Question

Va rog cine ma poate ajuta cu exercițiul 10 si 9 . Ofer coroana .

Answer 1

$9a). \frac{x+3}{7} -2 \frac{1}{2} = \frac{x}{2} -5 \\ \frac{x+3}{7} - \frac{5}{2} = \frac{x}{2} -5 \\ 2(x+3)-7*5=7x-14*5 \\ 2x+6-35=7x-70 \\ 2x-7x=-70-6+35 \\ -5x=-41 \\ x= \frac{41}{5}$
$b). \frac{x+7}{9} -1 \frac{1}{2} = \frac{x-3}{2} \\ \frac{x+7}{9} - \frac{3}{2} = \frac{x-3}{2} \\ 2(x+7)-9*3=9(x-3) \\ 2x+14-27=9x-27 \\ 2x-9x=-27-14+27 \\ -7x=-14 \\ x= \frac{-14}{-7} \\ x=2$
$c). \frac{3x-2}{5} - \frac{x-7}{2} = \frac{2x}{5} +3 \frac{4}{5} \\ \frac{3x-2}{5} - \frac{x-7}{2} = \frac{2x}{5} + \frac{19}{5} \\ 2(3x-2)-5(x-7)=2*2x+19*2 \\ 6x-4-5x+35=4x+38 \\ 6x-5x-4x=38+4-35 \\ -3x=7 \\ x=- \frac{7}{3}$
$d).x- \frac{3(x+3)}{5} -1 \frac{9}{10} = \frac{x}{5} +4 \frac{1}{2} \\ x- \frac{3x+9}{5} - \frac{19}{10} = \frac{x}{5} + \frac{9}{2} \\ 10x-2(3x+9)-19=2x+9*5 \\ 10x-6x-18-19=2x+45 \\ 10x-6x-2x=45+18+19 \\ 2x=82 \\ x= \frac{82}{2} \\ x=41$
$10a). \frac{1}{3} x-5= \frac{1}{2} (1-x)-3 \\ \frac{x}{3} -5= \frac{1-x}{2} -3 \\ 2x-5*6=3(1-x)-3*6 \\ 2x-30=3-3x-18 \\ 2x+3x=3-18+30 \\ 5x=15 \\ x= \frac{15}{5} \\ x=3$
$b).3(x-1)+1 \frac{1}{2} =x+ \frac{1}{2} \\ 3x-3+ \frac{3}{2} =x+ \frac{1}{2} \\ 3x*2-3*2+3=2x+1 \\ 6x-6+3=2x+1 \\ 6x-2x=1+6-3 \\ 4x=4 \\ x= \frac{4}{4} \\ x=1$
$c). \frac{3}{5} (x-2)+x=1 \frac{1}{5} (x+1) \\ \frac{3(x-2)}{5} +x= \frac{6}{5} (x+1) \\ \frac{3x-6}{5} +x= \frac{6(x+1)}{5} \\ \frac{3x-6}{5} +x= \frac{6x+6}{5} \\ 3x-6+5x=6x+6 \\ 3x+5x-6x=6+6 \\ 2x=12 \\ x= \frac{12}{2} \\ x=6$
$d). \frac{2}{7} (x+3)+ \frac{x}{2} = \frac{12}{14} (x-4) \\ \frac{2(x+3)}{7} + \frac{x}{2} = \frac{12(x-4)}{14} \\ \frac{2x+6}{7} + \frac{x}{2} = \frac{12x-48}{14} \\ 2(2x+6)+7x=12x-48 \\ 4x+12+7x=12x-48 \\ 4x+7x-12x=-48-12 \\ -x=-60 \\ x=60$
$e).1 \frac{2}{5} (x+3)-2 \frac{1}{3} = \frac{4}{5} (x-1)-x \\ \frac{7}{5} (x+3)- \frac{7}{3} = \frac{4(x-1)}{5} -x \\ \frac{7(x+3)}{5} - \frac{7}{3} = \frac{4x-4}{5} -x \\ \frac{7x+21}{5} - \frac{7}{3} = \frac{4x-4}{5} -x \\ 3(7x+21)-7*5=3(4x-4)-15x \\ 21x+63-35=12x-12-15x \\ 21x-12x+15x=-12-63+35 \\ 24x=-40 \\ x=- \frac{40}{24} \\ x=- \frac{5}{3}$
$f). \frac{x-3}{4} - \frac{2}{3} x= \frac{5}{6} (x+1)+ \frac{1}{2} \\ \frac{x-3}{4} - \frac{2x}{3} = \frac{5(x+1)}{6} + \frac{1}{2} \\ \frac{x-3}{4} - \frac{2x}{3} = \frac{5x+5}{6} + \frac{1}{2} \\ 3(x-3)-2x*4=2(5x+5)+6 \\ 3x-9-8x=10x+10+6 \\ 3x-8x-10x=10+6+9 \\ -15x=25 \\ x=- \frac{25}{15} \\ x=- \frac{5}{3}$