#(x+4)/(x-3)=(x+2)/(x-4)#
#"please note that : "a/b=c/d" than "a*d=b*c#
#"we can write the equation above as "#
#(x+4)(x-4)=(x-3)(x+2)#
#"Since "(x-y)(x+y)=x^2-y^2," we can write as the fallowing "#
#(x+4)(x-4)=x^2-4^2#
#x^2-4^2=(x-3)(x+2)#
#x^2-16=x^2+2x-3x-6#
#x^2-16=x^2-x-6#
#cancel(x^2)-16-cancel(x^2)+x+6=0#
#x-16+6=0#
#x-10=0#
#x=10#