Example 7. Solve the inequality
$$
\sqrt{\frac{1}{2} x+1}-1>\frac{1}{5}(2 x-7) .
$$
The set of permissible values for the unknown $x$ is $x \geqslant-2$. (1)
$$
\text { Let } y_{1}=\sqrt{\frac{1}{2} x+1}-1, y_{2}=\frac{1}{5}(2 x-7) \text {, }
$$

Then the graphs of $\mathrm{y}_{1}$ and $\mathrm{y}_{2}$ are shown in Figure 1.
Now find the x-coordinate of the intersection point $\mathrm{A}$ of the parabola $\mathrm{y}_{1}$ and the line $\mathrm{y}_{2}$. Solve the equation $\mathrm{y}_{1}=\mathrm{y}_{2}$, that is
$$
\sqrt{\frac{1}{2}} x+1-1=\frac{1}{5}(2 x-7) \text {, to get } x=6 \text {. }
$$

Considering (1), and combining with Figure 1, it can be seen that when $-2 \leqslant x \leqslant 6$, the solution to $y_{1} \leqslant y_{2}$ is $-2 \leqslant x<6$, which is the solution to the original inequality.