-5<= (x-2)/(-3)<3 — step-by-step solution

1) We have the following 2 conditions:

[1]:

-5 \leq \frac{x - 2}{-3}

[2]:

\frac{x - 2}{-3} < 3

2) Let's solve for x:

-5 \leq \frac{x - 2}{-3}

[1]

ⓘ The sign of a fraction goes in front of the bar, whether it comes from the numerator or the denominator:

\frac{x - 2}{-3}

- \frac{x - 2}{3}

- \frac{x - 2}{3} \geq -5

ⓘ The minus sign of the fraction goes into the numerator or the denominator, changing the sign of each of its terms:

- \frac{x - 2}{3}

\frac{- x + 2}{3}

\frac{- x + 2}{3} \geq -5

- x + 2 \geq -5 \cdot 3

ⓘ A single minus sign does not cancel out, the product of a negative number by a positive number is negative:

-5 \cdot 3

-15

- x + 2 \geq -15

ⓘ We gather the terms in

x

on the right and the other terms on the left

2 + 15 \geq x

ⓘ Start from

2

and move forward

15

on the number line:

2 + 15

17

17 \geq x

Solution found [1]:

x \leq 17

3) Let's solve for x:

\frac{x - 2}{-3} < 3

[2] using

x \leq 17

ⓘ The sign of a fraction goes in front of the bar, whether it comes from the numerator or the denominator:

\frac{x - 2}{-3}

- \frac{x - 2}{3}

- \frac{x - 2}{3} < 3

ⓘ The minus sign of the fraction goes into the numerator or the denominator, changing the sign of each of its terms:

- \frac{x - 2}{3}

\frac{- x + 2}{3}

\frac{- x + 2}{3} < 3

- x + 2 < 3 \cdot 3

ⅰ

3 \cdot 3

9

- x + 2 < 9

ⓘ We gather the terms in

x

on the right and the other terms on the left

2 - 9 < x

ⓘ Start from

2

and move back

9

on the number line, going below zero:

2 - 9

-7

-7 < x

Solution obtained [2]:

x > -7

Intersecting

x \leq 17

and

x > -7

gives

x \in (-7, 17]

▶Solution:

x \in (-7, 17]