Вирішіть будь-ласка. Буду вдячна))

[latex] \frac{5}{7} (7x-2)+ \frac{1}{7}(2-7x)+3= \frac{5}{7}*7x- \frac{5}{7}*2+ \frac{1}{7}*2+ \frac{1}{7}*(-7x)+3= [/latex] [latex]= \frac{5*7*x}{7*1}- \frac{5*2}{7}+ \frac{1*2}{7}- \frac{1*7*x}{1*1*7}+3= = \frac{5*x}{1}- \frac{10}{7}+ \frac{2}{7}- \frac{x}{1}+3= [/latex] [latex]= 5x+ \frac{-10+2}{7}-x+3 = 5x-x+ \frac{-8}{7}+3 = 4x- \frac{8}{7}+3= [/latex] [latex]= 4* \frac{3}{4} - \frac{8}{7}+3= \frac{4*3}{4*1} - \frac{8}{7} +3= \frac{3}{1} - \frac{1+7}{7} +3= 6 -1- \frac{1}{7}=[/latex] [latex]=5- \frac{1}{7}=4 \frac{6}{7} [/latex]