{5a+8/3–a≥2a {1–6–15a/4≥a
[m]\left\{\begin {matrix}\frac{5a+8}{3–a}≥2a\\1–\frac{6–15a}{4}≥a\end {matrix}\right.[/m] ⇒ [m]\left\{\begin {matrix}\frac{5a+8}{3–a}-2a≥0\\1–\frac{6–15a}{4}-a≥0\end {matrix}\right.[/m] ⇒ [m]\left\{\begin {matrix}\frac{5a+8-6a+2a^2}{3–a} ≥ 0\\\frac{4-6+15a-4a}{4}≥0\end {matrix}\right.[/m] ⇒
