Задание 5 стр 176

Question

Решите неравенства.
$$
\begin{align*}
&a) \ 5(x-2) + 12 > 7 - 3(x+4) & c) \ 4(x-1,5) + 1 \geq 6x+1,2\
&b) \ 3(a+8) - 5(a-1) < 7 & d) \ 4x-(x+6) - 3(2-x)\leq 4
\end{align*}
$$

Gimli · Accepted Answer

a) $$5(x-2)+12>7-3(x+4)$$ $$5 x-10+12>7-3 x-12$$ $$5 x+2>-3 x-5$$ $$8 x>-7$$ $$x>-\frac{7}{8}$$ b) $$3(a+8)-5(a-1)<7$$ $$3 a+24-5 a+5 < 7$$ $$-2 a+29<7$$ $$-2 a<7-29$$ $$-2 a<-2 2$$ $$a>11$$ c) $$ \begin{aligned} & 4(x-1,5) - 1 \geqslant 6 x+1,2 \ & 4 x-6-1 \geqslant 6 x+1,2 \ & 4 x-5 \geqslant 6 x+12^2 \ & 4 x-6 x \geqslant 12+5 \ & -2 x \geq 17(\div(-2)) \ & x \leq 8,5 \end{aligned} $$ d) $$ \begin{aligned} & 4 x-(x+6)-3(2-x) \leqslant 4 \ & 4 x-x-6-(6-3 x) \leqslant 4 \ & 3 x-6-6+3 x \leqslant 4 \ & 6 x-12 \leq 4 \ & 6 x \leqslant 16 \ & x \leqslant 2 \frac{2}{3} \end{aligned} $$

Задание 5 стр 176

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Задание 5 стр 176

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