Results 1 to 5 of 5

Math Help - Help!very eager to know the way to solve it!

  1. #1
    Member helloying's Avatar
    Joined
    Jul 2008
    Posts
    177

    Help!very eager to know the way to solve it!

    X is a real no. and -5< x ≤ 3 is the solution to the two simulyaneous inequalities ax+15>3x and 8-x ≥ bx-10 where a and b are positive contants such that a>3 and b>0, find the possible values of a and b
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Hello !
    Quote Originally Posted by helloying View Post
    X is a real no. and -5< x ≤ 3 is the solution to the two simulyaneous inequalities ax+15>3x and 8-x ≥ bx-10 where a and b are positive contants such that a>3 and b>0, find the possible values of a and b
    ax+15>3x \implies x(a-3)>-15
    Since a>3, a-3>0, then we can divise by (a-3) without changing the direction of the inequation :
    x>\tfrac{-15}{a-3}

    But x \le 3

    \implies \boxed{3} \ge x > \boxed{\tfrac{-15}{a-3}}

    3>\frac{-15}{a-3} \implies a-3>-5 \implies a>-2, i.e. a \in (-2; +\infty)

    But we wanted a \in (3; +\infty)

    Thus the solution is a \in (-2; +\infty) \cap (3; +\infty) \Leftrightarrow a \in (3; +\infty)

    Solution is a>3
    Last edited by Moo; August 31st 2008 at 05:14 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member helloying's Avatar
    Joined
    Jul 2008
    Posts
    177
    Quote Originally Posted by Moo View Post
    Hello !


    ax+15>3x \implies x(a-3)>15
    Since a>3, a-3>0, then we can divise by (a-3) without changing the direction of the inequation :
    x>\tfrac{15}{a-3}

    But x \ge 3

    \implies \boxed{3} \ge x > \boxed{\tfrac{15}{a-3}}

    3>\frac{15}{a-3} \implies a-3>5 \implies a>8

    The only condition for a was that a>3, which is the case here. So the solution is \boxed{a>8}


    Can you do it for y ? It's exactly the same reasoning.

    Shouldnt it be x(a-3)>-15 because it have been brought over? so the ans will be different but I couldnt get the ans if I made it to -15.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Quote Originally Posted by helloying View Post
    Shouldnt it be x(a-3)>-15 because it have been brought over? so the ans will be different but I couldnt get the ans if I made it to -15.
    I guess it wasn't my day at all
    I edited it

    Ok, it looks like there are problems with your exercises, can't you check if you've copied it right ?
    Last edited by Moo; August 31st 2008 at 06:13 AM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member
    Joined
    Dec 2007
    From
    Melbourne
    Posts
    428
    X is a real no. and -5< x ≤ 3 is the solution to the two simulyaneous inequalities ax+15>3x and 8-x ≥ bx-10 where a and b are positive contants such that a>3 and b>0, find the possible values of a and b
    I think what the question wants you to do is find values for a and b so that solving those inequalities would get you -5< x ≤ 3.
    So after you do this part

    Since a>3, a-3>0, then we can divise by (a-3) without changing the direction of the inequation :
    You just have to solve \frac{-15}{a-3} = -5 (because x>-5) and then repeat the process for b.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. need to solve summation equation to solve sum(x2)
    Posted in the Statistics Forum
    Replies: 2
    Last Post: July 16th 2010, 11:29 PM
  2. Replies: 1
    Last Post: June 9th 2009, 11:37 PM
  3. how can i solve this
    Posted in the Algebra Forum
    Replies: 2
    Last Post: August 6th 2008, 10:11 AM
  4. Solve
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: July 30th 2008, 10:57 AM
  5. solve for x
    Posted in the Math Topics Forum
    Replies: 3
    Last Post: July 27th 2008, 01:24 PM

Search Tags


/mathhelpforum @mathhelpforum