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

1. Originally Posted by Narcosys
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. Originally Posted by Narcosys
2a = ab + c

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

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

2= 2a/a

2=2
**** WHERE does that "2" come from?!!

3. Originally Posted by Narcosys
Usually pretty good at math, but I keep getting stumped and getting a variable equals a variable.

Problem:

2=(ab+c)/a

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 $r = 6q - \dfrac{3}{2p}.$

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

$r = 6q - \dfrac{3}{2p} \implies$

$2p * r = 2p * \left ( 6q - \dfrac{3}{2p} \right ) \implies$

$2pr = 12pq - 3 \implies$

$12pq - 3 = 2pr \implies$

$12pq - 3 + 3 = 2pr + 3 \implies$

$12pq = 3 + 2pr \implies$

$12pq - 2pr = 3 + 2pr - 2pr \implies$

$12pq - 2pr = 3$

$p(12q - 2r) = 3 \implies$

$\dfrac{p(12q - 2r)}{12q - 2r} = \dfrac{3}{12q - 2r } \implies$

$p = \dfrac{3}{12q - 2r} = \dfrac{3}{2(6q-r)}.$

$6q - \dfrac{3}{2p} = 6q - \dfrac{3}{2 * \dfrac{3}{2(6q - r)}} =$

$6q - \dfrac{\dfrac{3}{1}}{\dfrac{3}{6q - r}} = 6q - \dfrac{3}{1} * \dfrac{6q - r}{3} =$

$6q - (6q - r) = 6q - 6q + r = r.$

It checks.

Now you have a model.

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