|(x+5)/(x-3)|>1 — step-by-step solution

1) Let's solve in

ℝ

| \frac{x + 5}{x - 3} | > 1

| A | > B

means that:

A > B

A < - B

2) Let's solve in

ℝ

\frac{x + 5}{x - 3} > 1

3) Let's sum the fractions:

\frac{x + 5}{x - 3} - 1 > 0

ⓘ To add fractions, put them over the same denominator, then add the numerators:

\frac{x + 5}{x - 3} - 1

\frac{(x + 5) - 1 \cdot (x - 3)}{x - 3}

\frac{(x + 5) - 1 \cdot (x - 3)}{x - 3} > 0

ⓘ Multiplying by -1 gives the opposite:

-1 \cdot (x - 3)

- (x - 3)

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

- (x - 3)

- x + 3

\frac{x + 5 - x + 3}{x - 3} > 0

ⓘ The numerator and the denominator are computed before the division:

5 + 3

8

■Like terms are grouped by adding their coefficients:

x - x

0

\frac{0 + 8}{x - 3} > 0

ⓘ Adding 0 changes nothing:

0 + 8

8

\frac{8}{x - 3} > 0

⇥The expression

x - 3

is zero at

x = 3

(forbidden value), it is negative before and positive after

4) Let's draw up the sign table:

Candidates, pending the domain check:

x > 3

5) Let's solve in

ℝ

\frac{x + 5}{x - 3} < -1

6) Let's sum the fractions:

\frac{x + 5}{x - 3} + 1 < 0

ⓘ To add fractions, put them over the same denominator, then add the numerators:

\frac{x + 5}{x - 3} + 1

\frac{(x + 5) + (x - 3)}{x - 3}

\frac{(x + 5) + (x - 3)}{x - 3} < 0

ⓘ The numerator and the denominator are computed before the division:

5 - 3

2

■Like terms are grouped by adding their coefficients:

x + x

2 x

\frac{2 x + 2}{x - 3} < 0

We study the sign of the numerator

⇥The expression

2 x + 2

is zero at:

⇥

x = - \frac{2}{2}

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

2

, their greatest common divisor:

- \frac{2}{2}

-1

⇥

x = -1

⇥Its coefficient of

x

2

, being positive, it is negative before and positive after

-1

We study the sign of the denominator, which must not be zero

⇥The expression

x - 3

is zero at

x = 3

(forbidden value), it is negative before and positive after

7) Let's draw up the sign table:

Candidates, pending the domain check:

x \in (-1, 3)

8) Let's evaluate the domain of

| \frac{x + 5}{x - 3} | > 1

For the expression

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

to exist, you must check:

x - 3 \neq 0

ⓘ We add

3

to both sides

Domain constraints:

x \neq 3

▶Solution:

x \in (-1, 3) \cup (3, +∞)