​ Solving fractions

We can combine these fractions into one single fraction after getting a common denominator.

$\mathbf{a.}\qquad\frac{4}{x}-\frac{6}{x}\quad=\quad\frac{4-6}{x}\quad=\quad-\frac{2}{x}\\~\\~\\ \mathbf{b.}\qquad\frac{x}{2}+\frac{2}{3}\quad=\quad\frac{3\,\cdot \,x}{3\,\cdot\,2}+\frac{2\,\cdot\,2}{3\,\cdot\,2}\quad=\quad\frac{3x}{6}+\frac46\quad=\quad\frac{3x+4}{6}\\~\\~\\ \mathbf{c.}\qquad2x\cdot\frac{3}{10}+\frac{x}{10}\quad=\quad\frac{2x\,\cdot\,3}{10}+\frac{x}{10}\quad=\quad\frac{6x}{10}+\frac{x}{10}\quad=\quad\frac{6x+x}{10}\quad=\quad\frac{7x}{10}$

For this last one, we don't need to get a common denominator because we aren't adding fractions.

We just need to multiply them together and cancel common factors.

$\mathbf{d.}\qquad\frac{3}{5x^2}\cdot\frac{4x^2}{6x}\quad=\quad\frac{3\,\cdot\,4x^2}{5x^2\,\cdot\,6x}\quad=\quad\frac{3\,\cdot\,2\,\cdot\,2\,\cdot\, x^2}{5\,\cdot\,x^2\,\cdot\,3\,\cdot\,2\cdot\,x}\quad=\quad \frac{{\color{gray}3}\,\cdot\,{\color{gray}2}\,\cdot\,2\,\cdot\, {\color{gray}x^2}}{5\,\cdot\,{\color{gray}x^2}\,\cdot\,{\color{gray}3}\,\cdot\,{\color{gray}2}\cdot\,x}\quad=\quad\frac{2}{5x}$_