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OpenStudy (anonymous):
Subtract, and then simplify, if possible.
b − 3/b + 9
−
b + 2/b − 9
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OpenStudy (haleyelizabeth2017):
multiply the first whole fraction by b-9 and the whole second fraction by b+9, I'll show you what you should get :)
OpenStudy (anonymous):
okay
OpenStudy (haleyelizabeth2017):
\[\frac{(b-3)(b-9)}{(b+9)(b-9)}-\frac{(b+9)(b+2)}{(b+9)(b-9)}\] Okay? We needed to get a common denominator, which is \[(b+9)(b-9)\]
OpenStudy (haleyelizabeth2017):
We need to expand the numerators in order to subtract them.
OpenStudy (anonymous):
for the first one I got b^2
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OpenStudy (anonymous):
I got b^2 -12b +27 as the numerator
OpenStudy (haleyelizabeth2017):
the first one is \[b^2-12b+27\] you are correct
OpenStudy (haleyelizabeth2017):
And for the second?
OpenStudy (anonymous):
and the second one I got b^2 +11b+18
OpenStudy (haleyelizabeth2017):
right. Now, we have one fraction.\[\frac{(b^2-12b+27)-(b^2+11b+18)}{(b+9)(b-9)}\]
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OpenStudy (haleyelizabeth2017):
So, we then have \[\frac{-23b+9}{(b+9)(b-9)}\]
OpenStudy (anonymous):
that's what I got
OpenStudy (haleyelizabeth2017):
Awesome. I don't think we can simplify, but let me check.
OpenStudy (anonymous):
Thankyou
OpenStudy (haleyelizabeth2017):
You're welcome! That's as far as we can simplify haha
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