+128408 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)
+128408 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)
+118608 $$\begin{array}{rll}
\frac{5}{x-2} &<&\frac{3}{x-4}\\\\
(x-2)^2(x-4)^2\; \dfrac{5}{x-2} &<& (x-2)^2(x-4)^2\;\dfrac{3}{x-4}\\\\\\
5(x-2)(x-4)^2 &<& 3(x-2)^2(x-4)\\\\
5(x-2)(x^2-8x+16) &<& 3(x^2-4x+4)(x-4)\\\\
5(x^3-8x^2+16x-2x^2+16x-32) &<& 3(x^3-4x^2+4x-4x^2+16x-16)\\\\
5(x^3-10x^2+32x-32) &<& 3(x^3-8x^2+20x-16)\\\\
5x^3-50x^2+160x-160 &<& 3x^3-24x^2+60x-48\\\\
2x^3-26x^2+100x-112 &<& 0\\\\
x^3-13x^2+50x-56 &<& 0\\\\
\end{array}$$
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!!
+128408 As rosala would say......."You just made my head hurt, Melody."
(LOL).......
+11912 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!
+118608 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
+11912 Melosy i'll tell you that some time esle! a long story!
melody your an expert in LaTex i wish i could even be!
+128408 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 !!!
