Question

How do you solve and write the following in interval notation: `-3<3-2x<= 11`?

Answer 1

See a solution process below:

Explanation:

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

`-color(red)(3) - 3 < -color(red)(3) + 3 - 2x <= -color(red)(3) + 11`

`-6 < 0 - 2x <= 8`

`-6 < -2x <= 8`

Next, divide each segment by `color(blue)(-2)` to solve for `x` while keeping the system balanced. However, because we are multiplying or dividing inequalities by a negative number we need to reverse the inequality operators:

`(-6)/color(blue)(-2) color(red)(>) (-2x)/color(blue)(-2) color(red)(>=) 8/color(blue)(-2)`

`3 color(red)(>) (color(red)(cancel(color(black)(-2)))x)/cancel(color(blue)(-2)) color(red)(>=) -4`

`3 color(red)(>) x color(red)(>=) -4`

Or

`x >= -4` and `x < 3`

Or, in interval notation:

`[-4, 3)`

Answer 2

`[-4, 3)`

Explanation:

First we solve the inequality, then we put it into the notation we’d like.
−3 < 3 − 2x ≤ 11 Subtract the 3, divide by the -2 ( reverses the direction of the inequality)
`(-3 – 3)/(-2) > x >= (11 – 3)/(-2)`

`3 > x >= -4`

CHECK!: `−3 < 3 − 2x ≤ 11` ; `−3 < 3 − 2(2) ≤ 11` ; `−3 < 3 − 4 ≤ 11`

`−3 < − 1 ≤ 11` CORRECT!

In interval notation our result is written with an open bracket on the left, and a closed bracket on the right.
`(3, -4]`, or in increasing numberline order, `[-4, 3)`