a)
\frac{3x+3}{x-2}=\frac{9x+3}{3x} , x\ne 2 i x\ne0 , D=\mathbb R \backslash \{0,2\}
\frac{3x+3}{x-2}=\frac{3(3x+1)}{3x}
\frac{3x+3}{x-2}=\frac{3x+1}{x}
proporcja - mnożę “na krzyż”
(3x+3)*x=(3x+1)(x-2)
3x^2+3x=3x^2-6x+x-2
3x^2+3x=3x^2-5x-2
3x^2+3x-3x^2+5x+2=0
8x+2=0
8x=-2
x=-\frac{2}{8}
x=-\frac{1}{4}
b)
\frac{-3x+5}{x+3}+4=0 , D=\mathbb R\backslash \{-3\}
\frac{-3x+5}{x+3}=-4
-3x+5=-4(x+3)
-3x+5=-4x-12
-3x+4x=-12-5
x=-17