Answer:
[tex]\large\boxed{x=\dfrac{1}{2}}[/tex]
Step-by-step explanation:
[tex]\dfrac{x+3}{x-4}=\dfrac{x-5}{x+4}\\\\\text{Domain:}\ x-4\neq0\ \wedge\ x+4\neq0\\\\x\neq4\ \wedge\ x\neq-4\\\\D:x\in\mathbb{R}-\{-4,\ 4\}[/tex]
[tex]\dfrac{x+3}{x-4}=\dfrac{x-5}{x+4}\qquad\text{cross multiply}\\\\(x+3)(x+4)=(x-4)(x-5)\qquad\text{use FOIL}\ (a+b)(c+d)=ac+ad+bc+bd\\\\(x)(x)+(x)(4)+(3)(x)+(3)(4)=(x)(x)+(x)(-5)+(-4)(x)+(-4)(-5)\\\\x^2+4x+3x+12=x^2-5x-4x+20\qquad\text{combine like terms}\\\\x^2+(4x+3x)+12=x^2+(-5x-4x)+20\\\\x^2+7x+12=x^2-9x+20\qquad\text{subtract}\ x^2\ \text{from both sides}\\\\7x+12=-9x+20\qquad\text{subtract 12 from both sides}\\\\7x=-9x+8\qquad\text{add}\ 9x\ \text{to both sides}\\\\16x=8\qquad\text{divide both sides by 16}\\\\x=\dfrac{8}{16}\\\\x=\dfrac{8:8}{16:8}\\\\x=\dfrac{1}{2}\in D[/tex]
