Question

Решить неравенства:
1) 3х – 7<4 (х + 2)
2) 7 – 6x >= 1/3(9x — 1)
3) 1,5(х – 4) + 2,5x <x+6.
4) 1,4(х + 5) + 1,6x >9+х​

Answer 1

Ответ:

$1) 3x - 7 < 4(x + 2)\\\\3x-7<4x+8\\\\3x - 4x < 8 + 7\\\\-x < 15 \\\\x>-15\\\\x\in[-15;\infty)$

$2) 7 - 6x \geq \frac{9x-1}{3} \\\\21-18x \geq 9x - 1\\\\-18x - 9x \geq -1 - 21\\\\-27x \geq -22\\\\ x \leq \frac{22}{27}\\\\x\in [-\infty;\frac{22}{27}]$

$3) 1,5(x - 4) + 2,5x < x+ 6\\\\1,5x - 6 + 2,5x < x + 6\\\\1,5x + 2,5x - x < 6 + 6\\\\3x < 12\\\\x < 4\\\\x\in (-\infty; 4)$

$4) 1,4(x + 5) + 1,6x > 9 + x\\\\1,4x + 7 + 1,6x > 9 + x\\\\1,4x + 1,6x - x > 9 - 7\\\\2x > 2\\\\x\in (1; \infty)$

Answer 2

Решение во вложении: