Question 775978
{{{(x+4)/(3 x^2-7 x) = (1/x+4/x^2)/(3-7/x)}}}
 
Let's leave the left side as is and only deal with the right side:
{{{(1/x+4/x^2)/(3-7/x)}}}
 
For the numerator and denominator, find a common denominator:
{{{(x/x^2+4/x^2)/(3 x/x-7/x)}}}
 
We can rewrite it as:
{{{((x+4)/x^2)/((3x-7)/x)}}}
 
Division with fraction means taking the reciprocal of the second number and multiplying.
{{{((x+4)/x^2)(x/(3x-7))}}}
 
Simplify.
{{{(x+4)/(x(3x-7))}}}
{{{(x+4)/(3 x^2-7 x))}}}
 
Now we can see that the right side is equal to the left side.