# What are x and y? x(1+2i)+y(2-i)=4+3i

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

Let us expand the brackets.

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

==> We will combine real terms and complex terms together.

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

Now we will compare terms.

==> x + 2y = 4 ..............(1)

==> 2x - y = 3 ....................(2)

Now we will solve the system.

We will use the substitution method to solve.

==> We will rewrite y= 2x - 3.

==> x+ 2y = 4

==> x + 2(2x-3) = 4

==> x + 4x - 6 = 4

==> 5x = 10

**==> x= 2.**

==> y= 2x -3 = 2*2 - 3 = 4-3 =1

**==> y= 1**

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

To find x and y:

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

We open the brackets :

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

We collect the reals of the left together. And similarly collect the imaginary numbers of the left together.

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

We equate real parts on both sides: x+2y = 4......(1)

We equate the imaginary parts on both sides: 2x-y = 3....(2)

(1)+2*(2): x +2*2x= 4+2*3 = 10. So 5x = 10. So x= 10/5 = 2.

Put x = 2 in (2) : 2*2-y = 3. Or -y = 3-4 = -1. So y = 1.

Therefore x= 2 and y = 1.

To determine x and y, we'll have to re-write the left side as a complex number put in rectangular form:

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.**