Question

what are the values of x and y if this equation is true?
22(x + yi) + (28 + 4i) = 72 - 62i

a. x = 3 and y = 2
b. x = -3 and y = -2
c. x = 2 and y = -3
d. x = 2 and y = 3

Explanation:

Step1: Expand the left - hand side

First, we expand \(22(x + yi)\) using the distributive property. \(22(x+yi)=22x + 22yi\). Then the left - hand side of the equation \(22(x + yi)+(28 + 4i)\) becomes \(22x+22yi + 28+4i=(22x + 28)+(22y + 4)i\).
The right - hand side of the equation is \(72-62i\).

Step2: Equate the real and imaginary parts

For two complex numbers \(a + bi\) and \(c+di\) to be equal, we must have \(a = c\) (equating the real parts) and \(b = d\) (equating the imaginary parts).

Equate the real parts:

We set the real part of the left - hand side equal to the real part of the right - hand side. So, \(22x+28 = 72\).
Subtract 28 from both sides: \(22x=72 - 28=44\).
Divide both sides by 22: \(x=\frac{44}{22}=2\).

Equate the imaginary parts:

We set the imaginary part of the left - hand side equal to the imaginary part of the right - hand side. So, \(22y + 4=-62\).
Subtract 4 from both sides: \(22y=-62 - 4=-66\).
Divide both sides by 22: \(y=\frac{-66}{22}=-3\).

Answer:

C. \(x = 2\) and \(y=-3\)