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Mathematics 8 Online
OpenStudy (anonymous):

Solve the following inequality and write your answer using interval notation. Please show all of your work. 1/(8+ x)<_(8-5x)/43

OpenStudy (anonymous):

(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

OpenStudy (anonymous):

been a while could be wrong >.<

OpenStudy (anonymous):

ah i forget how to put it in interval notation

OpenStudy (zarkon):

\[−7≤x≤\frac{3}{5} \text{ or } x<−8\]

OpenStudy (anonymous):

gotta teach me Zarkon

OpenStudy (anonymous):

im just glad i was headed in the right direction maybe lol

OpenStudy (zarkon):

\[1/(8+ x)\le(8-5x)/43\] write as \[0\le(8-5x)/43-1/(8+ x)\] then write the RHS as one fraction

OpenStudy (anonymous):

Zarkom so how do I write it in interval notation?

OpenStudy (zarkon):

\[(-\infty,-8)\cup [-7,3/5]\]

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