6. Use the result from the previous question to find the integer solutions of the following equations:
(i) $43 x+15 y=8$.
(ii) $10 x-37 y=3$
6. (i) Solution: Converting $\frac{43}{15}$ into a continued fraction, we get
$$\frac{43}{15}=2+\frac{1}{1+\frac{\Gamma}{6+\frac{1}{2}}}$$

Therefore,
$$k=4 . \quad \frac{p_{1}}{q_{1}}=2, \quad \frac{p_{2}}{q_{2}}=3, \quad \frac{p_{3}}{q_{3}}=\frac{6 \times 3+2}{6 \times 1+1}=\frac{20}{7}$$

So,
$$\left\{\begin{array}{l}
x_{0}=(-1)^{4} \times 8 \times 7=56 \\
y_{0}=(-1)^{5} \times 8 \times 20=-160
\end{array}\right.$$

is a particular solution. By Theorem One of Chapter Three, its general solution is
$$\left\{\begin{array}{l}
x=56-15 t, \\
y=-160+43 t .
\end{array} \quad(t=0, \pm 1, \pm 2, \cdots)\right.$$
(ii) Solution: Converting $\frac{10}{37}$ into a continued fraction, we get
$$\frac{10}{37}=\frac{1}{3+\frac{1}{1+\frac{1}{2+\frac{1}{-3}}}}$$

Therefore,
$$\begin{aligned}
k=5 . \quad \frac{p_{1}}{q_{1}} & =\frac{0}{1}, \quad \frac{p_{2}}{q_{2}}=\frac{1}{3} \\
\frac{p_{3}}{q_{3}} & =\frac{1 \times 1+0}{1 \times 3+1}=\frac{1}{4} \\
\frac{p_{4}}{q_{4}} & =\frac{2 \times 1+1}{2 \times 4+3}=\frac{3}{11} \\
\frac{p_{5}}{q_{5}} & =\frac{3 \times 3+1}{3 \times 11+4}=\frac{10}{37^{\circ}}
\end{aligned}$$

So,
$$\left\{\begin{array}{l}
x_{0}=(-1)^{5} \times 3 \times 11=-33 \\
y_{0}=(-1)^{5} \times 3 \times 3=-9
\end{array}\right.$$

is a set of integer solutions. Its general solution is
$$\left\{\begin{array}{l}
x=-33+37 t, \\
y=-9+10 t
\end{array},(t=0, \pm 1, \pm 2, \cdots)\right.$$