$$\left[{\frac{\left({\mathtt{12}}\right)}{\left({\mathtt{3}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}\right)}}\right] = \left[{\frac{\left({\mathtt{10}}\right)}{\left({\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{4}}\right)}}\right]$$
+118725 $$\begin{array}{rlll}
\frac{12}{3x+3}&=&\frac{10}{2x+4}\qquad \mbox{}\\\\
\frac{12}{3(x+1)}&=&\frac{10}{2(x+2)}\qquad \mbox{simplify both sides}\\\\
\frac{4}{x+1}&=&\frac{5}{x+2}\qquad \mbox{simplify both sides}\\\\
\frac{4}{x+1}\times \frac{(x+1)(x+2)}{1}&=&\frac{5}{x+2}\times \frac{(x+1)(x+2)}{1}\qquad \mbox{simplify both sides}\\\\
\frac{4(x+2)}{1}&=&\frac{5(x+1)}{1}\qquad \mbox{simplify both sides}\\\\
4x+8&=&5x+5\\\\
4x+8-4x&=&5x+5-4x\\\\
8&=&x+5\\\\
8-5&=&x+5-5\\\\
3&=&x\\\\
x&=&3\\\\
\end{array}$$
+118725 $$\begin{array}{rlll}
\frac{12}{3x+3}&=&\frac{10}{2x+4}\qquad \mbox{}\\\\
\frac{12}{3(x+1)}&=&\frac{10}{2(x+2)}\qquad \mbox{simplify both sides}\\\\
\frac{4}{x+1}&=&\frac{5}{x+2}\qquad \mbox{simplify both sides}\\\\
\frac{4}{x+1}\times \frac{(x+1)(x+2)}{1}&=&\frac{5}{x+2}\times \frac{(x+1)(x+2)}{1}\qquad \mbox{simplify both sides}\\\\
\frac{4(x+2)}{1}&=&\frac{5(x+1)}{1}\qquad \mbox{simplify both sides}\\\\
4x+8&=&5x+5\\\\
4x+8-4x&=&5x+5-4x\\\\
8&=&x+5\\\\
8-5&=&x+5-5\\\\
3&=&x\\\\
x&=&3\\\\
\end{array}$$
