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Help, Math-Pa, how to solve |2x-1|+3|x+1|>5?

4 Answers

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  • cidyah
    Lv 7
    5 years ago

    2x-1 > 0 and x+1 > 0

    |2x-1| = 2x-1 and |x+1| = x+1

    2x-1 + 3(x+1) > 5

    2x-1+3x+3 > 5

    5x+2 > 5

    5x > 3

    x > 3/5 -----(1)

    2x-1 > 0 and x+1 < 0

    |2x-1| = 2x-1 and |x+1| = -(x+1)

    2x-1 + 3(-x-1) > 5

    2x-1-3x-3 > 5

    -x -4 > 5

    -(x+4) > 5

    (x+4) < -5

    x < - 9 -------(2)

    2x-1 < 0 and x+1 > 0

    |2x-1| = -(2x-1) and |x+1| = x+1

    -(2x-1)+3(x+1) > 5

    -2x+1+3x+3 > 5

    x + 4 > 5

    x > 1 -----(3)

    2x-1 < 0 and x+1 < 0

    |2x-1| = -(2x-1) and |x+1| = -(x+1)

    -2x+1 +3(-x-1) > 5

    -2x+1-3x-3 > 5

    -5x-2 > 5

    -5x > 7

    5x < -7

    x < -7/5 ------(4)

    from (1) & (3) x > 3/5

    from (2) & (4) x > -7/5

    x < -7/5 & x > 3/5

  • 5 years ago

    |2x-1|=2x-1, if x>=1/2

    .........=1-2x, if x<=1/2

    3|x+1|=3x+3, if x>=-1

    ..........=-3x-3, if x<=-1

    =>

    1-2x-3x-3>5=>x<-7/5

    3x+3+2x-1>5=>x>3/5

    =>

    |2x-1|+3|x+1|>5 has the

    solution as

    x<-7/5 or x>3/5

  • 5 years ago

    Think of the situations you have to cover...

    |2x -1| ==> 2x -1 and -(2x-1)

    |x+1| ==> (x+1) and -(x+1)

    So you could break it into four sets

    (1)...(2x-1) + 3(x+1) > 5

    ==> 2x -1 + 3x + 3 > 5

    ==> 5x > 3

    ==> x > 3/5

    (2)... (2x-1) -3(x+1) > 5

    ==> 2x -1 -3x -3 > 5

    ==> -x > 9

    ==> x > -9

    (3)... -(2x-1) + 3(x+1) >5

    ==> -2x + 1 + 3x + 3 > 5

    ==> x > 1

    (4)... -(2x-1) -3(x+1) > 5

    etc.

  • 5 years ago

    die

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