Solving:
\(\frac{3+4x}{2} + \frac{5-x}{3} = \frac{29}{6}\)
Make the Least Common Multiple (2,3,6)
\(2,3,6\:|2\)
\(1,3,3\:|3\)
\(1,1,1\:|\underline{2*3=6}\)
Replace denominators and resolve:
\(\frac{3+4x}{2} + \frac{5-x}{3} = \frac{29}{6}\)
\(\frac{3(3+4x)}{6} + \frac{2(5-x)}{6} = \frac{29}{6}\)
Cancel the dominators
\(\frac{3(3+4x)}{\diagup\!\!\!\!6} + \frac{2(5-x)}{\diagup\!\!\!\!6} = \frac{29}{\diagup\!\!\!\!6}\)
\(3(3+4x) + 2(5-x) = 29\)
\(9 + 12x + 10 - 2x = 29\)
\(12x - 2x = 29 - 9 - 10\)
\(10x = 20 - 10\)
\(10x = 10\)
\(x = \frac{10}{10}\)
\(\boxed{\boxed{x = 1}}\end{array}}\qquad\quad\checkmark\)
