Question

How do I get the term (4/(x-1)) <= x+2 to a single rational expression?

Answer 1

I bring it all over the the left side, to get \[(4/(x-1)) -x - 2 <= 0\], then I multiple the second part by (x-1)/(x-1) to get (x-2)(x-1)/(x-1)...then??

Answer 2

instead of multiplying by \(\frac{x-1}{x-1}\)
multiply by \(\frac{x+1}{x+1}\)

Answer 3

\[4/(x-1) - (x ^{2} -x - 2)/(x+1) \]?

Answer 4

\[\frac 4{x-1} \leq x+2\]\[\frac 4{x-1} - x+2\leq0\]
\[\left(\frac 4{x-1}\right)\times\frac{x+1}{x+1} - (x+2)\times\frac{x+1}{x+1}\leq0\]
\[\frac {4({x+1})}{(x-1)\times(x+1)}- \frac{(x+2)(x+1)}{x+1} \leq0\]
\[\frac {4({x+1})}{(x-1)(x+1)}- \frac{(x+2)(x+1)(x-1)}{(x+1)(x-1)} \leq0\]
\[\frac {4({x+1})-(x+2)(x+1)(x-1)}{(x+1)(x-1)} \leq0\]

Answer 5

I think you have taken it seriously @UnkleRhaukus

Answer 6

Thanks so much for the amazingly clear and helpful answer. You're my hero.

Answer 7

How did you know what to multiply everything by though?

Answer 8

when we are trying to get rid of a term
like this in the denominator like \((x+a)\),

multiply by its conjugate \((x-a)\)

Answer 9

well ,we really multiply by \(\frac{x-a}{x-a}=1\)

Answer 10

\[\frac 4{x-1} \leq x+2 \implies \frac{4}{x - 1} -x - 2 \le 0 \implies \frac{4 - (x+2)(x - 1)}{(x-1)} \le 0\]

\[\frac{4 - (x^2 + x - 2)}{(x-1)} \le 0 \implies \frac{x^2 + x + 2}{(x-1)} \ge 0\]

Answer 11

Ha ha ha..

Done something wrong..

Answer 12

how can we get a rational expression?

Answer 13

Meaning ??

Answer 14

my result wasent rational, and neither is yours @waterineyes

Answer 15

WolframAlpha says that it can be reduced to \[(x-2)(x+3)/(x-1)\] (How do you insert fractions?)

Answer 16

Sorry I have done there a mistake:

\[4 - x^2 - x + 2 \implies -x^2 -x + 6 \implies -(x^2 + x - 6) \implies -(x+3)(x-2)\]

Answer 17

So that will become:

\[\frac{-(x+3)(x-2)}{(x-1)} \le 0 \implies \frac{(x+3)(x-2)}{(x-1)} \ge 0\]

Answer 18

OH! Haha you make it look easy. Back to the books for me! :)