Question

resolva:

c) x+12 - 1 ≤ 1-x
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10 4

d) 1 (x-2) < x - 1
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3 2

e) -3 (x-2) - 3 ≥ 3 + 2x

f) x + 1 > x-2
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3 2

g) x+1 < x-2 + 2
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6 9 3
​

Answer 1

c)

$\frac{x + 12}{10} - 1 \leqslant \frac{1 - x}{4} \\ \frac{2(x + 12) - 20 \leqslant 5(1 - x)}{20} \\ 2(x + 12) - 20 \leqslant 5(1 - x)$

$2x + 24 - 20 \leqslant 5 - 5x \\ 2x + 5x \leqslant 5 + 20 - 24 \\ 7x \leqslant 1 \\ x \leqslant \frac{1}{7}$

d)

$\frac{1}{3}(x - 2) < \frac{x}{2} - 1 \\ \frac{2(x - 2) < 3x - 6}{6} \\ 2(x - 2) < 3x - 6$

$2x - 4 < 3x - 6 \\ 2x - 3x < 4 - 6 \\ - x < - 2 \times ( - 1) \\ x > 2$

e)

$- 3(x - 2) - 3 \geqslant 3 + 2x \\ - 3x + 6 - 3 \geqslant 3 + 2x \\ - 3x -2x \geqslant 3 + 3 - 6 \\ - 5x \geqslant 0 \times ( -1)$

$5x \leqslant 0 \\ x \leqslant \frac{0}{5} \\ x \leqslant 0$

f)

$\frac{x}{3} + 1 > \frac{x - 2}{2} \\ \frac{2x + 6 > 3(x - 2)}{6} \\ 2x + 6 > 3(x -2)$

$2x + 6 > 3x - 6 \\ 2x - 3x > - 6 - 6 \\ - x > - 12 \times ( - 1) \\ x < 12$

g)

$\frac{x + 1}{6} < \frac{x - 2}{9} + \frac{2}{3} \\ \frac{3(x + 1) < 2(x - 2) + 12}{18} \\ 3(x + 1) < 2(x - 2) + 12$

$3x + 3 < 2x - 4 + 12 \\ 3x - 2x < 12 - 4 - 3 \\ x < 5$