Results 1 to 3 of 3

Math Help - Half right, missed the other half...

  1. #1
    Junior Member
    Joined
    Jul 2007
    Posts
    43

    Half right, missed the other half...

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

    I solved it by first finding the slope \frac {2r + 1}{r - 1} =  5 which simplifies to r = 2. Then I used the point-slope form to derive the equation 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 (2, \infty). Unfortunately, the answer should have included also the set (-\infty, 1).

    Algebraically, what should I have done to get to that answer? What part did I skip?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,861
    Thanks
    742
    Hello, earachefl!

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


    You found the slope of AB . . . good!

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


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

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


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

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

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

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

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

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


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

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

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

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

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

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


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

    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Jul 2007
    Posts
    43
    Quote Originally Posted by Soroban View Post
    Hello, earachefl!


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

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

    Thanks!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Half-life.
    Posted in the Calculus Forum
    Replies: 6
    Last Post: February 25th 2011, 08:43 PM
  2. The # half-way between 2.1 and 2.11
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: December 2nd 2008, 12:09 AM
  3. half angles ..
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: November 13th 2008, 07:57 AM
  4. half life
    Posted in the Algebra Forum
    Replies: 2
    Last Post: October 31st 2008, 06:23 PM
  5. Half Width Half Maximum
    Posted in the Calculus Forum
    Replies: 1
    Last Post: May 27th 2008, 05:30 AM

Search Tags


/mathhelpforum @mathhelpforum