#1**+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

#1**+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**+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

#5**+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**+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**+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**+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