Question

How do I solve for x?

Answer 1

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
$\frac{3(3+4x)}{\diagup\!\!\!\!6} + \frac{2(5-x)}{\diagup\!\!\!\!6} = \frac{29}{\diagup\!\!\!\!6}$
$3(3+4x) + 2(5-x) = 29$
$9 + 12x + 10 - 2x = 29$
$12x - 2x = 29 - 9 - 10$
$10x = 20 - 10$
$10x = 10$
$x = \frac{10}{10}$
$\boxed{\boxed{x = 1}}\end{array}}\qquad\quad\checkmark$