Question

Can somebody please help me

Answer 1

`4-\cfrac{2}{3}b+\cfrac{1}{4}a~~ + ~~\cfrac{1}{2}a+\cfrac{1}{6}b-7\implies 4-\cfrac{2b}{3}+\cfrac{a}{4}~~ + ~~\cfrac{a}{2}+\cfrac{b}{6}-7 \\\\\\ \left( \cfrac{a}{4}+\cfrac{a}{2} \right)+\left(-\cfrac{2b}{3} + \cfrac{b}{6}\right)+(4-7)\implies \left( \cfrac{a}{4}+\cfrac{a}{2} \right)+\left( \cfrac{b}{6}-\cfrac{2b}{3} \right)+(4-7)`

`\left( \cfrac{(1)a+(2)a}{\underset{\textit{using this LCD}}{4}} \right)+\left( \cfrac{(1)b-(2)2b}{\underset{\textit{using this LCD}}{6}} \right)+(-3) \\\\\\ \left( \cfrac{a+2a}{4} \right)+\left( \cfrac{b-4b}{6} \right)-3\implies \left( \cfrac{3a}{4} \right)+\left( \cfrac{-3b}{6} \right)-3 \\\\[-0.35em] ~\dotfill\\\\ ~\hfill {\Large \begin{array}{llll} \cfrac{3}{4}a-\cfrac{1}{2}b-3 \end{array}}~\hfill`