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Thread: Half right, missed the other half...

  1. #1
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    Half right, missed the other half...

    "Find all values of $\displaystyle r$ such that the slope of the line through the points $\displaystyle (r, 4)$ and $\displaystyle (1, 3 - 2r)$ is less than 5."

    I solved it by first finding the slope $\displaystyle \frac {2r + 1}{r - 1} = 5$ which simplifies to $\displaystyle r = 2$. Then I used the point-slope form to derive the equation $\displaystyle y = 5x + 16$. Then, I got lazy, and plugged in a couple of test numbers into the equation, found that numbers above 2 seemed to qualify, and decided that the answer was the set $\displaystyle (2, \infty)$. Unfortunately, the answer should have included also the set $\displaystyle (-\infty, 1)$.

    Algebraically, what should I have done to get to that answer? What part did I skip?
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  2. #2
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    Hello, earachefl!

    Find all values of $\displaystyle r$ such that the slope of the line
    through the points $\displaystyle A(r,\,4)$ and $\displaystyle B(1,\,3\!-\!2r)$ is less than 5.
    You should have solved the inequality . . . tricky, but safe.


    You found the slope of $\displaystyle AB$ . . . good!

    . . Then we have: .$\displaystyle \frac{2r + 1}{r - 1} \:<\:5$ . [1]


    I'd like to multiply both sides by $\displaystyle (r-1)$
    . . but the result depends on whether $\displaystyle (r-1)$ is positive or negative.

    Recall: when multiplying or dividing an inequality by a negative quantity,
    . . . . . .reverse the inequality.


    Suppose $\displaystyle (r-1)$ is positive . . . that is: .$\displaystyle r > 1$

    Multiply [1] by by positive $\displaystyle (r-1)\!:\;\;2r + 1 \:<\:5(r - 1)$

    . . $\displaystyle 2r + 1 \:<\:5r-5\quad\Rightarrow\quad -3r \:<\:-6$

    Divide both sides by -3: .$\displaystyle r \:>\:2$

    We have two inequalities to satisfy: .$\displaystyle r > 1$ and $\displaystyle r > 2$

    We take the "stronger" of the two: .$\displaystyle \boxed{r > 2}$


    Suppose $\displaystyle (r-1)$ is negative . . . that is: .$\displaystyle r < 1$

    Multiply [1] by negative $\displaystyle (r-1)\!:\;\;2r + 1 \;\;{\color{red}>} \;\;5(r-1)$

    . . $\displaystyle 2r + 1 \:>\:5r - r\quad\Rightarrow\quad -3r \:>\:-6 $

    Divide both sides by -3: .$\displaystyle r \:<\:2$

    We have two inequalities to satisfy: .$\displaystyle r < 1$ and $\displaystyle r < 2$

    We take the "stronger" of the two: .$\displaystyle \boxed{r < 1}$


    Therefore: .$\displaystyle (r < 1) \vee (r > 2)$ .. . . .or: .$\displaystyle (-\infty,\,1) \,\cup \,(2,\,\infty)$

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  3. #3
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    Quote Originally Posted by Soroban View Post
    Hello, earachefl!


    We have two inequalities to satisfy: .$\displaystyle r < 1$ and $\displaystyle r < 2$

    We take the "stronger" of the two: .$\displaystyle \boxed{r < 1}$
    Interesting approach. I don't think my book ever put it in quite these terms.

    Thanks!
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