Respuesta :

[tex]\bf \cfrac{1}{4}n+12 \geqslant \cfrac{3}{4}n-4\implies \cfrac{1}{4}n+16 \geqslant \cfrac{3}{4}n\implies 16\geqslant \cfrac{3n}{4}-\cfrac{1n}{4}\implies 16\geqslant \cfrac{3n-1n}{4} \\\\\\ 16\geqslant\cfrac{2n}{4}\implies 16\geqslant \cfrac{n}{2}\implies 32 \geqslant n[/tex]

Answer:

n ≤ 32

Step-by-step explanation:

Given

[tex]\frac{1}{4}[/tex] n + 12 ≥ [tex]\frac{3}{4}[/tex] n - 4

Subtract [tex]\frac{1}{4}[/tex] n from both sides

12 ≥ [tex]\frac{1}{2}[/tex] n - 4 ( add 4 to both sides )

16 ≥ [tex]\frac{1}{2}[/tex] n ( multiply both sides by 2 )

32 ≥ n, hence

n ≤ 32

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