3/(|5-2x|)<2 — step-by-step solution

1) Let's solve in

ℝ

\frac{3}{| 5 - 2 x |} < 2

2) Let's sum the fractions:

\frac{3}{| 5 - 2 x |} - 2 < 0

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

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

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

\frac{3 - 2 \cdot | 5 - 2 x |}{| 5 - 2 x |} < 0

We study the sign of the numerator

⇥2.1) Let's study the sign of

3 - 2 \cdot | 5 - 2 x |

by solving in

ℝ

3 - 2 \cdot | 5 - 2 x | < 0

⇥We divide both sides by

-2

; since this number is negative, we reverse the inequality sign

⇥

| A | > B

means that:

A > B

A < - B

⇥2.2) Let's solve in

ℝ

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

⇥ⓘ We gather the terms in

x

on the right and the other terms on the left

⇥

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

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

5 - \frac{3}{2}

\frac{2 \cdot 5 - 3}{2}

⇥

\frac{2 \cdot 5 - 3}{2} > 2 x

⇥ⓘ Multiplications are done before additions and subtractions:

2 \cdot 5

10

⇥

\frac{10 - 3}{2} > 2 x

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

10 - 3

7

⇥

\frac{7}{2} > 2 x

⇥ⓘ Both sides are divided by

2

⇥

x < \frac{\frac{7}{2}}{2}

⇥ⓘ Dividing a fraction by

2

is multiplying its denominator by

2

\frac{\frac{7}{2}}{2}

\frac{7}{2 \cdot 2}

⇥

x < \frac{7}{2 \cdot 2}

⇥ⅰ

2 \cdot 2

4

⇥Candidates, pending the domain check:

x < \frac{7}{4}

⇥2.3) Let's solve in

ℝ

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

⇥ⓘ We gather the terms in

x

on the right and the other terms on the left

⇥

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

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

5 + \frac{3}{2}

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

⇥

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

⇥ⓘ Multiplications are done before additions and subtractions:

2 \cdot 5

10

⇥

\frac{10 + 3}{2} < 2 x

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

10 + 3

13

⇥

\frac{13}{2} < 2 x

⇥ⓘ Both sides are divided by

2

⇥

x > \frac{\frac{13}{2}}{2}

⇥ⓘ Dividing a fraction by

2

is multiplying its denominator by

2

\frac{\frac{13}{2}}{2}

\frac{13}{2 \cdot 2}

⇥

x > \frac{13}{2 \cdot 2}

⇥ⅰ

2 \cdot 2

4

⇥Candidates, pending the domain check:

x > \frac{13}{4}

⇥So the expression

3 - 2 \cdot | 5 - 2 x |

is negative on

(\frac{13}{4}, +∞)

and positive or zero otherwise

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

⇥2.4) Let's study the sign of

| 5 - 2 x |

by solving in

ℝ

| 5 - 2 x | < 0

⇥Solution:

x \in \emptyset

⇥So the expression

| 5 - 2 x |

is always positive or zero

3) Let's draw up the sign table:

x	$-∞$		$\frac{7}{4}$		$\frac{13}{4}$		$+∞$
$3 - 2 \cdot \| 5 - 2 x \|$		-	0	+	0	-
$\| 5 - 2 x \|$		+		+		+
$\frac{3}{\| 5 - 2 x \|} - 2$		-	0	+	0	-

Candidates, pending the domain check:

x \in (-∞, \frac{7}{4}) \cup (\frac{13}{4}, +∞)

4) Let's find the domain of

\frac{3}{| 5 - 2 x |} < 2

For the expression

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

to be defined, we must have:

| 5 - 2 x | \neq 0

5) Let's solve in

ℝ

5 - 2 x \neq 0

ⓘ We add

2 x

to both sides

5 \neq 2 x

ⓘ Both sides are divided by

2

Domain constraints:

x \neq \frac{5}{2}

▶Solution:

x \in (-∞, \frac{7}{4}) \cup (\frac{13}{4}, +∞)

6) The numeric value of:

x \in (-∞, \frac{7}{4}) \cup (\frac{13}{4}, +∞)

equals:

▷

x \in (-∞, 1.75) \cup (3.25, +∞)