Answer:
[tex]\dfrac{x}{2}+\dfrac{1}{3}[/tex]
Step-by-step explanation:
We need to find the value of [tex]((\dfrac{1}{4}x+\dfrac{3}{4}x-\dfrac{1}{2}x)+1+(\dfrac{-2}{3}))[/tex]
We can solve it as follows :
[tex]((\dfrac{1}{4}x+\dfrac{3}{4}x-\dfrac{1}{2}x)+1+(\dfrac{-2}{3}))\\\\=(\dfrac{x+3x-2x}{4})+1-\dfrac{2}{3}\\\\=\dfrac{2x}{4}+\dfrac{1}{3}\\\\=\dfrac{x}{2}+\dfrac{1}{3}[/tex]
Hence, the value of [tex]((\dfrac{1}{4}x+\dfrac{3}{4}x-\dfrac{1}{2}x)+1+(\dfrac{-2}{3}))[/tex] is equal to [tex]\dfrac{x}{2}+\dfrac{1}{3}[/tex].
