Question

$$\frac{ 9m+1 }{ 9m+4 } + \frac{ 7 }{ 3m+3 }$$

Answer

$$(27*m^2+93*m+31)/(3*(9*m+4)*(m+1))$$

Solution


Factor out the common term \(3\).
\[\frac{9m+1}{9m+4}+\frac{7}{3(m+1)}\]
Rewrite the expression with a common denominator.
\[\frac{(9m+1)\times 3(m+1)+7(9m+4)}{(9m+4)\times 3(m+1)}\]
Regroup terms.
\[\frac{3(9m+1)(m+1)+7(9m+4)}{(9m+4)\times 3(m+1)}\]
Expand.
\[\frac{27{m}^{2}+27m+3m+3+63m+28}{(9m+4)\times 3(m+1)}\]
Collect like terms.
\[\frac{27{m}^{2}+(27m+3m+63m)+(3+28)}{(9m+4)\times 3(m+1)}\]
Simplify  \(27{m}^{2}+(27m+3m+63m)+(3+28)\)  to  \(27{m}^{2}+93m+31\).
\[\frac{27{m}^{2}+93m+31}{(9m+4)\times 3(m+1)}\]
Regroup terms.
\[\frac{27{m}^{2}+93m+31}{3(9m+4)(m+1)}\]