Question 1086171
Solve for x where a, p > 0: p|ax + b|-q ≤ 0
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{{{p*abs(ax + b) <= q}}}

The left side is non-negative so q is non-negative.

Case 1:  q = 0 

{{{p*abs(ax + b) = 0}}}, and since p > 0, 

ax + b = 0 
    ax = -b
     x = -b/a

So if q=0, then x = -b/a

Case 2: q > 0

{{{p*abs(ax + b) - q <= 0}}}, 

{{{abs(apx + bp) <= q}}},

{{{-q <= apx+bp <= q}}}

{{{-q-bp <= apx <= q-bp}}}

{{{(-q-bp)/(ap) <=x<=(q-bp)/(ap)}}}

Edwin</pre></b></font>