Question

Допоможіть будь ласка с-ми

Answer 1

Ответ:

3) -12 < y < 17

4) -3 ≤ x ≤ -1

Объяснение:

$\displaystyle 3)\quad \left \{ {{2(3y-1)-4(2y+3) < 10} \atop {\displaystyle \frac{y-3}{2}^{\setminus 3}-\frac{y+4}{3} ^{\setminus 3} < 0 \hfill }} \right. \left \{ {{6y-2-8y-12 < 10} \atop {3y-9-2y-8 < 0\hfill}} \right. \left \{ {{-2y-14 < 10} \atop {y-17 < 0}} \right. \\\\\\ \left \{ {{-2y < 24} \atop {y < 17}} \right. \left \{ {{y > -12} \atop {y < 17}} \right. \quad \Rightarrow \boldsymbol {12 < y < 17}$

$\displaystyle 4)\quad \left \{ {{\displaystyle \frac{7x+1}{2} -2^{\setminus 2}\geq 5x^{\setminus 2} \hfill} \atop {(x+5)(x+3)\geq x^2+3x}} \right. \left \{ {{7x+1-4\geq 10x\hfill} \atop {x^2+3x+5x+15\geq x^2+3x}} \right. \\\\\\\left \{ {{-3x\geq 3\hfill} \atop {5x+15\geq 0}} \right. \left \{ {{x\leq -1} \atop {x\geq -3}} \quad \Rightarrow \boldsymbol {-3\leq x\leq -1}\right.$