Question

Find real value is x and y if {(1+I)x-2i/3+i}+{(2-3i)y+I}/3-i=i​

Answer 1

answer :- Consider the given expressions.

3+i

(1+i)x−2i

+

3−i

(2−3i)y+i

=i

3+i

x+(x−2)i

+

3−i

2y+(1−3y)i

=i

9−i

2

(x+(x−2)i)(3−i)+(2y+(1−3y)i)(3+i)

=i

x(3−i)+i(x−2)(3−i)+2y(3+i)+i(1−3y)(3+i)=(9+1)i

3x−ix+i(3x−ix−6+2i)+6y+2iy+i(3+i−9y−3yi)=10i

3x−ix+3xi−i

2

x−6i+2i

2

+6y+2iy+3i+i

2

−9yi−3yi

2

=10i

3x−ix+3xi+x−6i−2+6y+2iy+3i−1−9yi+3y=10i

4x+9y−3+2xi−7yi−13i=0

4x+9y−3+(2x−7y−13)i=0

On comparing real part and imaginary part, we get

4x+9y−3=0 …… (1)

2x−7y−13=0 …… (2)

On solving both equations, we get

x=3

y=−1

Hence, the value of x,y is 3,−1.

explanation :- The values of x and y satisfying the equation

3+i

(1+i)x−2i +

3−i

(2−3i)y+i

=i, are