How do you solve $- 8 \leq 8 x - 3 < 2$ $-8\leq 8x - 3< 2$ ?

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Answer 1 · 2017-02-20T23:29:01

See the entire solution process below:

Step 1) Add $3$ $color(red)(3)$ to each segment of the system of inequalities to isolate the $x$ $x$ term while keeping the system in balance:

$- 8 + 3 \leq 8 x - 3 + 3 < 2 + 3$ $-8 + color(red)(3) <= 8x - 3 + color(red)(3) < 2 + color(red)(3)$

$- 5 \leq 8 x - 0 < 5$ $-5 <= 8x - 0 < 5$

$- 5 \leq 8 x < 5$ $-5 <= 8x < 5$

Step 2) Divide each segment of the system of inequalities by $8$ $color(red)(8)$ to solve for $x$ $x$ while keeping the system in balance:

$- \frac{5}{8} \leq \frac{8 x}{8} < \frac{5}{8}$ $-5/color(red)(8) <= (8x)/color(red)(8) < 5/color(red)(8)$

$-5/8 <= (color(red)(cancel(color(black)(8)))x)/cancel(color(red)(8)) < 5/8$

$-5/8 <= x < 5/8$

How do you solve -8\leq 8x - 3< 2−8≤8x−3<2-8\leq 8x - 3< 2?