Question

Resolver, em R, as inequacoes.
$3x - 12 \leqslant 0$
B)
$5( - x + 1) + 2(3x - 4) > - 1$
C)
$- x + 3 \div 2 < 2x + 5 \div 3$


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Answer 1

Resposta:

a) $x\leq 4$

b) $x>2$

c) $-\frac{1}{9}<x$

Explicação passo a passo:

a)

$3x-12\leq 0\\3x\leq 0+12\\3x\leq 12\\x\leq \frac{12}{3}\\x\leq 4$

b)

$5(-x+1)+2(3x-4)>-1\\-5x+5+6x-8>-1\\x-3>-1\\x>-1+3\\x>2$

c)

$-x+3:2<2x+5:3\\\frac{3}{2}-\frac{5}{3}<2x+x\\\frac{9-10}{6}<3x\\\frac{-1}{6}<3x\\\frac{-1}{\frac{6}{3}}<x\\\frac{-1}{6}.\frac{1}{3}<x\\\frac{-1}{9}<x\\-\frac{1}{9}<x$