Question

gente me ajudem serio mesmo

Answer 1

$-3(x+2) \geq 5- \dfrac{x+2}{6} \to\\\\-3x-6\geq5- \dfrac{x+2}{6} \to~~mmc=6 \\\\ -18x-36\geq30-(x+2)\to\\\\ -18x-36\geq30-x-2\to\\\\ -18x+x \geq 30-2+36\to\\\\ -17x\geq64~(-1)\to\\\\ 17x \leq -64\to\\\\ \large\boxed{x\leq -\dfrac{64}{17} }$





$\Large\boxed{\boxed{S=\left\{x \epsilon \mathbb{R} ~|~x \leq -\dfrac{64}{17} ~\right\}}}$

Answer 2

1)

3.(x + 1) + 4.(x - 2) - (x - 4) > 7x - 4.(x - 8)
3x + 3 + 4x - 8 - x + 4 > 7x - 4x + 32
6x - 1 > 3x + 32
6x - 3x > 32 + 1
3x > 33
x > 33/3
x > 11

R.: {x ∈ R / x >11}
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- 3.(x + 2) ≥ 5 - x + 2
                          6

- 3x - 6 ≥ 5 - x + 2
                     6

6.(- 3x - 6) ≥ 30 - 1(x + 2)
     6                   6

- 18x - 36 ≥ 30 - x - 2
- 18x + x ≥ 28 + 36
- 17x ≥ 64  ( - 1)
17x ≤ - 64

x ≤ - 64
        17

R.: {x ∈ R / x ≤ - 64/17}