The inequality |ax + 5| <= b has solution set -4 <= x <= 6. Find a - b.

Step-by-step solution

1. 1

   Rewrite the inequality as a centre-and-radius statement. For $a\ne 0$,

   $|ax+5|\le b\iff |a|\left|x+\frac5a\right|\le b\iff \left|x-\left(-\frac5a\right)\right|\le\frac{b}{|a|}$

   So the solution set is always a closed interval centred at $-\tfrac5a$ with radius $\tfrac{b}{|a|}$. This is the shortcut that avoids case-splitting on the sign of $a$.

2. 2

   Read the centre and radius off the given interval. For $[-4,6]$,

   $\text{centre}=\frac{-4+6}{2}=1,\qquad \text{radius}=\frac{6-(-4)}{2}=5$

3. 3

   Match centre to find $a$.

   $-\frac{5}{a}=1\ \Longrightarrow\ a=-5$

4. 4

   Match radius to find $b$. With $|a|=5$,

   $\frac{b}{5}=5\ \Longrightarrow\ b=25$

   Note $b=25\ge 0$, which is required — a negative $b$ would make $|ax+5|\le b$ unsolvable.

5. 5

   Compute $a-b$ and verify the original inequality.

   $a-b=-5-25=-30$

   Check: with $a=-5$, $b=25$ the inequality reads $|-5x+5|\le 25$, i.e. $5|x-1|\le 25$, i.e. $|x-1|\le 5$, i.e. $-4\le x\le 6$ — exactly the given solution set. The endpoints give $|-5(-4)+5|=25$ and $|-5(6)+5|=25$, both attaining the bound, as equality demands.

6. 6

   The general technique. For any $|ax+c|\le b$: the solution interval's midpoint is $-c/a$ and its half-length is $b/|a|$. Equating those two numbers to the given interval's midpoint and half-length always determines $a$ and $b$ in one step, with no sign cases.