8.10 When $x$ takes positive rational values, if
$$
\frac{a}{10^{x}-1}+\frac{b}{10^{x}+2}=\frac{2 \cdot 10^{x}+3}{\left(10^{x}-1\right)\left(10^{x}+2\right)}
$$

is an identity, then the value of $a-b$ is
(A) $\frac{4}{3}$.
(B) $\frac{5}{3}$.
(C) 2 .
(D) $\frac{11}{4}$.
(E) 3 .
(18th American High School Mathematics Examination, 1967)
［Solution］After making the left side of the given equation have a common denominator, the numerator becomes
$$
a\left(10^{x}+2\right)+b\left(10^{x}-1\right)=(a+b) 10^{x}+2 a-b \text {, }
$$

By the property of identity, comparing the coefficients of the numerators on both sides, we have
$$
\left\{\begin{array} { l } 
{ a + b = 2 , } \\
{ 2 a - b = 3 . }
\end{array} \text { Solving, we get } \left\{\begin{array}{l}
a=\frac{5}{3}, \\
b=\frac{1}{3} .
\end{array}\right.\right.
$$

Therefore, the answer is $(A)$.