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Math Help - Help: Test tomorrow!

  1. #1
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    Help: Test tomorrow!

    Determine the solution set and mention what theorem you used to solve the problem.

    a) |2x - 7| > or = 3
    answer: ( - inf., 2] U [5, inf.)
    theorem: ?

    b) |3 - 5x| < or = 12
    answer: [-9/5, 9]
    theorem: ?

    c) |6 - 4x| < or = 0
    answer: (3,9)
    theorem: ?

    d) (3x^2 + 4) / (5 - x ) < 0
    answer: ?
    theorem: ?

    f) |3x - 2| = |5 + 2x|
    answer: {-3/5, 7, -3}
    theorem: ?

    help thanks! i have a feeling most are wrong
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  2. #2
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    Quote Originally Posted by NeedHelp18 View Post
    Determine the solution set and mention what theorem you used to solve the problem.

    a) |2x - 7| > or = 3
    answer: ( - inf., 2] U [5, inf.)
    theorem: ?

    b) |3 - 5x| < or = 12
    answer: [-9/5, 9]
    theorem: ?

    c) |6 - 4x| < or = 0
    answer: (3,9)
    theorem: ?

    d) (3x^2 + 4) / (5 - x ) < 0
    answer: ?
    theorem: ?

    f) |3x - 2| = |5 + 2x|
    answer: {-3/5, 7, -3}
    theorem: ?

    help thanks! i have a feeling most are wrong
    a) Do you remember what the definition of the mod function (absolute value function) is?

    |x|=x if x\geq 0 and |x|=-x if x<0.

    So if |2x-7|\geq 3 then, by definition of a mod...

    2x-7\geq 3 if 2x-7\geq 0 and so x\geq 5 if x\geq \frac{7}{2}

    AND

    -(2x-7)\geq 3 if 2x-7<0, and so

    2x-7 \leq -3 if 2x-7<0, thus x \leq 2 if x<\frac{7}{2}.

    So your answer is right, but you have to write the restrictions on x as well as the solution set. The theorem you would have used here is the definition of a mod.

    b) and c) follow the exact same principle.
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  3. #3
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    Quote Originally Posted by NeedHelp18 View Post
    Determine the solution set and mention what theorem you used to solve the problem.

    a) |2x - 7| > or = 3
    answer: ( - inf., 2] U [5, inf.)
    theorem: ?

    b) |3 - 5x| < or = 12
    answer: [-9/5, 9]
    theorem: ?

    c) |6 - 4x| < or = 0
    answer: (3,9)
    theorem: ?

    d) (3x^2 + 4) / (5 - x ) < 0
    answer: ?
    theorem: ?

    f) |3x - 2| = |5 + 2x|
    answer: {-3/5, 7, -3}
    theorem: ?

    help thanks! i have a feeling most are wrong
    For d), you can solve an inequality exactly the same way as you would an equality, except that if you multiply or divide by a negative number, the inequality changes direction. Also you have to take into account the IMPLIED domains.

    \frac{3x^2 + 4}{5 - x}<0.

    First of all, note that the denominator can not be 0. Thus x\neq 5.

    Now we solve exactly the same way as we would an equality...

    \frac{3x^2 + 4}{5 - x}<0. Multiply both sides by the denominator.

    3x^2+4 < 0 Subtract 4 from both sides...

    3x^2 < -4 Divide both sides by 3...

    x^2 < -\frac{4}{3} Take the square root of both sides...

    |x| < \sqrt{-\frac{4}{3}}.

    But here we have the square root of a negative number, which gives a non-real answer.

    So we say that there are not any real solutions for this inequality.
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  4. #4
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    Quote Originally Posted by NeedHelp18 View Post
    Determine the solution set and mention what theorem you used to solve the problem.

    a) |2x - 7| > or = 3
    answer: ( - inf., 2] U [5, inf.)
    theorem: ?

    b) |3 - 5x| < or = 12
    answer: [-9/5, 9]
    theorem: ?

    c) |6 - 4x| < or = 0
    answer: (3,9)
    theorem: ?

    d) (3x^2 + 4) / (5 - x ) < 0
    answer: ?
    theorem: ?

    f) |3x - 2| = |5 + 2x|
    answer: {-3/5, 7, -3}
    theorem: ?

    help thanks! i have a feeling most are wrong
    For f) we have to remember that there are 2 situations for each mod, so there can be up to 4 solutions to this equality.

    Case 1: 3x - 2 = 5 + 2x if 3x-2 \geq 0 and 5+2x \geq 0.

    This gives x = 7 if x \geq \frac{2}{3} and x \geq -\frac{5}{2}. Putting the inequalities together we get x = 7 if x \geq \frac{2}{3}.


    Case 2: 3x-2 = -(5+2x) if 3x-2 \geq 0 and 5+2x < 0.

    Just by looking at the inequalities, we see that x has to be both \geq \frac{2}{3} and \leq -\frac{5}[2}. This can not happen, so we can not find a solution for this case.


    Case 3: -(3x-2) = 5 + 2x if 3x - 2 < 0 and 5 + 2x \geq 0. This gives...

    x=-\frac{3}{5} if -\frac{5}{2}\leq x<\frac{2}{3}.


    Case 4: -(3x-2) = -(5+2x) if 3x-2 < 0 and 5+2x < 0.

    This gives x=7 if x<\frac{2}{3} and x<-\frac{5}{2}. For both these inequalities to occur, x<-\frac{5}{2}.


    So putting all the cases together we get...

    x=7 if x<-\frac{5}{2} or x \geq \frac{2}{3} and x=-\frac{3}{5} if -\frac{5}{2} \leq x<\frac{2}{3}.

    Once again the theorem used is the definition of a mod.
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