Verified answer

Объяснение:

3)

[tex]\left \{ {{2x^2+13x-7\leq 0} \atop {15-3x\geq 0}} \right.\ \ \ \ \left \{ {{2x^2+14x-x-7\leq 0} \atop {3x\leq 15\ |:3}} \right. \ \ \ \ \left \{ {{2x*(x+7)-(x+7)\leq 0} \atop {x\leq 5}} \right.\ \ \ \ \left \{ {{(x+7)(2x-1)\leq 0} \atop {x\leq 5}} \right. .[/tex]

(x+7)(2x-1)≤0

-∞__+__-7__-__1/2__+__+∞     ⇒

[tex]\left \{ {{x\in[-7;0,5].} \atop {x\in(-\infty;5]}} \right. \ \ \ \ \ \Rightarrow[/tex]

Ответ: x∈[-7;0,5].