inequalities help!!

**jvignacio** · October 4th 2008, 11:48 PM

hey guys,

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

any ideas? thanks guys

**mr fantastic** · October 4th 2008, 11:56 PM

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.

**jvignacio** · October 6th 2008, 01:30 AM

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) ???

**mr fantastic** · October 6th 2008, 02:20 AM

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.

**jvignacio** · October 6th 2008, 04:10 AM

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.

**mr fantastic** · October 6th 2008, 04:46 AM

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)

**jvignacio** · October 6th 2008, 05:34 AM

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

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