member
    Join Date
    Sep 2012
    Quote Originally Posted by Narcosys View Post
    Usually pretty good at math, but I keep getting stumped and getting a variable equals a variable.



    Solve for a.

    Been trying to solve for b and c cause I can isolate those, can't seem to isolate a.

    Thanks for any help.
    "Solve for a" does not necessarily mean finding a numeric value for a. It means finding an expression that equals a. The simplest type of expression is indeed a numeral, but it may not be possible to get the simplest type.

    A general rule is that you need at least n equations to get numeric answers if you have n variables. You have 3 variables and only one equation. So the best that you can do is to find the simplest algebraic expression that equals the indicated variable.

    Solve for p given [tex]r = 6q - \dfrac{3}{2p}.[/tex]

    Three variables, one equation means an algebraic solution, not a numeric one.

    [tex]r = 6q - \dfrac{3}{2p} \implies [/tex]

    [tex]2p * r = 2p * \left ( 6q - \dfrac{3}{2p} \right ) \implies [/tex]

    [tex]2pr = 12pq - 3 \implies [/tex]

    [tex]12pq - 3 = 2pr \implies [/tex]

    [tex]12pq - 3 + 3 = 2pr + 3 \implies [/tex]

    [tex]12pq = 3 + 2pr \implies [/tex]

    [tex]12pq - 2pr = 3 + 2pr - 2pr \implies[/tex]

    [tex]12pq - 2pr = 3[/tex]

    [tex]p(12q - 2r) = 3 \implies [/tex]

    [tex]\dfrac{p(12q - 2r)}{12q - 2r} = \dfrac{3}{12q - 2r } \implies[/tex]

    [tex]p = \dfrac{3}{12q - 2r} = \dfrac{3}{2(6q-r)}.[/tex]

    Now check your work.

    [tex]6q - \dfrac{3}{2p} = 6q - \dfrac{3}{2 * \dfrac{3}{2(6q - r)}} = [/tex]

    [tex]6q - \dfrac{\dfrac{3}{1}}{\dfrac{3}{6q - r}} = 6q - \dfrac{3}{1} * \dfrac{6q - r}{3} = [/tex]

    [tex]6q - (6q - r) = 6q - 6q + r = r.[/tex]

    It checks.

    Now you have a model.


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