OpenStudy (anonymous):

Solve for a: (a+3b)/(a-3b) = 10k+3

6 years ago
OpenStudy (anonymous):

\[\frac{a+3b}{a-3b}=10k+3\] \[a+3b=(10k+3)(a-3b)\] \[a-(10k+3)a=-3b(10k+3)-3b\] \[(1-(10k+3)a=-30bk-9b-3b\] \[(-9k-3)a=-30bk-12b\] \[a=\frac{-30bk-12b}{-9k-3}=\frac{30bk+12b}{9k+3}\]

6 years ago
OpenStudy (anonymous):

Thank you. You've become my new best friend haha

6 years ago
OpenStudy (anonymous):

can simplify a little by canceling the 3 top and bottom i guess you get \[a=\frac{10bk+4b}{3k+1}\]

6 years ago
OpenStudy (anonymous):

yw (check my algebra it is late)

6 years ago
OpenStudy (anonymous):

Thanks for the warning. Final answer was actually [6(5bk+2b)]/[2(5k+1)]

6 years ago