Answer:
The answer to your question
a) Values excluded = -2, 0
b) solutions = -2/3, 2
Step-by-step explanation:
Equation
[tex]\frac{4x}{2x + 4} = \frac{x+ 2}{2x}[/tex]
a) x is restricted when the denominators equal zero, so let's get them.
2x + 4 = 0 2x = 0
2x = - 4 x = 0 / 2
x = -4 / 2 x = 0
x = -2
x must be different to -2, 0
b) [tex]\frac{4x}{2x + 4} = \frac{x+ 2}{2x}[/tex]
[tex]\frac{2x}{x + 2} = \frac{x + 2}{2x}[/tex]
4x² = (x + 2)²
4x² = x² + 4x + 4
4x² - x² - 4x - 4 = 0
3x² - 4x - 4 = 0
3x² - 6x + 2x - 4 = 0
3x(x - 2) + 2(x - 2) = 0
(x - 2)(3x + 2) = 0
x₁ - 2 = 0 3x₂ + 2 = 0
x₁ = 2 x₂ = -2/3
