Question

Can somebody show me how to do this?

Answer 1

notice, we'll be using the LCD of all denominators, in this case the LCD is 3, that way we do away with the denominators, but first let's distribute that 4.

`\bf \cfrac{1}{3}h-4\left( \cfrac{2}{3}h-3 \right)=\cfrac{2}{3}h-6\implies \cfrac{h}{3}-4\left( \cfrac{2h}{3}-3 \right)=\cfrac{2h}{3}-6 \\\\\\ \cfrac{h}{3}-\stackrel{\textit{distributing}}{\left( \cfrac{8h}{3}-12 \right)}=\cfrac{2h}{3}-6\implies \cfrac{h}{3}-\cfrac{8h}{3}+12 =\cfrac{2h}{3}-6 \\\\\\ \textit{now, let's multiply both sides by }\stackrel{LCD}{3}\textit{ to do away with the denominators}`

`\bf 3\left( \cfrac{h}{3}-\cfrac{8h}{3}+12 \right)=3\left( \cfrac{2h}{3}-6 \right)\implies h-8h+36=2h-18 \\\\\\ -7h+36=2h-18\implies 36+18=2h+7h\implies 54=9h \\\\\\ \cfrac{54}{9}=h\implies 6=h`