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All x's that count for 5/(x-2) < 3/(x-4) ?

Guest Sep 14, 2014

Best Answer 

 #1
avatar+81115 
+13

All x's that count for 5/(x-2) < 3/(x-4) ?

Cross-multiplying, we have

5(x-4) < 3(x-2)     simplify

5x - 20 < 3x - 6     subtract 3x from both sides and add 20 to both sides

2x < 14                 divide by 2  on  both sides

x < 7

But we have to be careful with this answer......x cannot be equal to either 2  or 4  because it would make the denominator(s) in the original problem undefined.....also, when x < 2 or when  4 < x < 7 the original inequality is true, but it is not true when 2< x < 4

So, our answer is

(-∞, 2) U (4, 7)

 

CPhill  Sep 14, 2014
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12+0 Answers

 #1
avatar+81115 
+13
Best Answer

All x's that count for 5/(x-2) < 3/(x-4) ?

Cross-multiplying, we have

5(x-4) < 3(x-2)     simplify

5x - 20 < 3x - 6     subtract 3x from both sides and add 20 to both sides

2x < 14                 divide by 2  on  both sides

x < 7

But we have to be careful with this answer......x cannot be equal to either 2  or 4  because it would make the denominator(s) in the original problem undefined.....also, when x < 2 or when  4 < x < 7 the original inequality is true, but it is not true when 2< x < 4

So, our answer is

(-∞, 2) U (4, 7)

 

CPhill  Sep 14, 2014
 #2
avatar+91510 
+11

$$\begin{array}{rll}
\frac{5}{x-2} &<&\frac{3}{x-4}\\\\
(x-2)^2(x-4)^2\; \dfrac{5}{x-2} &<& (x-2)^2(x-4)^2\;\dfrac{3}{x-4}\\\\\\
5(x-2)(x-4)^2 &<& 3(x-2)^2(x-4)\\\\
5(x-2)(x^2-8x+16) &<& 3(x^2-4x+4)(x-4)\\\\
5(x^3-8x^2+16x-2x^2+16x-32) &<& 3(x^3-4x^2+4x-4x^2+16x-16)\\\\
5(x^3-10x^2+32x-32) &<& 3(x^3-8x^2+20x-16)\\\\
5x^3-50x^2+160x-160 &<& 3x^3-24x^2+60x-48\\\\
2x^3-26x^2+100x-112 &<& 0\\\\
x^3-13x^2+50x-56 &<& 0\\\\
\end{array}$$

 

Look for roots using factor theorum.  Any integer factors must be factors of 56

$$try\;\pm1\qquad \pm1-13\pm50-56$$      No they are going to be put together to get 0

$$try \pm2 \qquad \pm8-52\pm100-56=\pm8\pm100-52-56=\pm108-108$$      Great +2 is a factor.

 

Doing the polynomial division and struggling considerably with latex I get:

$$\begin{tabular}{cccccc}
&&\;x^2&-11&+28&
&&&||&||&||&||&
x&-2&\|\;x^3&-13x^2&+50x&-56\\
&&\;x^3&-2x^2&&&
&&||&||&||&||&
&&&-11x^2&+50x&-56&\\
&&&-11x^2&+22x&&
&&&||&||&||&
&&&&+28x&-56&\\
&&&&+28x&-56&
&&&&||&||&
&&&&&0&
\end{tabular}$$

 

SO Now I have

$$\begin{array}{rll}
x^3-13x^2+50x-56&<&0\\
(x-2)(x^2-11x+28)&<&0\\
(x-2)(x-4)(x-7)&<&0\\
\end{array}$$

 

 

I would have drawn this freehand if I was not doing it for forum reproduction.

 

It can be seen from the graph that  x

 

Otherwise stated as          $$(-\infty,2)\cup(4,7)$$

 

I'm not saying that this was the easiest or the best way but surely you have to be impressed!

 

I HAVE AN EXCELLENT GIFT FOR DOING THINGS THE LONG WAY !  IT IS A GOD GIVEN GIFT!!

 

Melody  Sep 15, 2014
 #3
avatar+81115 
+3

As rosala would say......."You just made my head hurt, Melody."

(LOL).......

 

CPhill  Sep 15, 2014
 #4
avatar+91510 
+3

I didn't have to 'check' my answer!   LOL

Melody  Sep 15, 2014
 #5
avatar+11762 
+3

Melody you seriously are gifted , i would go crazy writing that much and not only writing but reading aswell!And yes my head is really spinning after seeing your anwer!btw even i have got a unique gift!anyways what gift is yours CPhill!

rosala  Sep 15, 2014
 #6
avatar+91510 
+3

What is your gift Rosala?

Surely that is worth your 3 points Chris - even just for head hurting value.

That Latex division took me for ever, I even had to "invent" some of it !    LOL

Melody  Sep 15, 2014
 #7
avatar+11762 
+3

Melosy i'll tell you that some time esle! a long story!

 

melody your an expert in LaTex i wish i could even be!

rosala  Sep 15, 2014
 #8
avatar+81115 
+8

Notice one thing more about this one , if we want to use Melody's algebraic method....taking it from here

5(x-2)(x-4)2 < 3(x-2)2(x-4)

We can do this to simplify things.....

5(x-2)(x-4)2 - 3(x-2)2(x-4) < 0      factoring, we have

(x-2)(x-4)[5(x-4) - 3(x-2)] < 0

(x-2)(x-4)[5x- 20 -3x +6 ] < 0

(x-2)(x-2)[2x -14] < 0

2(x-2)(x-4)(x-7) < 0

(x-2)(x-4)(x-7) < 0

And notice that we don't have to use the Factor Theorem at all....the "roots" (i.e., critical interval values) are right in front of us !!!

 

CPhill  Sep 16, 2014
 #9
avatar+91510 
+3

What a show off!  Anyway,  my gift is to do things the   L  O  N  G     w  a  y    

Melody  Sep 16, 2014
 #10
avatar+81115 
+3

HAHAHAHAHAHAHAHAHAHAHAHAHAHA  !!!!!!!!!!!!!!!!!

(Ain't  I crafty ????????)

 

CPhill  Sep 16, 2014
 #11
avatar+91510 
+3

Yes you are very crafty!

Melody  Sep 18, 2014
 #12
avatar+91510 
+3

Hi Heureka

I was showing off my latex for the algebraic division and I wondered if you could improve upon it?

Melody  Sep 18, 2014

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