Question

find values of x and y:(1+i)(x+iy)=2-5i​

Answer 1

Solution:

LHS = (1+i)(x+iy)

=1(x+iy)+i(x+iy)

=x+iy+ix+i²y

= x+i(x+y)-y

=(x-y)+i(x+y)

= 2-5i [ RHS , given ]

x-y = 2 ----(1)

x+y = -5 ---(2)

Add (1)&(2), we get,

2x = -3

=> x = (-3)/2

Subtract (1) from (2) , we get

2y= -7

=> y = (-7)/2

Therefore,

x = -3/2 , y = -7/2

•••

Answer 2

(1+i)(x+iy)= 2--5i

=>x+iy+ix+i^y=2--5i

we know that

i^2 = --1

then

=>x+i(y+x)--y= 2--5i

compare the real and imagenary part

of both sides

then

x--y=2----------(1)

and

x+y= --5-------------(2)

equation(2) -(1)

y = -7

y = --7/2

put thvalue of y in equation(2)

x-7/2= --5

x= -3/2

.

.

.

HOPE IT HELP

MARK AS BRAINLIST PLZ