Zad. 1
[tex]\left \{ {{\frac{1}{2}x =4+y} \atop {x-\frac{1}{2}y =4}} \right.[/tex]
[tex]\left \{ {{\frac{1}{2}x *2 =4*2+y*2} \atop {x-\frac{1}{2}y =4}} \right.[/tex]
[tex]\left \{ {{x=8+2y} \atop {x-\frac{1}{2}y =4 }} \right.[/tex]
[tex]\left \{ {{x=8+2y} \atop {8+2y-\frac{1}{2}y =4 }} \right.[/tex]
[tex]\left \{ {{x=8+2y} \atop {8+1\frac{1}{2}y =4 }} \right.[/tex]
[tex]\left \{ {{x=8+2y} \atop {1\frac{1}{2}y =-4 }} \right.\\\frac{3}{2} y = -4\\\frac{3}{2} y *\frac{2}{3} = -4*\frac{2}{3}\\y=-\frac{8}{3} \\\left \{ {{y=-\frac{8}{3}} \atop {x=8+2y}} \right. \\\left \{ {{y=-\frac{8}{3}} \atop {x=8+2*(-\frac{8}{3})}} \right. \\\\\left \{ {{y=-\frac{8}{3}} \atop {x=8-\frac{16}{3}}} \right. \\\\\\\left \{ {{y=-\frac{8}{3}} \atop {x=8-5\frac{1}{3}}=2\frac{2}{3} } \right. \\\\\\\\y = -\frac{8}{3}} \\x=2\frac{2}{3}[/tex]
Zad. 2
[tex]x^{2} -8x+16-(x^{2} -4)=0\\x^{2} -8x+16-x^{2} +4=0\\-8x+20 = 0\\-8x=-20\\-8x :(-8) = -20 : (-8)\\x= \frac{20}{8} = 2 \frac{4}{8} =2 \frac{1}{2}[/tex]
