# How do I get $-2$ as a solution to $\frac{3x+1}{x+2}<2$?

I'm stuck at this seemingly simple problem again.

Solve the difference $$\frac{3x+1}{x+2}<2$$

I try to solve this in the intuitive way:

$$\frac{3x+1}{x+2}<2$$ $$=>3x+1<2(x+2)$$ $$=>3x+1<2x+4$$ $$=>x<3$$

I then read that the solution is $$-2

Where in the world did they get that $$-2$$ from?

• Make it homogeneous by moving $2$ to the left and simplifying to get: $\frac{x-3}{x+2}<0$. Now you see the zeros $-2$ and $3$. Jan 23 '19 at 18:17

In this case you can't multiply both sides by $$(x+2)$$ as it might be negative for certain values of $$x$$ and that would require reversing the inequality sign. Try multiplying both sides by $$(x+2)^2$$ instead and see if the resulting quadratic leads anywhere.
When you multiply by something negative, the inequality sign flips. So when you go from $$\frac{3x+1}{x+2}<2$$ to $$3x+1<2(x+2)$$ without changing the inequality sign, you're implicitly assuming that $$x+2>0$$. For $$x+2<0$$, we instead get $$3x+1>2(x+2)$$
You should notice that you could only multiply both sides by $$x+2$$ and preserve the inequality sign as $$3x+1<2(x+2)$$ if and only if $$x+2>0$$. Otherwise, the inequality sign will change since you multiply both sides by a negative number