Question

If x(1+3i)+y(2-i)-5+i=0 find x+y​

Answer 1

Given :

• x(1+3i)+y(2-i)-5+i=0

To find :

• x+y

Solution :

$\sf \implies x(1 + 3i) + y(2 - i) - 5 + i = 0$

Where , i = √(-1)

$\sf \implies x(1 + 3i) + y(2 - i) - 5 + i = 0 + 0i$

$\sf \implies x + 3xi + 2y- iy - 5 + i = 0 + 0i$

$\sf \implies i+ 3xi- iy + 2y - 5 +x = 0 + 0i$

$\sf \implies (1+ 3x- y )i+ (2y - 5 +x) = 0 + 0i$

$\sf \implies (2y - 5 +x) + (1+ 3x- y )i= 0 + 0i$

Now ,

$\longmapsto \sf 2y - 5 +x = 0$

$\longmapsto \sf 2y +x = 5$

$\longmapsto \sf x = 5 - 2y......(1)$

Also :-

$\longmapsto \sf ( 1+ 3x- y) i = 0i$

$\longmapsto \sf 1+ 3x- y= 0$

$\longmapsto \sf 3x- y= - 1......(2)$

Now from (1) put x = 5-2y in (2).

$\mapsto \sf 3(5 - 2y)- y= - 1$

$\mapsto \sf 15 - 6y- y= - 1$

$\mapsto \sf - 7y= - 1 - 15$

$\mapsto \sf y= \dfrac{16}{7}$

Now we will find the value of x :-

$\mapsto \sf x = 5 - 2( \frac{16}{7} )$

$\mapsto \sf x = 5 - \dfrac{32}{7}$

LCM =7

$\mapsto \sf x = \dfrac{35 - 32}{7}$

$\mapsto \sf x = \dfrac{3}{7}$

$\rightarrow\sf x + y= \dfrac{3}{7} + \dfrac{16}{7}$

$\rightarrow\sf x + y= \dfrac{3 + 16}{7} = \dfrac{19}{7}$

Answer : 19/7