Answer:
[tex]y = i[/tex]
[tex]x = 1[/tex]
Step-by-step explanation:
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Make [tex]x[/tex] the subject for [tex]ix-2y=-i[/tex] :
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Add 2y to both sides:
[tex]\implies ix=-i+2y[/tex]
Divide both sides by i:
[tex]\implies x=\dfrac{-i+2y}{i}[/tex]
To remove i from the denominator, multiply the numerator and denominator by its complex conjugate:
[tex]\implies x=\dfrac{-i+2y}{i} \times\dfrac{-i}{-i}[/tex]
[tex]\implies x=\dfrac{i^2-2iy}{-i^2}[/tex]
Apply the imaginary number rule [tex]i^2=-1[/tex] :
[tex]\implies x=\dfrac{-1-2iy}{-(-1)}[/tex]
[tex]\implies x=\dfrac{-1-2iy}{1}[/tex]
[tex]\implies x=-1-2iy[/tex]
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Substitute [tex]x=-1-2iy[/tex] into [tex](1+i)x-2iy=3+i[/tex] and make y the subject:
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[tex]\implies (1+i)(-1-2iy)-2iy=3+i[/tex]
Apply complex arithmetic rule [tex](a+bi)(c+di)=(ac-bd)+(ad+bc)i[/tex] :
[tex]\implies (-1+2y)+(-2y-1)i-2iy=3+i[/tex]
Simplify:
[tex]\implies -1+2y-2iy-i-2iy=3+i[/tex]
[tex]\implies 2y-4iy=4+2i[/tex]
Factor:
[tex]\implies 2(1-2i)y=2(2+i)[/tex]
Divide both sides by [tex]2(1-2i)[/tex] :
[tex]\implies \dfrac{2(1-2i)y}{2(1-2i)}=\dfrac{2(2+i)}{2(1-2i)}[/tex]
[tex]\implies y=\dfrac{(2+i)}{(1-2i)}[/tex]
Multiply by the complex conjugate [tex]\dfrac{1+2i}{1+2i}[/tex] to remove [tex](1 - 2i)[/tex] from the denominator :
[tex]\implies y=\dfrac{(2+i)(1+2i)}{(1-2i)(1+2i)}[/tex]
Apply complex arithmetic rule [tex](a+bi)(c+di)=(ac-bd)+(ad+bc)i[/tex] to numerator:
[tex]\implies y=\dfrac{(2-2)+(4+1)i}{(1-2i)(1+2i)}[/tex]
[tex]\implies y=\dfrac{5i}{(1-2i)(1+2i)}[/tex]
Apply complex arithmetic rule [tex](a+bi)(a-bi)=a^2+b^2[/tex] to the denominator :
[tex]\implies y=\dfrac{5i}{1^2+2^2}[/tex]
[tex]\implies y=\dfrac{5i}{5}[/tex]
[tex]\implies y=i[/tex]
Therefore, one solution is [tex]y=i[/tex]
For the other solution, substitute [tex]y=i[/tex] into [tex]x=-1-2iy[/tex]:
[tex]\implies x=-1-2ii[/tex]
[tex]\implies x=-1-2i^2[/tex]
Apply the imaginary number rule [tex]i^2=-1[/tex] :
[tex]\implies x=-1-2(-1)[/tex]
[tex]\implies x=-1+2[/tex]
[tex]\implies x=1[/tex]
Therefore, the second solution is [tex]x = 1[/tex]
