Solve the inequality: $2 < \frac{3x+1}{2x+4}$

Step 1: I simplified $\frac{3x+1}{2x+4}$ into: $3x+1-2x-4= x-3$.
Step 2: $2>x-3$ Here I subtracted $2$ from both sides into: $x>-5$ or $5<x$

However the answer in the book shows $-7>x>-2$. It's odd that $7-2=5$ but oke...

So, I also tried using a sign diagram and with this I did get: $-2 for 2x+4$. I couldn't get $-7$ of of $(3x+1)$. It has probably something to do with the $2$ on the lhs.

My questions are:
- Why is my first and second method wrong?
- How do I know if I should approach it using this method or with an sign diagram? Sometimes the first method I use works, sometimes it doesn't. Please explain why.
- What other methods can I use to solve inequality's.
- Finally, please explain how I can solve this inequality.

  • $\begingroup$ The user who edited it has probably changed it wrong, or it's my type mistake. Let me change it. $\endgroup$ – user160137 Jul 22 '14 at 18:54

Without making lots of conditional statements, the only way to solve this is to move everything to one side, so you are comparing some expression to zero. Then factor the expression as much as possible and determine where each factor is positive, negative, or zero. This includes factoring the denominator.

The usual approach is to determine where factors are zero or undefined, then the sign is constant on each of the complementary subintervals. Combine the signs into a single sign chart, and read your solution from there. Be careful about points where the sign is zero or undefined, as it could make the expression undefined (e.g., division by zero).

Your method isn't clear. I think you intended to "clear fractions" or cross-multiply. That's not valid in general. You can't multiply an inequality by a variable quantity whose sign is unknown, because if the quantity is negative, the inequality symbol must reverse (but it doesn't reverse if the sign is positive). If the quantity is zero, you lose most of your information, and if the inequality is strict you get no solution ($0<0$ has no solution).

  • $\begingroup$ Its starting to make sense :) Thank you! My first approach was totally wrong. I'm gonna reread to understand better. $\endgroup$ – user160137 Jul 22 '14 at 19:07
  • $\begingroup$ You're welcome. With a little practice, you will find this will become ridiculously easy. Then you can teach the method to someone else ;) I will add a simple example to my answer. $\endgroup$ – MPW Jul 22 '14 at 19:10
  • $\begingroup$ Let's hope so :DD I'm trying to redo it. $\endgroup$ – user160137 Jul 22 '14 at 19:21
  • $\begingroup$ Because you made me understand why method 1 was wrong I think I can solve it from now on. BIG thanks! I solved it now. $\endgroup$ – user160137 Jul 22 '14 at 19:31

$$\frac{3x+1}{2x+4} > 2$$ $$\frac{3x+1}{2x+4} - 2 > 0$$ $$-\frac{x+7}{2(x+2)} > 0$$

Now equate the numerator and denominator to zero and draw this numbers on number line.

enter image description here

Since sign of this equation is ">" we must determine intervals on which our function $f(x) = -\frac{x+7}{2(x+2)} $ takes positive values. To do this choose one number of each interval and substitute it to the function. For example, for number "-5": $f(-5) = \frac{1}{3}$ which is positive. For this equation it is one interval: $$ -7<x<-2$$

  • $\begingroup$ MPW made me understand what I did wrong and explained it clearly. That was my biggest question. If I could accept two answers I would. Great detailed explanation you did, thank you! +1 $\endgroup$ – user160137 Jul 22 '14 at 19:32

$$ \dfrac{3x + 1}{2x + 4} > 2 \quad \Rightarrow \quad \dfrac{3x + 1}{2x + 4} - 2 > 0 \quad \Rightarrow \quad \dfrac{-x - 7}{2x + 4} > 0 \quad \Rightarrow \quad \dfrac{x + 7}{2x + 4} < 0 $$ The other details are in the picture below: enter image description here

Thus, $S = \{x \in \mathbb{R} : -7 < x < -2\}$

  • $\begingroup$ Thus, S={x∈R:−7<x<−2} I haven't had most of these sign yet. When I get there I'll get back to if I understand it :) Very clear diagram! $\endgroup$ – user160137 Jul 22 '14 at 19:41
  • $\begingroup$ Can you explain the blue line? I thought between -7 and -2 was the answer. Should the blue plus and minus signs not be reversed? $\endgroup$ – user160137 Jul 22 '14 at 22:50
  • 1
    $\begingroup$ Since the inequality is made by dividing two linear terms, then the blue line is obtained by dividing the signals from the two red and green lines in each sector. $\endgroup$ – Mathsource Jul 22 '14 at 22:56
  • $\begingroup$ So [+ and + = +] and [- plus - = +]? You're saying that red line has plus in the middle, green has minus in the middle. Minus + plus= minus. Is that why blue is minus? $\endgroup$ – user160137 Jul 22 '14 at 23:05
  • $\begingroup$ This is exactly the procedure. $\endgroup$ – Mathsource Jul 22 '14 at 23:10

You have to multiply both sides with 2x+4.

Here you have to distinguish two cases:

Case 1: If $2x+4 > 0$, then the (un)equality sign doesen´t change.

Case 2: If $2x+4 < 0$, then the (un)equality sign changes from < to >.

Solve both equations seperately.

  • $\begingroup$ This is incorrect. You must exclude the case $2x+4=0$ in your case 1. $\endgroup$ – MPW Jul 22 '14 at 19:08
  • $\begingroup$ @MPW I have done it. Thanks for the hint. $\endgroup$ – callculus Jul 22 '14 at 19:11

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