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Find all integers n such that \(2n < n - 4 \le 3n + 8\)

 Jul 29, 2020
 #1
avatar+1042 
+6

Solving directly for n is pretty hard. Which is why we can divide this up into two inequalities:

\(2n

and

\(n-4\leq3n+8\)

Let's solve the first one:

\(2n

\(\text{Subtract n from both sides}\)

\(n<-4\)

Now, let's solve the second one:

\(n-4\leq3n+8\)

\(\text{Subtract n from both sides}\)

\(-4\leq2n+8\)

\(\text{Subtract 8 from both sides}\)

\(-12\leq2n\)

\(\text{Divide by 2}\)

\(-6\leq n\)

So, our inequality is now:

\(-6 \leq n < -4\)

Therefore, the integer solutions are:

\(\boxed{-6, -5}\)

 Jul 29, 2020
edited by ilorty  Jul 29, 2020
 #2
avatar+1042 
+6

It seems that my LaTeX has not loaded properly. Wherever it says \(2n, I mean 2n < n-4

ilorty  Jul 29, 2020

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