2/3 |x+5|-5/3 =-1 — step-by-step solution

1) Let's solve in

ℝ

\frac{2}{3 \cdot | x + 5 |} - \frac{5}{3} = -1

We multiply both sides by

3 \cdot | x + 5 |

, nonzero for the fractions to be defined

\frac{2}{3 \cdot | x + 5 |} \cdot 3 \cdot | x + 5 | - \frac{5}{3} \cdot 3 \cdot | x + 5 | = -1 \cdot 3 \cdot | x + 5 |

ⓘ Multiplying a fraction by a term multiplies its numerator:

\frac{2}{3 \cdot | x + 5 |} \cdot 3 \cdot | x + 5 |

\frac{2 \cdot 3 \cdot | x + 5 |}{3 \cdot | x + 5 |}

\frac{6 \cdot | x + 5 |}{3 \cdot | x + 5 |}

■Multiplications are done before additions and subtractions:

\frac{5}{3} \cdot 3

5

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

-1 \cdot 3 \cdot | x + 5 |

-3 \cdot | x + 5 |

\frac{6 \cdot | x + 5 |}{3 \cdot | x + 5 |} - 5 \cdot | x + 5 | = -3 \cdot | x + 5 |

ⓘ The original denominator must remain nonzero. The fraction is simplified by dividing the numerator and the denominator by their common factor:

\frac{6 \cdot | x + 5 |}{3 \cdot | x + 5 |}

\frac{6 \cdot | 1 |}{3 \cdot | 1 |}

\frac{6 \cdot | 1 |}{3 \cdot | 1 |} - 5 \cdot | x + 5 | = - 3 \cdot | x + 5 |

ⓘ A fraction is simplified by dividing the numerator and the denominator by the same number:

\frac{6 \cdot | 1 |}{3 \cdot | 1 |}

\frac{2 \cdot | 1 |}{| 1 |}

\frac{2 \cdot | 1 |}{| 1 |} - 5 \cdot | x + 5 | = - 3 \cdot | x + 5 |

\frac{2 \cdot 1}{1} - 5 \cdot | x + 5 | = - 3 \cdot | x + 5 |

2 - 5 \cdot | x + 5 | = - 3 \cdot | x + 5 |

The absolute-value arguments are zero at

-5

. Split ℝ at these points and solve on each interval

(-∞, -5]

x + 5 \leq 0

, so

| x + 5 | = - (x + 5)

;

x + 5 \leq 0

, so

| x + 5 | = - (x + 5)

(-∞, -5]

, the equation becomes

2 - 5 \cdot (- (x + 5)) = - 3 \cdot (- (x + 5))

2) Let's solve in

ℝ

2 - 5 \cdot (- (x + 5)) = - 3 \cdot (- (x + 5))

ⓘ Extract the minus signs: an even number of minus signs cancels out, an odd number leaves one minus sign:

- 5 \cdot (- (x + 5))

5 \cdot (x + 5)

- 3 \cdot (- (x + 5))

3 \cdot (x + 5)

2 + 5 \cdot (x + 5) = 3 \cdot (x + 5)

ⓘ Expand by multiplying each term inside the parentheses:

5 \cdot (x + 5)

5 x + 25

3 \cdot (x + 5)

3 x + 15

2 + 5 x + 25 = 3 x + 15

ⓘ Start from

2

and move forward

25

on the number line:

2 + 25

27

27 + 5 x = 3 x + 15

ⓘ We subtract

27

from both sides■Start from

15

and move back

27

on the number line, going below zero:

15 - 27

-12

5 x = 3 x - 12

ⓘ We subtract

3 x

from both sides

5 x - 3 x = -12

ⓘ Like terms are grouped by adding their coefficients:

5 x - 3 x

2 x

2 x = -12

ⓘ Both sides are divided by

2

x = - \frac{12}{2}

ⓘ We simplify the fraction by dividing the numerator and the denominator by

2

, their greatest common divisor:

- \frac{12}{2}

-6

Candidates, pending the domain check:

x = -6

-6 \leq -5

, we can keep the solution

-6

[-5, +∞)

x + 5 \geq 0

, so

| x + 5 | = x + 5

;

x + 5 \geq 0

, so

| x + 5 | = x + 5

[-5, +∞)

, the equation becomes

2 - 5 \cdot (x + 5) = - 3 \cdot (x + 5)

3) Let's solve in

ℝ

2 - 5 \cdot (x + 5) = - 3 \cdot (x + 5)

ⓘ Expand by multiplying each term inside the parentheses:

- 5 \cdot (x + 5)

- (5 x + 25)

- 3 \cdot (x + 5)

- (3 x + 15)

2 - (5 x + 25) = - (3 x + 15)

ⓘ A minus sign before a parenthesis changes the sign of each of its terms:

- (5 x + 25)

- 5 x - 25

- (3 x + 15)

- 3 x - 15

2 - 5 x - 25 = - 3 x - 15

ⓘ Start from

2

and move back

25

on the number line, going below zero:

2 - 25

-23

- 23 - 5 x = - 3 x - 15

ⓘ We gather the terms in

x

on the right and the other terms on the left

- 23 + 15 = - 3 x + 5 x

ⓘ Start from

-23

and move forward

15

on the number line, getting closer to zero:

- 23 + 15

-8

■Like terms are grouped by adding their coefficients:

- 3 x + 5 x

2 x

-8 = 2 x

ⓘ Both sides are divided by

2

x = - \frac{8}{2}

ⓘ We simplify the fraction by dividing the numerator and the denominator by

2

, their greatest common divisor:

- \frac{8}{2}

-4

Candidates, pending the domain check:

x = -4

-4 \geq -5

, we can keep the solution

-4

Candidates, pending the domain check:

x = {-6, -4}

4) Let's evaluate the domain of

\frac{2}{3 \cdot | x + 5 |} - \frac{5}{3} = -1

For the expression

\frac{2}{3 \cdot | x + 5 |}

to exist, you must check:

3 \cdot | x + 5 | \neq 0

We simplify the expression by

3

Let's solve in

ℝ

| x + 5 | \neq 0

Let's solve in

ℝ

x + 5 \neq 0

ⓘ We subtract

5

from both sides

Domain constraints:

x \neq -5

▶Solution:

x = {-6, -4}