Question

Poate rezolva cineva exercitiul 5, multumesc mult!!!

Answer 1

Explicație pas cu pas:

$4(x + 3) < 3x + 7 \\ 4x + 12 < 3x + 7 \\ 4x - 3x < 7 - 12 \\ x < - 5 \\ \implies x \in \Big(-\infty ; - 5\Big)$

$8(x + 2) < 5x + 4 \\ 8x + 16 < 5x + 4 \\ 8x - 5x < 4 - 16 \\ 3x < - 12 \iff x < - 4 \\ \implies x \in \Big(-\infty ; - 4 \Big)$

$6(x - 3) - 10 \leqslant 2(2x + 1) \\ 6x - 18 - 10 \leqslant 4x + 2 \\ 6x - 4x \leqslant 2 + 28 \\ 2x \leqslant 30 \iff x \leqslant 15 \\ \implies x \in \Big(-\infty ; 15 \Big]$

$5(2x - 3) - 3(x + 1) > 4(x - 9) \\ 10x - 15 - 3x - 3 > 4x - 36 \\ 7x - 4x > - 36 + 18 \\ 3x > - 18 \iff x > - 6 \\ \implies x \in \Big(-6 ; +\infty \Big)$

$3(x - 1) - 6 < 2(x + 7) \\ 3x - 3 - 6 < 2x + 14 \\ 3x - 2x < 14 + 9 \\ x < 23 \implies x \in \Big(-\infty ; 23 \Big)$

$5(x - 1) - 3 < 3(x + 4) \\ 5x - 5 - 3 < 3x + 12 \\ 5x - 3x < 12 + 8 \\ 2x < 20 \iff x < 10 \\ \implies x \in \Big(-\infty ;10 \Big)$

$6(x - 2) + x \leqslant 5(x + 4) \\ 6x - 12 + x \leqslant 5x + 20 \\ 7x - 5x \leqslant 20 + 12 \\ 2x \leqslant 32 \iff x \leqslant 16 \\ \implies x \in \Big(-\infty ; 16 \Big]$

$4(x + 3) - 9 \leqslant 3(x - 1) \\ 4x + 12 - 9 \leqslant 3x - 3 \\ 4x - 3x \leqslant- 3 - 3 \\ x \leqslant - 6 \implies x \in \Big(-\infty ; - 6 \Big]$

Answer 2

Răspuns:

Explicație pas cu pas: