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    Write the inequality -3 < x < 8 in the form |x - a| < b
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    Hello i have got very stuck trying to solve the question, any help would be appreciated.

    Write the inequality -3 < x < 8 in the form |x-a| < b
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    (Original post by znx7)
    Write the inequality -3 < x < 8 in the form |x - a| < b
    From looking at the title of this thread I thought it was about geography/politics.
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    |x-5/2 |< 11/2
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    -3 < x < 8 in the form |x-a| < b

    if x-a>0 then
    x-a<b
    x<a+b

    if x-a<0 the
    a-x<b
    a-b<x

    that's too much for a hint
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    (Original post by znx7)
    Hello i have got very stuck trying to solve the question, any help would be appreciated.

    Write the inequality -3 < x < 8 in the form |x-a| < b
    |x-a|<b is equivalent to saying a-b<x<a+b
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    (Original post by Kotsiozz)
    |x-5/2 |< 11/2
    could you explain how you got that answer please?
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    |x-a|<b is equal to -b<x-a<b
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    |x-a|<b is equal to -b<x-a<b and then solve simultaneous eqs
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    (Original post by znx7)
    Write the inequality -3 < x < 8 in the form |x - a| < b
    This is easy if you understand that |x-a| means the distance from a variable point x to a fixed point a.

    So e.g. |x-2| &lt; 3 means that x is always less than 3 units away from the +ve number 2 on the number line. That means it can go 3 less than 2, to -1, or 3 more than 2, to +5, i.e. -1 &lt; x &lt; 5. Here, x can vary over an interval of length 6, but it can only go a distance of half that, 3, from the mid-point, 2.

    The last inequality is not symmetrical, but note that an inequality like |x-a| &lt; b *is* symmetrical - we've lost symmetry somehow. Note that we can get back to a nice symmetrical form by subtracting 2 from all sides:

    -1 &lt; x &lt; 5 \Leftrightarrow -1-2 &lt; x-2 &lt; 5-2 \Leftrightarrow -3 &lt; x-2 &lt; 3 \Leftrightarrow |x-2| &lt; 3

    That is now symmetrical since 2 is the midpoint (or average) of -1 and 5 i.e. 2 = \frac{1}{2}(5+(-1)), and half the length of the interval is 3 = \frac{1}{2}(5-(-1))

    So given a &lt; x &lt; b we can rewrite it symmetrically as:

    |x-\frac{b+a}{2}| &lt; \frac{b-a}{2}
 
 
 
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