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Solve the following inequality and write your answer using interval notation. Please show all of your work. 1/(8+ x)<_(8-5x)/43
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(1/(8+ x)<_(8-5x)/43)43 (43/(8+x)<_(8-5x))(8+x) 43=(8-5x)(8+x) 43<_ 5x^2-32x+64 0<_5x^2-32x+21 0<_ (5x+3 )(x-7 ) x|x<_-3/5, 7
been a while could be wrong >.<
ah i forget how to put it in interval notation
\[−7≤x≤\frac{3}{5} \text{ or } x<−8\]
gotta teach me Zarkon
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im just glad i was headed in the right direction maybe lol
\[1/(8+ x)\le(8-5x)/43\] write as \[0\le(8-5x)/43-1/(8+ x)\] then write the RHS as one fraction
Zarkom so how do I write it in interval notation?
\[(-\infty,-8)\cup [-7,3/5]\]
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