Answer :
[tex]x+\frac{4}{5}=\frac{2}{3}-x-\frac{\not3^1}{\not6_2}\\\\x+\frac{4}{5}=\frac{2}{3}-x-\frac{1}{2}\\\\find\ LCD(\frac{4}{5};\ \frac{2}{3};\ \frac{1}{2})=2\times3\times5=30\\\\multiply\ both\ sides\ by\ LCD=30\\\\30x+30\cdot\frac{4}{5}=30\cdot\frac{2}{3}-30x-30\cdot\frac{1}{2}\\\\30x+6\cdot4=10\cdot2-30x-15\cdot1\\\\30x+24=20-30x-15[/tex]
[tex]30x+24=5-30x\ \ \ \ |subtract\ 24\ from\ both\ sides\\\\30x=-19-30x\ \ \ \ \ |add\ 30x\ to\ both\ sides\\\\60x=-19\ \ \ \ \ |divide\ both\ sides\ by\ 60\\\\\boxed{x=-\frac{19}{60}}[/tex]
[tex]\frac{x+4}{5}=\frac{2}{3}-\frac{x-3}{6}\\\\\frac{x+4}{5}=\frac{2\cdot2}{3\cdot2}-\frac{x-3}{6}\\\\\frac{x+4}{5}=\frac{4}{6}-\frac{x-3}{6}\\\\\frac{x+4}{5}=\frac{4-(x-3)}{6}\\\\\frac{x+4}{5}=\frac{4-x+3}{6}\\\\\frac{x+4}{5}=\frac{7-x}{6}\ \ \ \ |cross\ multiply[/tex]
[tex]6(x+4)=5(7-x)\\\\6(x)+6(4)=5(7)-5(x)\\\\6x+24=35-5x\ \ \ \ \ |subtract\ 24\ from\ both\ sides\\\\6x=11-5x\ \ \ \ \ |add\ 5x\ to\ both\ sides\\\\11x=11\ \ \ \ \ |divide\ both\ sides\ by\ 11\\\\\boxed{x=1}[/tex]
[tex]30x+24=5-30x\ \ \ \ |subtract\ 24\ from\ both\ sides\\\\30x=-19-30x\ \ \ \ \ |add\ 30x\ to\ both\ sides\\\\60x=-19\ \ \ \ \ |divide\ both\ sides\ by\ 60\\\\\boxed{x=-\frac{19}{60}}[/tex]
[tex]\frac{x+4}{5}=\frac{2}{3}-\frac{x-3}{6}\\\\\frac{x+4}{5}=\frac{2\cdot2}{3\cdot2}-\frac{x-3}{6}\\\\\frac{x+4}{5}=\frac{4}{6}-\frac{x-3}{6}\\\\\frac{x+4}{5}=\frac{4-(x-3)}{6}\\\\\frac{x+4}{5}=\frac{4-x+3}{6}\\\\\frac{x+4}{5}=\frac{7-x}{6}\ \ \ \ |cross\ multiply[/tex]
[tex]6(x+4)=5(7-x)\\\\6(x)+6(4)=5(7)-5(x)\\\\6x+24=35-5x\ \ \ \ \ |subtract\ 24\ from\ both\ sides\\\\6x=11-5x\ \ \ \ \ |add\ 5x\ to\ both\ sides\\\\11x=11\ \ \ \ \ |divide\ both\ sides\ by\ 11\\\\\boxed{x=1}[/tex]
