1. ## Solve the Inequality

Solve the inequality and express the solution in terms of intervals whenever possible.

(a) [ (x + 1) / (2x - 3) ] > 2

(b) x(2x + 3) >= 5

2. ## Re: Solve the Inequality

have you made any attempts? Do you have any work to show us?

3. ## Re: Solve the Inequality

When I solve for x in (a), I get x > 7/3. I think that the interval is (7/3, +infinite).

When I solve for x in (b), I get x >= 1, x >= -5/2. I cannot figure out how to write the interval. Maybe, it is, [1, +infinite), [-5/2, +infinite).

4. ## Re: Solve the Inequality

Originally Posted by joshuaa
Solve the inequality and express the solution in terms of intervals whenever possible.

(a) [ (x + 1) / (2x - 3) ] > 2
$\dfrac{x+1}{2x-3}>2\\\dfrac{x+1}{2x-3}-2>0\\\dfrac{-3x+7}{2x-3}>0$

5. ## Re: Solve the Inequality

Originally Posted by joshuaa
When I solve for x in (a), I get x > 7/3. I think that the interval is (7/3, +infinite).

When I solve for x in (b), I get x >= 1, x >= -5/2. I cannot figure out how to write the interval. Maybe, it is, [1, +infinite), [-5/2, +infinite).
I'm seeing a different answer so let's take a look.

$\frac{x+1}{2x-3}>2$

if $(2x-3)>0 \Rightarrow x>\frac{3}{2}$ this becomes

$x+1>4x-6$

$7>3x$

$\frac{7}{3}>x$

so this yields $\frac{3}{2} < x < \frac{7}{3}$

if $2x-3<0 \Rightarrow x<\frac{3}{2}$

then you have to flip the inequality sign since you are multiplying by a negative number.

$x+1<4x-6$

$7 < 3x$

$\frac{7}{3} < x$

clearly x can't be both less than $\frac{3}{2}$ and greater than $\frac{7}{3}$ so there are no solutions for $x < \frac{3}{2}$

thus the final answer for (a) is $\frac{3}{2} < x < \frac{7}{3}$ or $\left(\frac{3}{2}, \frac{7}{3}\right)$

for (b)

$x(2x+3) \geq 5$

$2x^2+3x-5 \geq 0$

$(2x+5)(x-1) \geq 0$

so these factors are either both postive or both negative or 0

$2x+5 \geq 0 \Rightarrow x \geq -\frac{5}{2}$

$x-1 \geq 0 \Rightarrow x \geq 1$

these two result in $x \geq 1$

$2x+5 \leq 0 \Rightarrow x \leq -\frac{5}{2}$

$x-1 \leq 0 \Rightarrow x \leq 1$

these two result in $x \leq -\frac{5}{2}$

so the intervals the inequality is satisfied on are

$x \leq -\frac{5}{2}$ and $1 \leq x$ or $\left(-\infty, -\frac{5}{2}\right] \cup \left[1, \infty\right)$

6. ## Re: Solve the Inequality

I am a little confused. In (a), how did you get, (2x - 3) > 0?

When you get the factors of x, why do you solve the inequality twice? Like this, (2x - 3) > 0, (2x - 3) < 0.

7. ## Re: Solve the Inequality

when you multiply the sides of an inequality by a negative number you must switch which way the inequality points. So you have to consider both cases.

In this case we multiplied by (2x-3) so we had to deal with the case when this is positive vs. when it is negative separately.

8. ## Re: Solve the Inequality

Originally Posted by joshuaa
I am a little confused. In (a), how did you get, (2x - 3) > 0?
When you get the factors of x, why do you solve the inequality twice? Like this, (2x - 3) > 0, (2x - 3) < 0.
$\dfrac{x+1}{2x-3}>2\\\dfrac{x+1}{2x-3}-2>0\\\dfrac{-3x+7}{2x-3}>0$

I always strongly discourage students from "multiplying" sides.
I encourage that they work by comparing all to zero, as I did above.

In the last line you need $\dfrac{+}{+}~\text{ or }~\dfrac{-}{-}$

9. ## Re: Solve the Inequality

Plato,

Usually, if we want to factor, we get, for example, (x + 5) (x - 4) > 0. Then I say, x > -5, x > 4.

If I have the same factors, but with division, like this, (x + 5) / (x - 4) > 0, can I say, x > -5, x > 4?

Also, if this is true, why I did not directly say in example (a), (x + 1) > 2, (2x - 3) > 2?

romsek

Now, I got it why we did the inequality twice, but confused about, how directly, you got that (2x - 3) > 0?

10. ## Re: Solve the Inequality

It has to be either positive or negative since division by 0 isn't defined. I tried both. (2x-3) < 0 lead to a solution that couldn't exist thus it must be that (2x-3) > 0.

11. ## Re: Solve the Inequality

Thank you a Lot. It is clear now, but I need some practice to master this subject and eliminate confusion.

12. ## Re: Solve the Inequality

Originally Posted by joshuaa
Usually, if we want to factor, we get, for example, (x + 5) (x - 4) > 0. Then I say, x > -5, x > 4.
I don't think that you understand the logic of problem solving.
In the case $(x+5)(x-4)>0$ so we need $(+)(+)\text{ OR }(-)(-)$.

If $(x+5)>0\text{ AND }(x-4)>0$
$x>-5\text{ AND }x>4$
$\text{so }x>4$.

If $(x+5)<0\text{ AND }(x-4)<0$
$x<-5\text{ AND }x<4$
$\text{so }x<-5$.

What is the union of those two?

No Union.

14. ## Re: Solve the Inequality

I am sorry. Maybe, (-infinite, -5) U (4, +infinite).

15. ## Re: Solve the Inequality

Originally Posted by joshuaa
Thank you a Lot. It is clear now, but I need some practice to master this subject and eliminate confusion.
I'd pay attention to what Plato is saying. If he doesn't like multiplying out in problems like this there's certainly a good reason for it.

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