Question

Answer 1

\(e,3-\left(\dfrac{1}{6}-x\right).\dfrac{2}{3}=\dfrac{2}{3}\)

 \(\left(\dfrac{1}{6}-x\right).\dfrac{2}{3}=3-\dfrac{2}{3}\)

\(\left(\dfrac{1}{6}-x\right).\dfrac{2}{3}=\dfrac{9}{3}-\dfrac{2}{3}\)

\(\left(\dfrac{1}{6}-x\right).\dfrac{2}{3}=\dfrac{7}{3}\)

\(\dfrac{1}{6}-x=\dfrac{7}{3}:\dfrac{2}{3}\)

\(\dfrac{1}{6}-x=\dfrac{7}{3}.\dfrac{3}{2}\)

\(\dfrac{1}{6}-x=\dfrac{7}{2}\)

\(x=\dfrac{1}{6}-\dfrac{7}{2}\)

\(x=\dfrac{1}{6}+\dfrac{-21}{6}\)

\(x=\dfrac{-20}{6}\)

\(f,8x-4x=1208\)

\(\left(8-4\right).x=1208\)

  4       . x =1208

 x=1208 : 4

 x= 302

g,0,3. x +0,6.x=9

  (0,3+0,6).x=9

    0,9 .x=9

          x =9:0,9

          x= 10 

\(h,\dfrac{1}{2}x+\dfrac{2}{5}x=\dfrac{-18}{25}\)

 \(\left(\dfrac{1}{2}+\dfrac{2}{5}\right).x=\dfrac{-18}{25}\)

\(\left(\dfrac{5}{10}+\dfrac{4}{10}\right)x=\dfrac{-18}{25}\)

\(\dfrac{9}{10}x=\dfrac{-18}{25}\)

\(x=\dfrac{-18}{25}:\dfrac{9}{10}\)

\(x=\dfrac{-18}{25}.\dfrac{10}{9}\)

x= \(\dfrac{-4}{5}\)

Answer 2

e ] 3 - [ \(\dfrac{1}{6}\) - x ] . \(\dfrac{2}{3}\)  = \(\dfrac{2}{3}\)  

     3 - [ \(\dfrac{1}{6}\) - x ] = \(\dfrac{2}{3}\) : \(\dfrac{2}{3}\)

     3 - [ \(\dfrac{1}{6}\) - x ] = 1

          [ \(\dfrac{1}{6}\) - x ] = 3 - 1

            \(\dfrac{1}{6}\) - x = 2

                  x = \(\dfrac{1}{6}\) - 2

                  x = \(\dfrac{-11}{6}\)

f ] 8x - 4x = 1208

    [ 8 - 4 ]x = 1208

             4x = 1208

               x = 1208 : 4

               x = 302

g ] 0,3x + 0,6x = 9

    [ 0,3 + 0,6 ]x = 9

                 0,9x = 9

                      x = 9 : 0,9

                      x = 10

h ] \(\dfrac{1}{2}\)x + \(\dfrac{2}{5}\)x = \(\dfrac{-18}{25}\)

    [ \(\dfrac{1}{2}\) + \(\dfrac{2}{5}\) ]x = \(\dfrac{-18}{25}\)

              \(\dfrac{9}{10}\)x = \(\dfrac{-18}{25}\) 

                   x = \(\dfrac{-18}{25}\)  : \(\dfrac{9}{10}\)

                   x = \(\dfrac{-4}{5}\)