Question

If x(1+3i)+y(2-i)-5+i^3=0 then find x+y​.
plzz ans fast guys

Answer 1

Answer:  The required value of x+y is 3.

Step-by-step explanation:  We are given the following equality :

$x(1+3i)+y(2-i)-5+i^3=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)$

We are to find the value of x+y.

We know that

$i=\sqrt{-1},~~i^2=-1,~~i^3=i\times(-1)=-i.$

From equation (i), we have

$x(1+3i)+y(2-i)-5+i^3=0\\\\\Rightarrow x+3ix+2y-yi=5-i^3\\\\\Rightarrow (x+2y)+(3x-y)i=5-(-i)\\\\\Rightarrow (x+2y)+(3x-y)i=5+i.$

Equating the real and imaginary parts on both sides of the above, we get

$x+2y=5~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(ii)\\\\3x-y=1\\\\\Rightarrow 2(3x-y)=2\times1\\\\\Rightarrow 6x-2y=2~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(iii)$

Adding equations (ii) and (iii), we get

$(x+2y)+(6x-2y)=5+2\\\\\Rightarrow 7x=7\\\\\Rightarrow x=\dfrac{7}{7}\\\\\Rightarrow x=1.$

Substituting the value of x in equation (ii), we get

$1+2y=5\\\\\Rightarrow 2y=5-1\\\\\Rightarrow 2y=4\\\\\Rightarrow y=\dfrac{4}{2}\\\\\Rightarrow y=2.$

Therefore, we get

$x+y=1+2=3.$

Thus, the required value of x+y is 3.

Answer 2

Step-by-step explanation:

x(1+3i)+y(2-i)-5+i³=0

x+3xi+2y-yi-5-i=0

(x+2y-5)+(3x-y-1)i=0+0i

equating real and impairing part,we get

x+2y-5=0 (1)

3x-y-1=0 (2)

multiplying by equation 2 by 2

6x-2y=2 (3)

adding equation 1 and question 2

7x=7

x=1

putting x = 1 in equation 1

1+2y=5

2y=4

y=2

x+y=1+2

x+y=3