# inequalities help!!

• Oct 4th 2008, 11:48 PM
jvignacio
inequalities help!!
hey guys,

(3x+5)/(2x-1) (greater than or equal to) 8

any ideas? thanks guys
• Oct 4th 2008, 11:56 PM
mr fantastic
Quote:

Originally Posted by jvignacio
hey guys,

(3x+5)/(2x-1) (greater than or equal to) 8

any ideas? thanks guys

Note that $\frac{3x+5}{2x-1} = \frac{13/2}{2x-1} + \frac{3}{2} = \frac{1}{2} \left( \frac{13}{2x-1} + 3\right)$.

Substitute this expression and solve the inequality for x.
• Oct 6th 2008, 01:30 AM
jvignacio
Quote:

Originally Posted by mr fantastic
Note that $\frac{3x+5}{2x-1} = \frac{13/2}{2x-1} + \frac{3}{2} = \frac{1}{2} \left( \frac{13}{2x-1} + 3\right)$.

Substitute this expression and solve the inequality for x.

ok i have just solved it but not sure if its correct. is the answer

all x's in the intervals (1, negative infinity) and (1/2, positive infinity) ???
• Oct 6th 2008, 02:20 AM
mr fantastic
Quote:

Originally Posted by jvignacio
ok i have just solved it but not sure if its correct. is the answer

all x's in the intervals (1, negative infinity) and (1/2, positive infinity) ???

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.
• Oct 6th 2008, 04:10 AM
jvignacio
Quote:

Originally Posted by mr fantastic
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 = 1 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...

$
\frac{3x+5}{2x-1}\geq 8
$

$
\frac{3x+5}{2x-1}-8\geq 0
$

$
\frac{3x+5}{2x-1}-\frac{8(2x-1)}{2x-1}\geq 0
$

$
\frac{3x+5 - 8(2x-1)}{2x-1}\geq 0
$

$
\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.
• Oct 6th 2008, 04:46 AM
mr fantastic
Quote:

Originally Posted by jvignacio
whats bump?

ok this is what i did...

$
\frac{3x+5}{2x-1}\geq 8
$

$
\frac{3x+5}{2x-1}-8\geq 0
$

$
\frac{3x+5}{2x-1}-\frac{8(2x-1)}{2x-1}\geq 0
$

$
\frac{3x+5 - 8(2x-1)}{2x-1}\geq 0
$

$
\frac{-13x+13}{2x-1}\geq 0
$

[snip]

Case 1: $-13 x + 13 \geq 0$ and $2x - 1 > 0$, that is, $x \leq 1$ and $x > \frac{1}{2}$, that is, $\frac{1}{2} < x \leq 1$.

Case 2: $-13 x + 13 \leq 0$ and $2x - 1 < 0$, that is, $x \geq 1$ and $x < \frac{1}{2}$, which has no solution.

So the solution is $\frac{1}{2} < x \leq 1$.

By the way: http://en.wikipedia.org/wiki/Bump_(Internet)
• Oct 6th 2008, 05:34 AM
jvignacio
Quote:

Originally Posted by mr fantastic
Case 1: $-13 x + 13 \geq 0$ and $2x - 1 > 0$, that is, $x \leq 1$ and $x > \frac{1}{2}$, that is, $\frac{1}{2} < x \leq 1$.

Case 2: $-13 x + 13 \leq 0$ and $2x - 1 < 0$, that is, $x \geq 1$ and $x < \frac{1}{2}$, which has no solution.

So the solution is $\frac{1}{2} < x \leq 1$.

By the way: Bump (Internet) - Wikipedia, the free encyclopedia

i get it! thank u