(x-1)/(x+3)>(x-4)/(x+2) — step-by-step solution

1) Let's solve in

ℝ

\frac{x - 1}{x + 3} > \frac{x - 4}{x + 2}

2) Let's sum the fractions:

\frac{x - 1}{x + 3} - \frac{x - 4}{x + 2} > 0

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

- \frac{x - 4}{x + 2}

\frac{- x + 4}{x + 2}

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

\frac{x - 1}{x + 3} + \frac{- x + 4}{x + 2}

\frac{(x - 1) \cdot (x + 2) + (- x + 4) \cdot (x + 3)}{(x + 3) \cdot (x + 2)}

\frac{(x - 1) \cdot (x + 2) + (- x + 4) \cdot (x + 3)}{(x + 3) \cdot (x + 2)} > 0

We study the sign of the numerator

⇥2.1) Let's study the sign of

(x - 1) \cdot (x + 2) + (- x + 4) \cdot (x + 3)

by solving in

ℝ

(x - 1) \cdot (x + 2) + (- x + 4) \cdot (x + 3) > 0

⇥ⓘ Expand by multiplying each term inside the parentheses:

(x - 1) \cdot (x + 2)

x \cdot x

+ x \cdot 2

- 1 \cdot x

- 1 \cdot 2

(- x + 4) \cdot (x + 3)

- x \cdot x

- x \cdot 3

+ 4 x

+ 4 \cdot 3

⇥

x \cdot x

+ x \cdot 2

- 1 \cdot x

- 1 \cdot 2

- x \cdot x

- x \cdot 3

+ 4 x

+ 4 \cdot 3 > 0

⇥ⓘ Multiplying powers with the same base adds the exponents:

x \cdot x

x^{2}

■Multiplying by 1 changes nothing:

1 \cdot x

x

■Multiplications are done before additions and subtractions:

1 \cdot 2

2

4 \cdot 3

12

■Like terms are grouped by adding their coefficients:

x \cdot 2

- x \cdot x

- x \cdot 3

+ 4 x

3 x - x^{2}

⇥

x^{2}

+ 3 x

- x

- 2

- x^{2}

+ 12 > 0

⇥ⓘ Like terms are grouped by adding their coefficients:

3 x - x

2 x

x^{2} - x^{2}

0

■Start from

-2

and move forward

12

on the number line, going past zero:

- 2 + 12

10

⇥

0 + 2 x + 10 > 0

⇥ⓘ Adding 0 changes nothing:

0 + 2 x

2 x

⇥

2 x + 10 > 0

⇥ⓘ We subtract

10

from both sides

⇥

2 x > -10

⇥ⓘ Both sides are divided by

2

⇥

x > - \frac{10}{2}

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

2

, their greatest common divisor:

- \frac{10}{2}

-5

⇥Candidates, pending the domain check:

x > -5

⇥So the expression

(x - 1) \cdot (x + 2) + (- x + 4) \cdot (x + 3)

is positive on

(-5, +∞)

and negative or zero otherwise

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

⇥2.2) Let's solve in

ℝ

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

⇥We study the sign of each factor of the product

⇥⇥The expression

x + 3

is zero at

x = -3

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

⇥⇥The expression

x + 2

is zero at

x = -2

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

3) Let's draw up the sign table:

x	$-∞$		$-5$		$-3$		$-2$		$+∞$
$(x - 1) \cdot (x + 2) + (- x + 4) \cdot (x + 3)$		-	0	+		+		+
$x + 3$		-		-	0	+		+
$x + 2$		-		-		-	0	+
$\frac{x - 1}{x + 3} - \frac{x - 4}{x + 2}$		-	0	+	❌	-	❌	+

Candidates, pending the domain check:

x \in (-5, -3) \cup (-2, +∞)

4) Let's evaluate the domain of

\frac{x - 1}{x + 3} > \frac{x - 4}{x + 2}

For the expression

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

to exist, you must check:

x + 3 \neq 0

For the expression

\frac{x - 4}{x + 2}

to be defined, we must have:

x + 2 \neq 0

5) Let's solve in

ℝ

x + 3 \neq 0

ⓘ We subtract

3

from both sides

Solution obtained:

x \neq -3

6) Let's solve in

ℝ

x + 2 \neq 0

ⓘ We subtract

2

from both sides

Solution obtained:

x \neq -2

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

x	$-∞$		$-3$		$-2$		$+∞$
$x + 3 \neq 0$		✅	❌	✅	✅	✅
$x + 2 \neq 0$		✅	✅	✅	❌	✅
Domain of definition:		✅	❌	✅	❌	✅

Domain constraints:

x \in ℝ \ {-3, -2}

▶Solution:

x \in (-5, -3) \cup (-2, +∞)