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Question : 7 of 160
Marks:
+1,
-0
Solution:
+=i ⇒((1+i)x−2i)(3−i)+((2−3i)y)(3+i)=10i ⇒(x+ix−2i)(3−i)+(2y−3iy)(3+i)=10i ⇒3x+3ix−6i−ix+x−2+ 6y−9iy+2iy+3y=10i ⇒(3x+x−2+6y+3y)+ (3x−6−x+2Y−9y)i=10i ⇒4x+9y−2=0,2x−7y−16=0 ⇒y=,x= ∴ x+y=‌−=‌
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