Question

розв'яжіть систему нерівностей

1) {х>3,
2х-3<7;


2) {5-2(х+1)>6,
4+3(х-1)>2.


3) -6≤2х+4≤6​

Answer 1

Ответ:

1)

$x > 3 \\ 2x - 3 < 7$

$x > 3 \\ 2x - 3 < 7 = 2x < 7 + 3 = 2x < 10 \div 2 = x < 5$

$x > 3 \\ x < 5$

$x∈ < 3.5 >$

2)

$5 - 2(x + 1) > 6 \\ 4 + 3(x - 1) > 2$

$5 - 2(x + 1) > 6 = \\ 5 - 2x - 2 > 6 = \\ 3 - 2x > 6 = \\ - 2x > 6 - 3 = \\ - 2x > 3 \\ \div - 2 \\ x < - \frac{3}{2}$

$4 + 3(x - 1) > 2 = \\ 4 + 3x - 3 > 2 = \\ 1 + 3x > 2 = \\ 3x > 2 - 1 = \\ 3x > 1 \\ \div 3 \\ x > \frac{1}{3}$

$x < - \frac{3}{2} \\ x > \frac{1}{3}$

$Ø$

3)

$- 6 \leqslant 2x + 4 \leqslant 6$

$2x + 4 \geqslant - 6 \\ 2x + 4 \leqslant 6$

$2x + 4 \geqslant - 6 = \\ 2x \geqslant - 6 - 4 = \\ 2x \geqslant - 10 \\ \div 2 \\ x \geqslant - 5$

$2x + 4 \leqslant 6 = \\ 2x \leqslant 6 - 4 = \\ 2x \leqslant 2 \\ \div 2 \\ x \leqslant 1$

$x \geqslant - 5 \\ x \leqslant 1$

$x∈[ - 5.1]$

Answer 2

$\displaystyle\bf\\1)\\\\\left \{ {{x > 3} \atop {x-3 < 7}} \right. \\\\\\\left \{ {{x > 3} \atop {x < 7+3}} \right. \\\\\\\left \{ {{x > 3} \atop {x < 10}} \right. \\\\\\Otvet \ : \ x\in\Big(3 \ ; \ 10\Big)\\\\\\2)\\\\\\\left \{ {{5-2\cxdot(x+1) > 6} \atop {4+3\cdot(x-1) > 2}} \right. \\\\\\\left \{ {{5-2x-2 > 6} \atop {4+3x-3 > 2}} \right.\\\\\\\left \{ {{3-2x > 6} \atop {1+3x > 2}} \right. \\\\\\\left \{ {{-2x > 6-3} \atop {3x > 2-1}} \right.\\\\\\\left \{ {{-2x > 3} \atop {3x > 1}} \right.$

$\displaystyle\bf\\\left \{ {{x < -1,5} \atop {x > \dfrac{1}{3} }} \right. \\\\\\Otvet \ : \ x\in \oslash\\\\\\3)\\\\-6\leq 2x+4\leq 6\\\\-6-4\leq 2x+4-4\leq 6-4\\\\-10\leq 2x\leq 2\\\\-10:2\leq 2x:2\leq 2:2\\\\-5\leq x\leq 1\\\\Otvet \ : \ x\in\Big[-5 \ ; \ 1\Big]$