Answer
$\frac{2}{7}x^3-\frac{1}{4}x+2-\frac{1}{2}x^3+\frac{3}{8}x=\frac{3}{14}x^3+\frac{1}{8}x+2$
Work Step by Step
$\frac{2}{7}x^3-\frac{1}{4}x+2-\frac{1}{2}x^3+\frac{3}{8}x=(\frac{2}{7}-\frac{1}{2})x^3+(-\frac{1}{4}+\frac{3}{8})x+2=(\frac{4}{14}-\frac{7}{14})x^3+(-\frac{2}{8}+\frac{3}{8})x+2=\frac{-3}{14}x^3+\frac{1}{8}x+2=-\frac{3}{14}x^3+\frac{1}{8}x+2$
