Question

Determine the x and y please

Answer 1

x = 7 and y = 3.2

Explanation:

`\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>(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}`

`|AD|=|BD|\ \text{and}\ |AE|=|EC|\\\\\text{therefore}\ DE\ \text{and}\ BC\ \text{are parallel}.\\\\\text{Corresponding segments are in proportion}.\\\\(AD)/(AE)=(BD)/(CE)\qquad/|AE|=|EC|/,\ \text{therefore}\\\\(AD)/(AE)=(BD)/(AE)\Rightarrow AD=BD\\\\\text{First equation:}\ 2x+4=5y+2\qquad(1)\\\\(AD)/(DE)=(AB)/(BC)\\\\(2x+4)/(3x-1)=(2x+4+5y+2)/(x^2-9)\qquad(2)\\\\\text{Substitute (1) to (2)}`

`(5y+2)/(3x-1)=(5y+2+5y+2)/(x^2-9)\\\\(5y+2)/(3x-1)=(2(5y+2))/(x^2-9)\\\\(5y+2)/(3x-1)=(5y+2)/((1)/(2)(x^2-9))\iff3x-1=(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}`

`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\\otin D`

`\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`