<u>Answer:</u>
(x, y) = (-2, 2)
<u>Step-by-step explanation:</u>
We are given the following two equations which we are to solve using the substitution method:
--- (1)
--- (2)
From equation (1):
Substituting this value of x in (2) to get:
![x = [tex]x - 2y = -6\\\\(y-4)-2y=-6\\\\-y=-6+4\\\\-y=-2](https://tex.z-dn.net/?f=x%20%3D%20%5Btex%5Dx%20-%202y%20%3D%20-6%5C%5C%5C%5C%28y-4%29-2y%3D-6%5C%5C%5C%5C-y%3D-6%2B4%5C%5C%5C%5C-y%3D-2)
[/tex]
y = 2
Substituting the value of y in (1) to get x:
x = -2
Therefore, (x, y) = (-2, 2).
Answer:
Step-by-step explanation:
The question is to simplify the expression 
First, we can eliminate the 2nd parenthesis expression by changing signs of all the terms since there is a "-" sign in front. Then we add the "Like Terms". The process is shown below:
The first answer choices (has some typos) is correct. The correct answer is
<h3>
Answer:</h3>
x² -6x = -13 ⇒ x ∈ {3-2i, 2+2i}
<h3>
Step-by-step explanation:</h3>
To make a=1, divide the equation by the coefficient of x², which is 8.
... x² -6x = -13 . . . . . . your blanks are filled with -6 and -13
Now, to complete the square, add the square of half the x-coefficient:
... (-6/2)² = 9.
... x² -6x +9 = -4 . . . 9 added to both sides
... (x -3)² = -4 . . . . . rewrite as a square
... x -3 = ±2i . . . . . . take the square root
... x = 3 ±2i . . . . . . . add 3
The solutions are the complex numbers x = 3 ±2i.
