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Math Help - Solve the Inequality

  1. #1
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    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
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  2. #2
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    Re: Solve the Inequality

    have you made any attempts? Do you have any work to show us?
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  3. #3
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    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).
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    Re: Solve the Inequality

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

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

    Rethink your answer.
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  5. #5
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    Re: Solve the Inequality

    Quote Originally Posted by joshuaa View Post
    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 $\left(-\infty, -\frac{5}{2}\right] \cup \left[1, \infty\right)$
    Last edited by romsek; February 27th 2014 at 11:50 AM.
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  6. #6
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    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.
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  7. #7
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    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.
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  8. #8
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    Re: Solve the Inequality

    Quote Originally Posted by joshuaa View Post
    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.
    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 $\dfrac{+}{+}~\text{ or }~\dfrac{-}{-}$
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  9. #9
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    Re: Solve the Inequality

    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?
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  10. #10
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    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.
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  11. #11
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    Re: Solve the Inequality

    Thank you a Lot. It is clear now, but I need some practice to master this subject and eliminate confusion.
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  12. #12
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    Re: Solve the Inequality

    Quote Originally Posted by joshuaa View Post
    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?
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  13. #13
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    Re: Solve the Inequality

    No Union.
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  14. #14
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    Re: Solve the Inequality

    I am sorry. Maybe, (-infinite, -5) U (4, +infinite).
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  15. #15
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    Re: Solve the Inequality

    Quote Originally Posted by joshuaa View Post
    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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