How do you solve the compound inequality $3 < 2 x - 3 < 15$ $3<2x-3<15$ ?

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Answer 1 · 2017-05-21T11:33:12

See a solution process below:

First, add $3$ $color(red)(3)$ to each segment of the system of inequalities to isolate the $x$ $x$ term while keeping the system balanced:

$3 + 3 < 2 x - 3 + 3 < 15 + 3$ $3 + color(red)(3) < 2x - 3 + color(red)(3) < 15 + color(red)(3)$

$6 < 2 x - 0 < 18$ $6 < 2x - 0 < 18$

$6 < 2 x < 18$ $6 < 2x < 18$

Now, divide each segment by $2$ $color(red)(2)$ to solve for $x$ $x$ while keeping the system balanced:

$\frac{6}{2} < \frac{2 x}{2} < \frac{18}{2}$ $6/color(red)(2) < (2x)/color(red)(2) < 18/color(red)(2)$

$3 < (color(red)(cancel(color(black)(2)))x)/cancel(color(red)(2)) < 9$

$3 < x < 9$

Or

$x > 3$ ; $x < 9$

Or, in interval notation:

$(3, 9)$

Answer 2 · 2017-05-21T11:34:18

Solution: $3 < x < 9$ or in interval notation: $(3,9)$

$3 < 2x-3 <15$ Adding $3$ in all sides we get

$6 < 2x <18$ Multiplying by $1/2$ in all sides we get

$3 < x < 9$ .

Solution: $3 < x < 9$ or in interval notation: $(3,9)$ [Ans]

How do you solve the compound inequality 3<2x-3<153<2x−3<153<2x-3<15?