Answer:
[tex]=\frac{4y}{y+3}[/tex]
Step-by-step explanation:
[tex]\frac{2y}{y-3}\cdot \frac{4y-12}{2y+6}\\\frac{4y-12}{2y+6}=\frac{2\left(y-3\right)}{y+3}\\\frac{4y-12}{2y+6}\\\mathrm{Factor}\:4y-12:\quad 4\left(y-3\right)\\4y-12\\\mathrm{Rewrite\:as}\\=4y-4\cdot \:3\\\mathrm{Factor\:out\:common\:term\:}4\\=4\left(y-3\right)\\=\frac{4\left(y-3\right)}{2y+6}\\\mathrm{Factor}\:2y+6:\quad 2\left(y+3\right)\\2y+6\\\mathrm{Rewrite\:as}\\=2y+2\cdot \:3\\\mathrm{Factor\:out\:common\:term\:}2\\=2\left(y+3\right)\\=\frac{4\left(y-3\right)}{2\left(y+3\right)}[/tex]
[tex]\mathrm{Divide\:the\:numbers:}\:\frac{4}{2}=2\\=\frac{2\left(y-3\right)}{\left(y+3\right)}\\\mathrm{Remove\:parentheses}:\quad \left(a\right)=a\\=\frac{2\left(y-3\right)}{y+3}\\=\frac{2y}{y-3}\cdot \frac{2\left(y-3\right)}{y+3}\\\mathrm{Multiply\:fractions}:\quad \frac{a}{b}\cdot \frac{c}{d}=\frac{a\:\cdot \:c}{b\:\cdot \:d}\\=\frac{2y\cdot \:2\left(y-3\right)}{\left(y-3\right)\left(y+3\right)}\\\mathrm{Cancel\:the\:common\:factor:}\:y-3\\=\frac{2y\cdot \:2}{y+3}[/tex]
[tex]\mathrm{Multiply\:the\:numbers:}\:2\cdot \:2=4\\=\frac{4y}{y+3}[/tex]
