How do you solve $2 (n + 1) + 1 = (n - 5) + (n + 2)$ $2(n+1)+1=(n-5)+(n+2)$ ?

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Answer 1 · 2015-12-01T17:32:52

No solution.

Distribute the 2 to the terms in the parentheses:

$2 (n + 1) = 2 n + 2$ $2(n+1) = 2n+2$ .

Now you have $2 n + 2 + 1 = n - 5 + n + 2$ $2n+2+1=n−5+n+2$ .

Combine like terms and you're left with $2 n + 3 = 2 n - 3$ $2n+3=2n-3$ .

Isolate terms with variable $n$ $n$ to one side (take opposite sign):

$2 n - 2 n = - 3 + 3$ $2n-2n=-3+3$

$0 n = 0$ $0n=0$

So no solution.

How do you solve 2(n+1)+1=(n-5)+(n+2)2(n+1)+1=(n−5)+(n+2)2(n+1)+1=(n-5)+(n+2)?