Question 127409
{{{((q+9))/2}}} + {{{((q-3))/3}}} = 6 
We can get rid of the denominators by multiplying the equation by 6:
6*{{{((q+9))/2}}} + 6*{{{((q-3))/3}}} = 6(6)
Cancel out the denominators and you have:
3(q+9) + 2(q-3) = 36
:
3q + 27 + 2q - 6 = 36
:
5q + 21 = 36
:
5q = 36 - 21
:
5q = 15
q = {{{15/5}}}
q = 3
:
:
Check in original equation, substitute 3 for q:
{{{((3+9))/2}}} + {{{((3-3))/3}}} = 6
{{{12/2}}} + 0 = 6; confirms our solution