Chapter 9 - Cumulative Review - Page 600: 2

$x=11$

Using properties of equality, then, \begin{array}{l} \dfrac{1}{3}(x-2)=\dfrac{1}{4}(x+1) \\\\ 12\left(\dfrac{1}{3}(x-2) \right)=\left( \dfrac{1}{4}(x+1) \right)12 \\\\ 4(x-2)=3(x+1) \\\\ 4x-8=3x+3 \\\\ 4x-3x=3+8 \\\\ x=11 .\end{array}