I have a question when multiplying fractions in this case (assuming): $$x>0$$
1.$$\frac{x+4}{3x+2}>\frac{1}{x}$$ $$\frac{x+4}{3x+2}-\frac{1}{x}>0$$ $$\frac{x^2+x-2}{x(3x+2)}>0$$ 2.$$\frac{x+4}{3x+2}>\frac{1}{x}$$ $$x^2+4x>3x+2$$ $$x^2+x-2>0$$
Why can't I just simplify as in case 2.?Because if we had = instead of >, there would be no discussion what to do. Obviously the right solution is 1.solving example. Why aren't both cases equivalent?
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Because $x>0$ we also have $x(3x+2)>0$ and hence $$\tfrac{x^2+x-2}{x(3x+2)}>0 \qquad\text{ if and only if }\qquad x^2+x-2>0.$$ So the two cases are equivalent, and both are correct.
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Actually, approach 2. is OK, as long as you take care to keep in mind the assumption $x\gt0$ (which implies $2x+3\gt0$ as well). The difference between the two approaches is easy to see if you write out explicitly the correct logical statements that are actually being made. Here's approach 1.:
$${x+4\over3x+2}\gt{1\over x}\iff {x+4\over3x+2}-{1\over x}\gt0 \iff{x^2+x-2\over x(3x+2)}\gt0$$
And here's approach 2.:
$$\left(x\gt0\land{x+4\over3x+2}\gt{1\over x} \right)\iff\left(x\gt0\land x^2+4x\gt3x+2 \right)\iff\left(x\gt0\land x^2+x-2\gt0 \right)$$
Hint: Write $$\frac{x+4}{3x+2}-\frac{1}{x}=\frac{x^2+x-2}{x(3x+2)}$$ and you have to solve $$\frac{(x-1)(x+2)}{x(3x+2)}>0$$ now you must ditinguish several cases: 1) $$x>1$$ 2)$$-\frac{2}{3}<x<0$$ 3)$$x<-2$$