Answer:
Step-by-step explanation:
[tex]\displaystyle \Large \boldsymbol{} \tt a) \ 3-2x=3(x+4)-x-1\\\\3-2x=3x+12-x-1 \\\\3-12+1=3x+2x-x \\\\4x=-8 \\\\\boxed{ \tt x=-2} \\\\\\b) \ 5x\underline{(x-1)}-4\underline{(x-1)}=0 \\\\\\(5x-4)(x-1)=0 \Longrightarrow \boxed{ \tt x_1=0.8 \ \ ; \ \ x_2=1}[/tex]
