Question 775978

I need to prove that

((x+4)/(3x^2-7x))=(((1/x)+(4/x^2))/(3-(7/x)))

By showing the steps,

So far I have tried working from the right side of the equation to get to the left, and made it into a complex fraction and combined it, then multiplied the numerator by the denominators reciprocal in order to make it a normal fraction and ended up with ((1-x)(1-4)/(3-7)(x^2-x))


Simplifying the right-side: {{{(1/x + 4/x^2)/(3-(7/x))}}}, we get:


{{{(1/x + 4/x^2)}}} ÷ {{{(3 - 7/x)}}}


{{{(1/x + 4/x^2)}}} ÷ {{{(3/1 - 7/x)}}}


{{{(x + 4)/x^2}}} ÷ {{{(3x - 7)/x}}} ----- Multiplying 1st expression by LCD, {{{x^2}}}, and 2nd, by LCD, x


{{{(x + 4)/x^2}}} * {{{x/(3x - 7)}}} ------ Changing  ÷ to * and inverting DIVISOR


{{{(x + 4)/cross(x^2)x}}} * {{{cross(x)/(3x - 7)}}} ------- {{{(x + 4)/(x(3x - 7))}}} ------- {{{highlight_green((x + 4)/(3x^2 - 7x))}}}