|(2-3x)/(1+x)|>1 — step-by-step solution

1) Let's solve in

ℝ

| \frac{2 - 3 x}{1 + x} | > 1

| A | > B

means that:

A > B

A < - B

2) Let's solve in

ℝ

\frac{2 - 3 x}{1 + x} > 1

3) Let's add fractions:

\frac{2 - 3 x}{1 + x} - 1 > 0

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

\frac{2 - 3 x}{1 + x} - 1

\frac{(2 - 3 x) - 1 \cdot (1 + x)}{1 + x}

\frac{(2 - 3 x) - 1 \cdot (1 + x)}{1 + x} > 0

ⓘ Multiplying by -1 gives the opposite:

-1 \cdot (1 + x)

- (1 + x)

\frac{2 - 3 x - (1 + x)}{1 + x} > 0

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

- (1 + x)

- 1 - x

\frac{2 - 3 x - 1 - x}{1 + x} > 0

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

2 - 1

1

■Like terms are grouped by adding their coefficients:

- 3 x - x

-4 x

\frac{1 - 4 x}{1 + x} > 0

We study the sign of the numerator

⇥The expression

1 - 4 x

is zero at:

⇥

x = \frac{-1}{-4}

⇥ⓘ The minus signs of the numerator and the denominator cancel out:

\frac{-1}{-4}

\frac{1}{4}

⇥

x = \frac{1}{4}

⇥Its coefficient of

x

-4

, being negative, it is positive before and negative after

\frac{1}{4}

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

⇥The expression

1 + x

is zero at

x = -1

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

4) Let's draw up the sign table:

x	$-∞$		$-1$		$\frac{1}{4}$		$+∞$
$1 - 4 x$		+		+	0	-
$1 + x$		-	0	+		+
$\frac{2 - 3 x}{1 + x} - 1$		-	❌	+	0	-

Candidates, pending the domain check:

x \in (-1, \frac{1}{4})

5) Let's solve in

ℝ

\frac{2 - 3 x}{1 + x} < -1

6) Let's sum the fractions:

\frac{2 - 3 x}{1 + x} + 1 < 0

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

\frac{2 - 3 x}{1 + x} + 1

\frac{(2 - 3 x) + (1 + x)}{1 + x}

\frac{(2 - 3 x) + (1 + x)}{1 + x} < 0

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

2 + 1

3

■Like terms are grouped by adding their coefficients:

- 3 x + x

-2 x

\frac{3 - 2 x}{1 + x} < 0

We study the sign of the numerator

⇥The expression

3 - 2 x

is zero at:

⇥

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

⇥ⓘ The minus signs of the numerator and the denominator cancel out:

\frac{-3}{-2}

\frac{3}{2}

⇥

x = \frac{3}{2}

⇥Its coefficient of

x

-2

, being negative, it is positive before and negative after

\frac{3}{2}

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

⇥The expression

1 + x

is zero at

x = -1

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

7) Let's draw up the sign table:

x	$-∞$		$-1$		$\frac{3}{2}$		$+∞$
$3 - 2 x$		+		+	0	-
$1 + x$		-	0	+		+
$\frac{2 - 3 x}{1 + x} + 1$		-	❌	+	0	-

Candidates, pending the domain check:

x \in (-∞, -1) \cup (\frac{3}{2}, +∞)

8) Let's evaluate the domain of

| \frac{2 - 3 x}{1 + x} | > 1

For the expression

\frac{2 - 3 x}{1 + x}

to exist, you must check:

1 + x \neq 0

ⓘ We subtract

1

from both sides

Domain constraints:

x \neq -1

▶Solution:

x \in (-∞, -1) \cup (-1, \frac{1}{4}) \cup (\frac{3}{2}, +∞)

9) The numeric value of:

x \in (-∞, -1) \cup (-1, \frac{1}{4}) \cup (\frac{3}{2}, +∞)

is:

▷

x \in (-∞, -1) \cup (-1, 0.25) \cup (1.5, +∞)