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# Tricky Algebra

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Solve for n: (2-n/n+1) + (2n-4/2-n)

Aug 2, 2022

#2
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Answer: $$\mathbb{R} - (-1, 2)$$ (with what I have, anyway...)

I assume this is what you mean: $$(\frac{2-n}{n+1})+(\frac{2n-4}{2-n})$$.

Of course, n could be literally anything other than -1 or 2 because there isn't an equality present (maybe you forgot it????? Or maybe I went blind????)

Aug 2, 2022
#3
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2-n/n+1 + 2n-4/2-n = 1

Aug 2, 2022
#4
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We have the equation: $${2 - n \over n+1} + {2n - 4 \over 2 - n} = 1$$

Make a common denominator: $${(2 -n)(2-n) \over (n + 1)(2-n)} + {(2n-4)(n+1) \over (2-n)(n+1)} = 1$$

Simplify: $${n^2−4n+4 \over −n^2+n+2} + { 2n^2−2n−4\over -n^2 + n + 2} = 1$$

Add the fractions: $${3n^2 - 6n \over -n^2 + n + 2} = 1$$

Cross multiply: $$3n^2 - 6n = -n^2 + n + 2$$

Move everything to the left-hand side: $$4n^2 - 7n - 2 = 0$$

Factor: $$(4n+1)(n−2)=0$$

So, the solutions are $$-{1 \over 4}$$ and $$2$$.

But, to verify their validity, we need to plug them back into the original equation.

Doing so, eliminates $$n = 2$$, so $$n = \color{brown}\boxed{-{1 \over 4}}$$

Aug 2, 2022