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September 22nd, 2019, 01:01 PM   #1
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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
(x-1)/(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 :

[(x-1)(2x+3)]/(2x+3) < (1)(2x+3)

(x-1) < (2x+3)

Then we solve for x:

x-2x < 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.
PaoAndreCM is offline  
September 22nd, 2019, 01:24 PM   #2
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Multiplying by (2x + 3)² (so that the multiplier is non-negative)
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, +∞).
skipjack is offline  
September 22nd, 2019, 01:32 PM   #3
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Math Focus: Area of Circle
(x-1)/(2x+3) < 1

1. Assume $x>-3/2$
$x-1<2x+3 \Rightarrow x>-4$ therefore x>-3/2

2. Assume x<-3/2
$x-1>2x+3 \Rightarrow x<-4$ therefore x<-4
tahirimanov19 is offline  
September 22nd, 2019, 01:41 PM   #4
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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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