<s>[INST] Partial fractions of $\frac{-5x+19}{(x-1/2)(x+1/3)}$ Alright, I need to find the partial fractions for the expression above. I have tried writing this as $$\frac{a}{x-1/2}+\frac{b}{x+1/3}$$ but the results give me $a=25.8$ and $b=-20.8$, which are slightly wrong because they give me $5x+19$ instead of $-5x+19$. Can you please help? Thanks a lot
 [/INST] Either you've started from the wrong equations, or you've made a calculation mistake (don't worry, happens to all of us sometimes). When you write
$$\frac{a}{x-\frac{1}{2}}+\frac{b}{x+\frac{1}{3}}=\frac{-5x+19}{(x-\frac{1}{2})(x+\frac{1}{3})}$$
you get
$$\frac{a(x+\frac{1}{3})}{(x-\frac{1}{2})(x+\frac{1}{3})}+\frac{b(x-\frac{1}{2})}{(x-\frac{1}{2})(x+\frac{1}{3})}=\frac{ax+bx+\frac{a}{3}-\frac{b}{2}}{(x-\frac{1}{2})(x+\frac{1}{3})}=\frac{-5x+19}{(x-\frac{1}{2})(x+\frac{1}{3})}$$
so that the correct equations to solve are
$$\begin{align*}
a+b&=-5\\[0.1in]
\tfrac{a}{3}-\tfrac{b}{2}&=19
\end{align*}$$
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