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September 22nd, 2019, 01:01 PM  #1 
Newbie Joined: Sep 2019 From: Bogota Posts: 1 Thanks: 0  What's the fallacy here in the way I'm solving this inequation?
Hi everyone! I'm learning inequalities. I do exercises and then use photomath app (fantastic app!) to verify my solutions. There's one inequality the app gives me a different answer to the one I'm getting at, but I can't see the mistake with the way I'm solving my inequality. Can anyone help me understand, please? Here's the inequality (x1)/(2x+3) < 1 And here's my thinking: The best way to get rid of the denominator (2x+3) would be to multiply by it on both sides of the inequality. So : [(x1)(2x+3)]/(2x+3) < (1)(2x+3) (x1) < (2x+3) Then we solve for x: x2x < 3 + 1 x < 4 x > 4 So the solution, following my line of thought, would be x ∈ (4,+∞) (which doesn't satisfy the inequality) However, photomath app's solution is: x ∈ (∞, 4) ∪ (3/2, +∞) (which sastisfies the inequality). When I go to "show solving steps", instead of multiplying by (2x+3), they subtract 1 from both sides of the inequality. I understand why and I can get their solution if I do that first too. However, I still can't understand why multiplying by (2x+3) to get rid of the fraction is wrong, and why it gives out a different solution. What's the fallacy? What am I not factoring in? I thank you all! Last edited by skipjack; September 22nd, 2019 at 01:10 PM. 
September 22nd, 2019, 01:24 PM  #2 
Global Moderator Joined: Dec 2006 Posts: 21,035 Thanks: 2271 
Multiplying by (2x + 3)² (so that the multiplier is nonnegative) gives (x  1)(2x + 3) < (2x + 3)², which simplifies to 0 < (x + 4)(2x + 3). Hence x < 4 or x > 3/2, i.e. x ∈ (∞, 4) ∪ (3/2, +∞). 
September 22nd, 2019, 01:32 PM  #3 
Senior Member Joined: Mar 2015 From: Universe 2.71828i3.14159 Posts: 132 Thanks: 49 Math Focus: Area of Circle 
(x1)/(2x+3) < 1 1. Assume $x>3/2$ $x1<2x+3 \Rightarrow x>4$ therefore x>3/2 2. Assume x<3/2 $x1>2x+3 \Rightarrow x<4$ therefore x<4 
September 22nd, 2019, 01:41 PM  #4 
Senior Member Joined: Jun 2019 From: USA Posts: 310 Thanks: 162 
To explain it more simply, if you multiply (or divide) an inequality by a negative number, then the inequality sign changes direction. $x \leq 4 \rightarrow x \geq 4$ Your method works iff (2x+3) is positive. If it is negative, then you have to flip the inequality. This is why the solution consists of two distinct regions. 

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