the equation is:
[tex]\begin{gathered} \frac{7}{6}(v+2)+\frac{8}{9}(v+3)=\frac{11}{12}(v+1)+5 \\ \frac{7}{6}v+\frac{2\cdot7}{6}+\frac{8}{9}v+\frac{8\cdot3}{9}=\frac{11}{12}v+\frac{11}{12}+5 \\ \frac{7}{6}v+\frac{14}{6}+\frac{8}{9}v+\frac{24}{9}=\frac{11}{12}v+\frac{11}{12}+5 \end{gathered}[/tex]
until now I only multiply the factor for the values inside the parenthesis, now I will move the terms with "v" on the left side of the equal and the independent terms to the right side:
[tex]\begin{gathered} \frac{7}{6}v+\frac{8}{9}v-\frac{11}{12}v=\frac{11}{12}+5-\frac{14}{6}-\frac{24}{9} \\ (\frac{7}{6}+\frac{8}{9}-\frac{11}{12})v=\frac{11}{12}+5-\frac{14}{6}-\frac{24}{9} \end{gathered}[/tex]
and now we just sum the fractions
[tex]\begin{gathered} \frac{41}{36}v=\frac{11}{12} \\ v=\frac{11}{12}\cdot\frac{36}{41} \\ v=\frac{33}{41} \end{gathered}[/tex]
so the answer is v=33/41
