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OpenStudy (anonymous):
OpenStudy (anonymous):
ick
OpenStudy (anonymous):
what
OpenStudy (anonymous):
it is ugly
you have to add on the right first
OpenStudy (anonymous):
\[\frac{4x+1}{12}+\frac{2x+2}{2x+1}\]
\[\frac{(4x+1)(2x+1)+12(2x+2)}{12(2x+1)}\]
\[\frac{8x^2+30x+25}{12(2x+1)}\] then
\[\frac{2x-1}{6}=\frac{8x^2+30x+25}{12(2x+1)}\]
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OpenStudy (anonymous):
at least we can multiply both sides by 6 and get
\[2x-1=\frac{8x^2+30x+25}{4x+2}\] then
\[(2x-1)(4x+2)=8x^2+30x+25\] and now we are ready to solve
OpenStudy (anonymous):
\[8x^2-2=8x^2+30x+25\]
\[-2=30x+25\]
\[-28=30x\]
\[x=-\frac{28}{30}=-\frac{14}{15}\] hhmm i wonder if there is an easier way, or if i made mistake