How can I get from step $3$ to step $4$ in the problem?

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I've tried this:

$6x^2 - 9x + 8x -12 + 2\Delta x - 6x^2 -8x +9x + 12 + 3\Delta x$

It doesn't cancel out to $17\Delta x$, though.

I've also tried this:

$(3x+4)(2x-3) - (2x-3)(3x+4) = 0$, then the only things left are $2\Delta x$ and $3\Delta x$. That adds up to $5\Delta x$, which isn't the answer.

How can I solve this problem?



Step 3: $$\frac{(3x+4)[(2x-3)+2\Delta x]-(2x-3)[(3x+4)+3\Delta x])}{(3x+4)(3x+3\Delta x+4)}$$

$$=\frac{(3x+4)(2x-3)+2(3x+4)\Delta x-(2x-3)(3x+4)-3(2x-3)\Delta x)}{(3x+4)(3x+3\Delta x+4)}$$

Now first and third term cancel out and you will get

$$=\frac{2(3x+4)\Delta x-3(2x-3)\Delta x)}{(3x+4)(3x+3\Delta x+4)}$$

$$=\frac{17\Delta x}{(3x+4)(3x+3\Delta x+4)}$$

which is your step 4


I think you could make life easier writing $$f(x)=\frac{2 x-3}{3 x+4}$$ So $$f(y)-f(x)=\frac{2 y-3}{3 y+4}-\frac{2 x-3}{3 x+4}$$ Reduce to same denominator $$f(y)-f(x)=\frac{(2 y-3)(3x+4)-(2x-3)(3y+4)}{(3 y+4)(3 x+4)}$$ Expand the numerator and simplify to get $$f(y)-f(x)=\frac{17 (y-x)}{(3 x+4) (3 y+4)}$$ Now, make $y=x+\Delta x$ to get $$f(x+\Delta x)-f(x)=\frac{17 \Delta x}{(3 x+4) (3 (x+\Delta x)+4)}$$


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