How do you solve $1 < 3 x - 2 \leq 10$ $1< 3x - 2\leq 10$ ?

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Answer 1 · 2018-06-16T12:04:05

See a solution process below:

First, add $2$ $color(red)(2)$ to each segment of the system of inequalities to isolate the $x$ $x$ term while keeping the system balanced:

$1 + 2 < 3 x - 2 + 2 \leq 10 + 2$ $1 + color(red)(2) < 3x - 2 + color(red)(2) <= 10 + color(red)(2)$

$3 < 3 x - 0 \leq 12$ $3 < 3x - 0 <= 12$

$3 < 3 x \leq 12$ $3 < 3x <= 12$

Now, divide each segment by $3$ $color(red)(3)$ to solve for $x$ $x$ while keeping the system balanced:

$\frac{3}{3} < \frac{3 x}{3} \leq \frac{12}{3}$ $3/color(red)(3) < (3x)/color(red)(3) <= 12/color(red)(3)$

$1 < (color(red)(cancel(color(black)(3)))x)/cancel(color(red)(3)) <= 4$

$1 < x <= 4$

Or

$x > 1$ ; $x <= 4$

Or, in interval notation:

$(1, 4]$

How do you solve 1< 3x - 2\leq 101<3x−2≤101< 3x - 2\leq 10?