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 \(\displaystyle r = 6q - \dfrac{3}{2p}.\)
Three variables, one equation means an algebraic solution, not a numeric one.
\(\displaystyle r = 6q - \dfrac{3}{2p} \implies \)
\(\displaystyle 2p * r = 2p * \left ( 6q - \dfrac{3}{2p} \right ) \implies \)
\(\displaystyle 2pr = 12pq - 3 \implies \)
\(\displaystyle 12pq - 3 = 2pr \implies \)
\(\displaystyle 12pq - 3 + 3 = 2pr + 3 \implies \)
\(\displaystyle 12pq = 3 + 2pr \implies \)
\(\displaystyle 12pq - 2pr = 3 + 2pr - 2pr \implies\)
\(\displaystyle 12pq - 2pr = 3\)
\(\displaystyle p(12q - 2r) = 3 \implies \)
\(\displaystyle \dfrac{p(12q - 2r)}{12q - 2r} = \dfrac{3}{12q - 2r } \implies\)
\(\displaystyle p = \dfrac{3}{12q - 2r} = \dfrac{3}{2(6q-r)}.\)
Now check your work.
\(\displaystyle 6q - \dfrac{3}{2p} = 6q - \dfrac{3}{2 * \dfrac{3}{2(6q - r)}} = \)
\(\displaystyle 6q - \dfrac{\dfrac{3}{1}}{\dfrac{3}{6q - r}} = 6q - \dfrac{3}{1} * \dfrac{6q - r}{3} = \)
\(\displaystyle 6q - (6q - r) = 6q - 6q + r = r.\)
It checks.
Now you have a model.