## Precalculus: Mathematics for Calculus, 7th Edition

(a) $\frac{3+a}{3}=\frac{1}{3}(3+a)=1+\frac{a}{3}$ hence $\frac{3+a}{3}=1+\frac{a}{3}$ (b) When $x=2$, $\frac{2}{4+x}=\frac{2}{6}=\frac{1}{3}$ however $\frac{1}{2}+\frac{2}{x}=\frac{1}{2}+\frac{2}{2}=\frac{1}{2}+1=\frac{3}{2}$ hence$\frac{1}{3}\ne \frac{3}{2}$ so LHS $\ne$ RHS hence $\frac{2}{4+x}\ne \frac{1}{2}+\frac{2}{x}$
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