\(\displaystyle {{a}})\;0,5{{x}} - {2 \over 3}x = {7 \over {12}} \)
\(\displaystyle x.\left( {{1 \over 2} - {2 \over 3}} \right) = {7 \over {12}} \)
\(\displaystyle x.\left( {{3 \over 6} - {4 \over 6}} \right) = {7 \over {12}} \)
\(\displaystyle x.{{ - 1} \over 6} = {7 \over {12}} \)
\(\displaystyle x = {7 \over {12}}:{{ - 1} \over 6} \)
\(\displaystyle \eqalign{
& b)\;x:4{1 \over 3} = - 2,5 \cr
& x:{{13} \over 3} = {{ - 5} \over 2} \cr
& x = {{ - 5} \over 2}.{{13} \over 3} \cr
& x = {{ - 65} \over 6} = - 10{5 \over 6} \cr
& x = {7 \over {12}}.{{ - 6} \over 1} = {{ - 7} \over 2} \cr} \)
\(\displaystyle \eqalign{
& c)\;5,5{{x}} = {{13} \over {15}} \cr
& {{55} \over {10}}x = {{13} \over {15}} \cr
& x = {{13} \over {15}}:{{55} \over {10}} \cr
& x = {{13} \over {15}}.{{10} \over {55}} \cr
& x = {{26} \over {165}} \cr} \)
\(\displaystyle \eqalign{
& {\rm{d}})\;\left( {{{3{{x}}} \over 7} + 1} \right):\left( { - 4} \right) = {{ - 1} \over {28}} \cr
& {{3{{x}}} \over 7} + 1 = {{ - 1} \over {28}}.\left( { - 4} \right) \cr
& {{3{{x}}} \over 7} = {1 \over 7} - 1 \cr
& {{3{{x}}} \over 7} = {{ - 6} \over 7} \cr
& 3{{x}} = - 3 \cr
& x = - 6:3 = - 2 \cr} \)
