3/(x+2)> 3/(x-5) — step-by-step solution

1) Let's solve in

ℝ

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

2) Let's sum the fractions:

\frac{3}{x + 2} - \frac{3}{x - 5} > 0

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

- \frac{3}{x - 5}

\frac{3}{- x + 5}

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

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

\frac{3 \cdot (- x + 5) + 3 \cdot (x + 2)}{(x + 2) \cdot (- x + 5)}

\frac{3 \cdot (- x + 5) + 3 \cdot (x + 2)}{(x + 2) \cdot (- x + 5)} > 0

We study the sign of the numerator

⇥2.1) Let's study the sign of

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

by solving in

ℝ

3 \cdot (- x + 5) + 3 \cdot (x + 2) > 0

⇥ⓘ Expand by multiplying each term inside the parentheses:

3 \cdot (- x + 5)

3 \cdot (- x) + 15

3 \cdot (x + 2)

3 x + 6

⇥

3 \cdot (- x)

+ 15

+ 3 x

+ 6 > 0

⇥ⓘ The minus sign is placed in front of the product:

3 \cdot (- x)

- 3 x

■Start from

15

and move forward

6

on the number line:

15 + 6

21

⇥

- 3 x + 21 + 3 x > 0

⇥ⓘ Like terms are grouped by adding their coefficients:

- 3 x + 3 x

0

⇥

0 + 21 > 0

⇥ⓘ Adding 0 changes nothing:

0 + 21

21

⇥

21 > 0

⇥The simplified relation is always true; the original expression's domain must still be respected

⇥So the expression

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

is always positive

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

⇥2.2) Let's solve in

ℝ

(x + 2) \cdot (- x + 5) > 0

⇥We study the sign of each factor of the product

⇥⇥The expression

x + 2

is zero at

x = -2

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

⇥⇥The expression

- x + 5

is zero at:

⇥⇥

x = 5

⇥⇥Its coefficient of

x

-1

, being negative, it is positive before and negative after

5

(forbidden value)

3) Let's draw up the sign table:

x	$-∞$		$-2$		$5$		$+∞$
$3 \cdot (- x + 5) + 3 \cdot (x + 2)$		+		+		+
$x + 2$		-	0	+		+
$- x + 5$		+		+	0	-
$\frac{3}{x + 2} - \frac{3}{x - 5}$		-	❌	+	❌	-

Candidates, pending the domain check:

x \in (-2, 5)

4) Let's find the domain of

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

For the expression

\frac{3}{x + 2}

to be defined, we must have:

x + 2 \neq 0

For the expression

\frac{3}{x - 5}

to exist, you must check:

x - 5 \neq 0

5) Let's solve in

ℝ

x + 2 \neq 0

ⓘ We subtract

2

from both sides

Solution found:

x \neq -2

6) Let's solve in

ℝ

x - 5 \neq 0

ⓘ We add

5

to both sides

Solution found:

x \neq 5

Let's aggregate the different solutions to find the overall solution:

x	$-∞$		$-2$		$5$		$+∞$
$x + 2 \neq 0$		✅	❌	✅	✅	✅
$x - 5 \neq 0$		✅	✅	✅	❌	✅
Domain of definition:		✅	❌	✅	❌	✅

Domain constraints:

x \in ℝ \ {-2, 5}

▶Solution:

x \in (-2, 5)