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August 16th, 2017, 10:16 AM   #1
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Inequations

Hi, I'm having a really bad time with inequations; can someone help me with this one step by step? Thanks.

x+2>5x/2-(8x-5)/4

Last edited by skipjack; August 16th, 2017 at 11:53 AM.
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August 16th, 2017, 10:34 AM   #2
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$\displaystyle x+2>\frac {5x}{2}-\frac{8x-5}{4}$
$\displaystyle x+2>\frac {10x}{4}-\frac{8x-5}{4}$
$\displaystyle x+2>\frac{10x-8x+5}{4}$
$\displaystyle (x+2)×4>2x+5$
$\displaystyle 4x+8>2x+5$
$\displaystyle 4x-2x>5-8$
$\displaystyle 2x > -3$
$\displaystyle x>-\frac {3}{2}$
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August 16th, 2017, 10:39 AM   #3
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Quote:
x+2>5x/2-(8x-5)/4
$x+2 > \dfrac{5x}{2} - \dfrac{8x-5}{4}$

$0 > -\dfrac{x}{2} - \dfrac{3}{4}$

$0 < \dfrac{2x+3}{4} \implies 2x+3 > 0 \implies x > -\dfrac{3}{2}$
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August 16th, 2017, 10:58 AM   #4
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Thanks so much, this really help me understand, just one thing, why does the -5 turns into +5?
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August 16th, 2017, 11:28 AM   #5
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Thanks so much, this really help me understand, just one thing, why does the -5 turns into +5?
$- \dfrac{8x-5}{4} = \dfrac{-(8x-5)}{4} = \dfrac{-8x+5}{4} = -\dfrac{8x}{4} + \dfrac{5}{4}$
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August 16th, 2017, 11:52 AM   #6
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Putting all the terms on the left-hand side gives x + 2 - 5x/2 + (8x - 5)/4 > 0,
i.e. x + 2 - 5x/2 + 2x - 5/4 > 0.
Hence x/2 + 3/4 > 0, which implies x > -3/2.
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August 17th, 2017, 03:20 AM   #7
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Remember that if you multiply by a negative number, you need to swap the sign of the inequality to keep the inequality consistent.

For example:

$\displaystyle 5 > 3$

Multiply both sides by -1:

$\displaystyle -5 < -3$

you need to swap the > to a < because -5 is actually less than -3.
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August 17th, 2017, 06:02 AM   #8
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Another way to do it: first solve the related equation:

I would start by multiplying both sides by 4:
$\displaystyle 4x+ 8= 10x- 8x+ 5= 2x+ 5$

Subtract 8 and 2x from both sides:
$\displaystyle 2x= -3$

So $\displaystyle x= -3/2$ is where the two sides are equal. The point is that this x separates "<" from ">". Now we just need to check the value for one x on each side.

x= -2< -3/2. With x= -2, x+ 2= -2+ 2= 0 and $\displaystyle \frac{5x}{2}- \frac{8x- 5}{4}= -5+ \frac{21}{4}= \frac{-20+ 21}{4}= \frac{1}{4} which is greater than 0.

x= 0> -3/2. With x= 0, x+ 2= 2 and $\displaystyle \frac{5x}{2}- \frac{8x- 5}{4}= 0- \frac{5}{4}= -\frac{5}{4}$ which is less than 2$

$\displaystyle x+ 2> \frac{5x}{2}- \frac{8x- 5}{4}$ for all x> -3/2.
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