Question

Buna, am si eu nevoie de rezolvarea exercitiului 6. Raman recunoscatoare si veti primi coroana si puncte. Va multumesc, si va rog sa ma ajutati <3

Answer 1

a)

$2-3iy-3x=14i-5y-5xi\\ 2-3x-3iy=-5y+i(14-5x)\\ \left \{ {{2-3x=-5y} \atop {-3y=14-5x}} \right =>\left \{ {{-3x+5y=-2}/*5 \atop {5x-3y=14}/*3} \right. =>\left \{ {{-15x+25y=-10} \atop {15x-9y=42}} \right. => 16y=32 => y=2\\ 5x-3y=14 => 5x-6=14 => 5x=20 => x=4$

b)

$\frac{i}{x}+\frac{i}{y}+\frac{1}{6}=\frac{1}{x}-\frac{1}{y}+\frac{5i}{y}\\ i(\frac{1}{x}+\frac{1}{y})+\frac{1}{6}=i*\frac{5}{y}+\frac{1}{x}-\frac{1}{y}\\ \left \{ {{\frac{1}{x}+\frac{1}{y}=\frac{5}{y}} \atop {\frac{1}{6}=\frac{1}{x}-\frac{1}{y}}} \right. =>\left \{ {{\frac{1}{x}=\frac{5}{y}-\frac{1}{y}} \atop {\frac{1}{y}=\frac{1}{x}-\frac{1}{6}}} \right. =>\left \{ {{\frac{1}{x}=\frac{5}{x}-\frac{5}{6}-\frac{1}{x}+\frac{1}{6}} \atop {{\frac{1}{y}=\frac{1}{x}-\frac{1}{6}}} \right. =>$

$\left \{ {{\frac{4}{6}=\frac{3}{x}} \atop {{\frac{1}{y}=\frac{1}{x}-\frac{1}{6}}} \right. =>\left \{ {{x=\frac{9}{2}} \atop {\frac{1}{y}=\frac{2}{9}-\frac{1}{6}}} \right. =>\left \{ {{x=\frac{9}{2}} \atop {\frac{1}{y}=\frac{1}{18}}} \right. =>\left \{ {{x=\frac{9}{2}} \atop {y=18}} \right.$

c)

$(3x+2yi)+(2y-xi)=2-3i\\ 3x+2y+i(2y-x)=2-3i\\ \left \{ {{3x+2y=2} \atop {2y-x=-3}} \right. =>\left \{ {{3x+2y=2} \atop {x-2y=3}} \right. =>4x=5 =>x=\frac{5}{4}\\ 2y-x=-3=>2y-\frac{5}{4}=-3 =>2y=-\frac{12}{4}+\frac{5}{4} => 2y=-\frac{7}{4}=>y=-\frac{7}{8}$

d)

$\frac{x-2}{1+i}+\frac{y+1}{1-i}=3+i\\ \frac{(x-2)(1-i)}{(1+i)(1-i)}+\frac{(y+1)(1+i)}{(1-i)(1+i)}=3+i\\ \frac{x-2-xi+2i}{1-i^2}+\frac{y+1+yi+i}{1-i^2}=3+i\\ \frac{x-2-xi+2i+y+1+yi+i}{1-i^2}=3+i\\ \frac{x-2+y+1+i(-x+2+y+1)}{2}=3+i\\ x+y-1+i(-x+y+3)=6+2i\\ \left \{ {{x+y-1=6} \atop {-x+y+3=2}} \right. => \left \{ {{x+y=7} \atop {-x+y=-1}} \right. => 2y=6 => y=3\\ x+y=7 => x+3=7 => x=4$