Rozwiąż dane układy metodą podstawiania, może ktoś to rozwiązać krok po kroku, ponieważ nie wiem jak to zrobić, będę bardzo wdzięczna, proszę jak najszybciej!!

Odpowiedź:
[tex]\left \{ {{3(2x-y)-2(x+4y)=6-10y} \atop {0,5(x-3y)-(4y-x)+8=0}/*2} \right. \\ \left \{ {{6x-3y-2x-8y=6-10y} \atop {x-3y-2(4y-x)+16=0}} \right. \\ \left \{ {{4x-11y=6-10y} \atop {x-3y-8y+2x+16=0}} \right. \\ \left \{ {{4x-y=6} \atop {3x-11y=-16}} \right. \\ \left \{ {{y=4x-6} \atop {3x-11(4x-6)=-16}} \right. \\ \left \{ {{y=4x-6} \atop {3x-44x+66=-16}} \right. \\ \left \{ {{y=4x-6} \atop {-41x=-82}/:(-41)} \right. \\ \left \{ {{y=4*2-6} \atop {x=2}} \right. \\ \left \{ {{y=2} \atop {x=2}} \right. \\[/tex]
[tex]\left \{ {{\frac{3x}{4} -\frac{y}{4} =-1} /*4\atop {\frac{5x}{2} +\frac{3y}{8} =16}/*8} \right. \\ \left \{ {{3x-y=-4} \atop {20x+3y=128}} \right. \\ \left \{ {{y=3x+4} \atop {20x+3(3x+4)=128}} \right. \\ \left \{ {{y=3x+4} \atop {20x+9x+12=128}} \right. \\ \left \{ {{y=3x+4} \atop {29x=116}/:29} \right. \\ \left \{ {{y=3x+4} \atop {x=4}} \right. \\ \left \{ {{y=3*4+4} \atop {x=4}} \right. \\ \left \{ {{y=16} \atop {x=4}} \right.[/tex]
[tex]\left \{ {{\frac{x-3}{2} -\frac{2x+y}{4} =\frac{x-y}{8} /*8} \atop {x+y+12=0}} \right. \\ \left \{ {{4(x-3)-2(2x+y)=x-y} \atop {x=-y-12}} \right. \\ \left \{ {{4x-12-4x-2y-x+y=0} \atop {x=-y-12}} \right. \\ \left \{ {{-x-y=12} \atop {x=-y-12}} \right. \\ \left \{ {{-2-y=12} \atop {x=-y-12}} \right. \\ \left \{ {{y=-14} \atop {x=14-12}} \right. \\ \left \{ {{y=-14} \atop {x=2}} \right.[/tex]
[tex]\left \{ {{(2x-y)(y+5)=y(2x-y-4)} \atop {3(8x-y)=2(11x-y)}} \right. \\ \left \{ {{2xy-y^{2} +10x-5y=2xy-y^{2}-4y } \atop {24x-3y=22x-2y}} \right. \\ \left \{ {{10x-5y+4y=0} \atop {24x-22x-3y+2y=0}} \right. \\ \left \{ {{y=10x} \atop {2x-y=0}} \right. \\ \left \{ {{y=10x} \atop {2x-10x=0}} \right. \\ \left \{ {{y=10x} \atop {x=0}} \right. \\ \left \{ {{y=0} \atop {x=0}} \right.[/tex]