Question

type the correct answer in each box.
what are the values of x and y that make the equations true?

equation a	$(x + yi) + (4 - 7i) = 3 - 4i$
equation b	$(x + yi) - (-6 + 14i) = 18 + 2i$

in equation a, $x = \square$ and $y = \square$
in equation b, $x = \square$ and $y = \square$

Explanation:

Equation A

Step1: Combine like terms

For the real parts: \( x + 4 \), for the imaginary parts: \( y - 7 \). So the equation becomes \( (x + 4) + (y - 7)i = 3 - 4i \).

Step2: Equate real and imaginary parts

Real parts: \( x + 4 = 3 \), solving for \( x \): \( x = 3 - 4 = -1 \).
Imaginary parts: \( y - 7 = -4 \), solving for \( y \): \( y = -4 + 7 = 3 \).

Equation B

Step1: Simplify the left side

Distribute the negative sign: \( x + yi + 6 - 14i = (x + 6) + (y - 14)i \). The equation is \( (x + 6) + (y - 14)i = 18 + 2i \).

Step2: Equate real and imaginary parts

Real parts: \( x + 6 = 18 \), solving for \( x \): \( x = 18 - 6 = 12 \).
Imaginary parts: \( y - 14 = 2 \), solving for \( y \): \( y = 2 + 14 = 16 \).

Answer:

In equation A, \( x = -1 \) and \( y = 3 \)
In equation B, \( x = 12 \) and \( y = 16 \)