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    https://8dedc505ac3fba908c50836f5905...%20Edexcel.pdf

    for q3 part b, how would I solve this?

    I worked out the eq (x+5)(x+2) > 0 but is this wrong?

    I'm confused as to what they expect me to do to solve it
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    Whenever you have inequalities questions with this one mark 'hence or otherwise' bit, I always find it best to do it graphically.

    The inequality is saying: "when is the line x+4 greater than the reciprocal graph 2/|X+3|"

    Note that the reciprocal graph is always positive (above the X-axis).

    Sketching the graphs leads to X>-2.
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    (Original post by Katiee224)
    https://8dedc505ac3fba908c50836f5905...%20Edexcel.pdf

    for q3 part b, how would I solve this?

    I worked out the eq (x+5)(x+2) > 0 but is this wrong?

    I'm confused as to what they expect me to do to solve it
    I'm not sure a sketch would be sufficient for this (as in I don't know).

    We can note that |x+3| = x+3 if x >-3. So any solution to part a, is also a solution to part b, as long as x > -3.

    Now suppose x < -3
    Then part b becomes x+4 > -2/(x+3)
    Whence (x+4)(x+3) < -2
    Sign reverses as x+3 < 0
    Then x^2+7x + 14 < 0
    which has no solution.

    Hence the only solutions are the overlap of the solutions to part a with x>-3, which gives x>-2.
 
 
 
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