have you made any attempts? Do you have any work to show us?
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 $\left(-\infty, -\frac{5}{2}\right] \cup \left[1, \infty\right)$
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.
Did you read my reply?
$\dfrac{x+1}{2x-3}>2\\\dfrac{x+1}{2x-3}-2>0\\\dfrac{-3x+7}{2x-3}>0$
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 $\dfrac{+}{+}~\text{ or }~\dfrac{-}{-}$
Plato,
I have seen your 1st reply, and I am confused.
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?
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?