smc

Given the equation: $$\frac{a}{10^x-1}+\frac{b}{10^x+2}=\frac{2\cdot 10^x+3}{(10^x-1)(10^x+2)}$$ Let's simplify by letting $y=10^x$. The equation becomes: $$\frac{a}{y-1}+\frac{b}{y+2}=\frac{2y+3}{(y-1)(y+2)}$$ Multiplying each term by the common denominator $(y-1)(y+2)$, we obtain: $$a(y+2)+b(y-1)=2y+3$$ For the equation to be an identity, the coefficients of like terms on both sides must be equal. Equating the coefficients of $y$ and the constant terms, we get the system of equations: $$\begin{array}{rl}a+b& =2\text{(from coefficients of }y\text{)}\\ 2a-b& =3\text{(from constant terms)}\end{array}$$ Solving this system, we find: $$a=\frac{5}{3}\text{and}b=\frac{1}{3}$$ Thus, the difference is: $$a-b=\boxed{\frac{4}{3}}$$