We have to find x and y given x(1+2i)+y(2-i)=4+3i

x(1+2i)+y(2-i)=4+3i

=> x + 2xi + 2y - yi = 4 + 3i

equate the real and imaginary coefficients

x + 2y = 4 and 2x - y = 3

x = 4 - 2y

substitute in 2x - y = 3

=> 8 - 4y - y = 3

=> y = 1

x = 2

**The value of x = 2 and y = 1**

To find out x and y, we'll have to write the rectangular form of the complex number:

z = a + b*i

We'll have to identify the real part a and the imaginary part b.

We'll start by removing the brackets:

x + 2ix + 2y - iy = 4 + 3i

We'll combine the real parts and the imaginary parts from the left side:

(x+2y) + i(2x - y) = 4 + 3i

We'll compare both sides and we'll get:

x + 2y = 4 (1)

2x - y = 3 (2)

We'll add (1) + 2*(2):

x + 2y + 4x - 2y = 4 + 6

We'll combine and eliminate like terms:

5x = 10

We'll divide by 5:

x = 2

We'll substitute x in (1):

2 + 2y = 4

2y = 2

y = 1

**The valid values for x and y are: x = 2 and y = 1.**