The answer is [tex] \frac{5x-12}{3x-12} [/tex]
The expression is: [tex] \frac{ \frac{1}{x-3}+ \frac{4}{x} }{ \frac{4}{x} - \frac{1}{x-3} } [/tex]
Let's factorize it:
[tex] \frac{ \frac{1}{x-3}+ \frac{4}{x} }{ \frac{4}{x} - \frac{1}{x-3} } = \frac{ \frac{x*1}{x(x-3)}+ \frac{4*(x-3)}{(x-3)*x}}{ \frac{(x-3)*4}{(x-3)*x}- \frac{x*1}{x*(x-3)}}= \frac{ \frac{x+4(x-3)}{x(x-3)} }{ \frac{4(x-3)-x}{x(x-3)} } [/tex]
Since [tex] \frac{ \frac{a}{b} }{ \frac{c}{d} }= \frac{a}{b}/ \frac{c}{d}= \frac{a*d}{b*c} [/tex], then:
[tex]\frac{ \frac{x+4(x-3)}{x(x-3)} }{ \frac{4(x-3)-x}{x(x-3)} }= \frac{(x+4(x-3))*x(x-3)}{(4(x-3)-x)*x(x-3)} [/tex]
Cancel x(x-3) and simplify:
[tex] \frac{(x+4(x-3))*x(x-3)}{(4(x-3)-x)*x(x-3)} = \frac{x+4(x-3)}{4(x-3)-x} = \frac{x+4x-12}{4x-12 -x} = \frac{5x-12}{3x-12} [/tex]
