hey guys,
(3x+5)/(2x-1) (greater than or equal to) 8
any ideas? thanks guys
What is (1, negative infinity) meant to mean?
And by (1/2, positive infinity) do you mean 1/2 < x < +oo ?
Have you taken some random values of x from your solution and checked to see if they satisfy the original inequality? I have. I took x = 2 from (1/2, positive infinity) (assuming it means what I think it means - see above) and it fails the test. So your solution cannot be correct.
If you show all of your working the mistakes you've made in getting this solution can be pointed out.
By the way - don't bump.
whats bump?
ok this is what i did...
$\displaystyle
\frac{3x+5}{2x-1}\geq 8
$
$\displaystyle
\frac{3x+5}{2x-1}-8\geq 0
$
$\displaystyle
\frac{3x+5}{2x-1}-\frac{8(2x-1)}{2x-1}\geq 0
$
$\displaystyle
\frac{3x+5 - 8(2x-1)}{2x-1}\geq 0
$
$\displaystyle
\frac{-13x+13}{2x-1}\geq 0
$
then i got -13(1) + 13 = 0 so 1 and every other negative number below 1 for x will satisfy its equal to or greater than 0.
and for 2x-1, i got 2(1/2)-1 = 0 so 1/2 and every other positive number above 1/2 will satisfy its equal to or greater than 0.
Case 1: $\displaystyle -13 x + 13 \geq 0$ and $\displaystyle 2x - 1 > 0$, that is, $\displaystyle x \leq 1$ and $\displaystyle x > \frac{1}{2}$, that is, $\displaystyle \frac{1}{2} < x \leq 1$.
Case 2: $\displaystyle -13 x + 13 \leq 0$ and $\displaystyle 2x - 1 < 0$, that is, $\displaystyle x \geq 1$ and $\displaystyle x < \frac{1}{2}$, which has no solution.
So the solution is $\displaystyle \frac{1}{2} < x \leq 1$.
By the way: http://en.wikipedia.org/wiki/Bump_(Internet)