\(x =  \dfrac{1}{b}\)

$$
\begin{aligned}
\frac{b}{x+a}+\frac{b}{x-a}&=\frac{2}{x^{2}-a^{2}} \\
\frac{b(x-a)+b(x+a)}{x^{2}-a^{2}}&=\frac{2}{x^{2}-a^{2}} \\
b x-a b+b x+a b&=2 \\
2 b x&=2 \\
x&=\frac{1}{b}
\end{aligned}
$$

\(x = \dfrac{{ab}}{{b - a}}\)

$$
\begin{aligned}
\frac{x}{a}+1&=\frac{x}{b}+2 \\
\frac{x}{a}-\frac{x}{b}&=1 \\
\frac{b x-a x}{a b}&=1 \\
x(b-a)&=a b \\
x&=\frac{a b}{b-a}
\end{aligned}
$$

\(x = \dfrac{{a + b}}{{{a^2} + {b^2}}},\,\,y = \dfrac{{b-a}}{{{a^2} + {b^2}}}\)

\begin{align}
&ax+by=1\quad \color{red}\ldots \text {(1)}\\
&bx-ay=1\quad \color{red}\cdots \text { (2) }\\
&{\color{red}\text {(1)}} \times a \quad a^{2} x+aby=a\quad \color{red}\cdots (3) \\
&{\color{red}\text { (2) }} \times b \quad b^{2} x-a b y=b\quad \color{red}\cdots (4) \\
&\begin{aligned} {\color{red}(3)+ (4)}\quad \left(a^{2}+b^{2}\right)x&=a+b\\
x&=\frac{a+b}{a^{2}+b^{2}}
\end{aligned}\\
&\text{In}\ {\color{red}(1)}\quad \frac{a(a+b)}{a^{2}+b^{2}}+b y=1 \\
&\begin{aligned}by &=1-\frac{a(a+b)}{a^{2}+b^{2}} \\ &=\frac{a^{2}+b^{2}-a^{2}-a b}{a^{2}+b^{2}} \\ &=\frac{b(b-a)}{a^{2}+b^{2}} \\ \therefore\ y &=\frac{b-a}{a^{2}+b^{2}} \end{aligned}
\end{align}

\(x = \dfrac{a}{{2 - a}},\,\,y = \dfrac{{a(1 - a)}}{{2-a}}\)

$$
\begin{align}
&\begin{aligned}
a x &=a-2 y \quad \ {\color{red}\cdots(1)} \\
y &=a-x \quad \ {\color{red}\cdots(2)}
\end{aligned}\\
&\text{Sub}\ {\color{red}(2)}\ \text{into}\ {\color{red}(1)}\\
&\begin{aligned}
a x &=a-2(a-x) \\
a x &=a-2 a+2 x \\
a x &=-a+2 x \\
a &=2 x-a x \\
a &=x(2-a) \\
x &=\frac{a}{2-a}\ {\color{red}\cdots(3)}
\end{aligned}\\
&\text{Sub}\ {\color{red}(3)}\ \text{into}\ {\color{red}(2)}\\
&\begin{aligned}y&=a-\frac{a}{2-a}\\
y&=\frac{a(2-a)-a}{2-a}\\
y&=\frac{2 a-a^{2}-a}{2-a}\\
y&=\frac{a-a^{2}}{2-a}\\
y&=\frac{a(1-a)}{2-a}\end{aligned}
\end{align}
$$