Simplify. Assume that all variables result in nonzero denominators. Enter the expression in simplest form. The numerator and denominator must be in explanded form.

remolatg 2021-08-05 Answered
Simplify. Assume that all variables result in nonzero denominators.
Enter the expression in simplest form. The numerator and denominator must be in explanded form (i.e. not a product of factors).
\(\displaystyle\Rightarrow{\frac{{{6}{q}}}{{{q}+{1}}}}-{\frac{{{q}-{4}}}{{{q}}}}+{\frac{{{6}}}{{{q}+{1}}}}=\)

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Expert Answer

ensojadasH
Answered 2021-08-06 Author has 6835 answers
\(\displaystyle\Rightarrow{\frac{{{6}{q}}}{{{q}+{1}}}}-{\frac{{{q}-{4}}}{{{q}}}}+{\frac{{{6}}}{{{q}+{1}}}}\)
\(\displaystyle\Rightarrow{\frac{{{6}{q}^{{{2}}}-{\left({q}-{4}\right)}{\left({q}+{1}\right)}+{6}{q}}}{{{q}{\left({q}+{1}\right)}}}}\)
\(\displaystyle\Rightarrow{\frac{{{6}{q}^{{{2}}}-{\left({q}^{{{2}}}+{q}-{4}{q}-{4}\right)}+{6}{q}}}{{{q}{\left({q}+{1}\right)}}}}\)
\(\displaystyle\Rightarrow{\frac{{{6}{q}^{{{2}}}-{\left({q}^{{{2}}}-{3}{q}-{4}\right)}+{6}{q}}}{{{q}{\left({q}+{1}\right)}}}}\)
\(\displaystyle\Rightarrow{\frac{{{6}{q}^{{{2}}}-{q}^{{{2}}}+{3}{q}+{4}+{6}{q}}}{{{q}{\left({q}+{1}\right)}}}}\)
\(\displaystyle\Rightarrow{\frac{{{5}{q}^{{{2}}}+{9}{q}+{4}}}{{{q}{\left({q}+{1}\right)}}}}\)
\(\displaystyle\Rightarrow{\frac{{{5}{q}^{{{2}}}+{5}{q}+{4}{q}+{4}}}{{{q}{\left({q}+{1}\right)}}}}\)
\(\displaystyle\Rightarrow{\frac{{{5}{q}{\left({q}+{1}\right)}+{4}{\left({q}+{1}\right)}}}{{{q}{\left({q}+{1}\right)}}}}\)
\(\displaystyle\Rightarrow{\frac{{{\left({5}{q}+{4}\right)}{\left({q}+{1}\right)}}}{{{q}{\left({q}+{1}\right)}}}}\)
\(\displaystyle\Rightarrow{\frac{{{5}{q}+{4}}}{{{q}}}}\)
So, the simplified expression is:
\(\displaystyle\Rightarrow{\frac{{{5}{q}+{4}}}{{{q}}}}\)
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