O să îți propun 2 variante:
1)
3x+8≤x-4
3x+8-x≤-4
2x+8≤-4
2x≤-4-8
2x≤-12
x≤[tex] \frac{12}{2} [/tex]
x≤6
2)
3x+8≤x-4
(3x+8)+(-8-x)≤(x-4)+(-8-x)
(3x+8)+(-x-8)≤(x-4)+(-x-8)
3x+8-x-8≤x-4-x-8
3x-x+8-8≤x-x-4-8
2x≤-12
[tex] \frac{2x}{2} [/tex]≤[tex] - \frac{12}{2} [/tex]
x≤[tex] - \frac{ {2}^{2} \times 3 }{2} [/tex]
x≤[tex] - ( {2}^{2 - 1} \times 3)[/tex]
x≤-(2×3)
x≤-6
x € (-∞,-6]
