Zad. 4
a)
[tex]2x + 5 - \frac{3x-12}{3} =2x + 5- \frac{\not{3}^1 \cdot (x-4)}{\not{3}_1} =2x+5-(x-4)=2x+5-x+4=x +9[/tex]
b)
[tex]\frac{12x+16}{4} -\frac{10x+15}{5} = \frac{\not{4}^1 \cdot (3x+4)}{\not{4}_1} -\frac{\not{5}^1 \cdot(2x+3)}{\not{5}_1} = 3x+4 -(2x+3) = 3x +4 -2x -3= \\ = x +1[/tex]
c)
[tex]\frac{9-6x}{3} -\frac{21-14x}{7} =\frac{\not{3}^1 \cdot (3-2x)}{\not{3}_1} -\frac{\not{7}^1 \cdot (3-2x)}{\not{7}_1} =3-2x - (3 - 2x) = 3 - 2x - 3 + 2x = 0[/tex]
d)
[tex]\frac{4x-8}{2} -3 \cdot \frac{10x-50}{5} =\frac{\not{2}^1 \cdot (2x-4)}{\not{2}_1} -3 \cdot \frac{\not{5}^1 \cdot (2x-10)}{\not{5}_1} = 2x - 4 -3 \cdot(2x-10) = \\ = 2x - 4 - 6x + 30 = -4x +26[/tex]
