Question

Розв'яжіть систему нерівностей: 1) [8.x-32<0, -3х +15>0; 2) [6x-5<13, 28+ 4x > 20.

Answer 1

Ответ:

1) x ∈ (-∞; 4)

2) x ∈ (-2; 3)

Объяснение:

Требуется решить неравенства

$\tt \displaystyle \large \boldsymbol {} 1) \left \{ {{8 \cdot x-32 < 0} \atop {-3 \cdot x+15 > 0}} \right. ; \;\;\;\;\; 2) \left \{ {{6 \cdot x-5 < 13} \atop {28+4 \cdot x > 20}} \right. .$

Свойства неравенств.

1) Если a > b, то b < a;

2) Если a < b, k > 0, a·k < b·k.

Решение.

$\tt \displaystyle \large \boldsymbol {} 1) \left \{ {{8 \cdot x-32 < 0} \atop {-3 \cdot x+15 > 0}} \right. \\\\\\\left \{ {{8 \cdot x < 32 \;\;\; |:8} \atop {15 > 3 \cdot x \;\;\; |:3}} \right. \\\\\\\left \{ {{x < 4 } \atop {5 > x}} \right. \\\\\\\left \{ {{x < 4 } \atop {x < 5}} \right. \\\\\\x < 4 \\\\x \in (-\infty; 4).$

$\tt \displaystyle \large \boldsymbol {} 2) \left \{ {{6 \cdot x-5 < 13} \atop {28+4 \cdot x > 20}} \right. \\\\\\\left \{ {{6 \cdot x < 13+5} \atop {4 \cdot x > 20-28}} \right. \\\\\\\left \{ {{6 \cdot x < 18 \;\;\; |:6} \atop {4 \cdot x > -8 \;\;\; |:4}} \right. \\\\\\\left \{ {{x < 3} \atop {x > -2}} \right. \\\\\\\left \{ {{x < 3} \atop {-2 < x}} \right. \\\\-2 < x < 3 \\\\x \in (-2;3).$

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