Answer:
x = 7 and y = 3.2
Step-by-step explanation:
[tex]\bold{DOMAIN}\\\\2x+4>0\qquad\text{subtract 4 from both sides}\\2x>-4\qquad\text{divide both sides by 2}\\x>-2\qquad(a)\\\\3x-1>0\qquad\text{add 1 to both sides}\\3x>1\qquad\text{divide both sides by 3}\\x>\frac{1}{3}\qquad(b)\\\\x^2-9>0\qquad\text{add 9 to both sides}\\x^2>9\to x <-\sqrt9\ \vee\ x>\sqrt9\to x<-3\ \vee\ x>3\qquad(c)\\\\5y+2>0\qquad\text{subtract 2 from both sides}\\5y>-2\qquad\text{divide both sides by 5}\\y>-0.4\qquad(d)\\\\\text{From}\ (a),\ (b),\ (c)\ \text{and}\ (d)\ \text{we have:}\\\\\boxed{D:x>3\ \wedge\ y>-0.4}[/tex]
[tex]|AD|=|BD|\ \text{and}\ |AE|=|EC|\\\\\text{therefore}\ DE\ \text{and}\ BC\ \text{are parallel}.\\\\\text{Corresponding segments are in proportion}.\\\\\dfrac{AD}{AE}=\dfrac{BD}{CE}\qquad/|AE|=|EC|/,\ \text{therefore}\\\\\dfrac{AD}{AE}=\dfrac{BD}{AE}\Rightarrow AD=BD\\\\\text{First equation:}\ 2x+4=5y+2\qquad(1)\\\\\dfrac{AD}{DE}=\dfrac{AB}{BC}\\\\\dfrac{2x+4}{3x-1}=\dfrac{2x+4+5y+2}{x^2-9}\qquad(2)\\\\\text{Substitute (1) to (2)}[/tex]
[tex]\dfrac{5y+2}{3x-1}=\dfrac{5y+2+5y+2}{x^2-9}\\\\\dfrac{5y+2}{3x-1}=\dfrac{2(5y+2)}{x^2-9}\\\\\dfrac{5y+2}{3x-1}=\dfrac{5y+2}{\frac{1}{2}(x^2-9)}\iff3x-1=\dfrac{1}{2}(x^2-9)\qquad\text{multiply both sides by 2}\\\\6x-2=x^2-9\qquad\text{subtract}\ 6x\ \text{from both sides}\\\\-2=x^2-6x-9\qquad\text{add 2 to both sides}[/tex]
[tex]x^2-6x-7=0\\\\x^2-7x+x-7=0\\\\x(x-7)+1(x-7)=0\\\\(x-7)(x+1)=0\iff x-7=0\ \vee\ x+1=0\\\\x-7=0\qquad\text{add 7 to both sides}\\\\x=7\in D\\\\x+1=0\qquad\text{subtract 1 from both sides}\\\\x=-1\notin D[/tex]
[tex]\text{Put}\ x=7\ \text{to (1):}\\\\5y+2=2(7)+4\\\\5y+2=14+4\\\\5y+2=18\qquad\text{subtract 2 from both sides}\\\\5y=16\qquad\text{divide both sides by 5}\\\\y=3.2\in D[/tex]
