Odpowiedź :
[tex]8-\frac{x+3}{2} = 5+\frac{3x+1}{4}\\\\-\frac{x+3}{2} = \frac{3x+1}{4}+5-8\\\\-\frac{x+3}{2} = \frac{3x+1}{4}-3 \ \ |\cdot4\\\\-2(x+3) = 3x+1 - 12\\\\-2x-6 = 3x - 11\\\\-2x-3x = -11+6\\\\-5x = -5 \ \ /:(-5)\\\\x = 1[/tex]
[tex]8-\frac{x+3}{2} = 5+\frac{3x+1}{4}\\\\-\frac{x+3}{2} = \frac{3x+1}{4}+5-8\\\\-\frac{x+3}{2} = \frac{3x+1}{4}-3 \ \ |\cdot4\\\\-2(x+3) = 3x+1 - 12\\\\-2x-6 = 3x - 11\\\\-2x-3x = -11+6\\\\-5x = -5 \ \ /:(-5)\\\\x = 1[/tex]