(4x)/(x-1)<= x/(x+6)+4 — step-by-step solution

1) Let's solve in

ℝ

\frac{4 x}{x - 1} \leq \frac{x}{x + 6} + 4

2) Let's sum the fractions:

\frac{4 x}{x - 1} - \frac{x}{x + 6} - 4 \leq 0

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

\frac{4 x}{x - 1} - \frac{x}{x + 6}

\frac{4 x \cdot (x + 6) - x \cdot (x - 1)}{(x - 1) \cdot (x + 6)}

\frac{4 x \cdot (x + 6) - x \cdot (x - 1)}{(x - 1) \cdot (x + 6)} - 4 \leq 0

ⓘ What the terms have in common is factored out, so that the fraction can be simplified:

\frac{4 x \cdot (x + 6) - x \cdot (x - 1)}{(x - 1) \cdot (x + 6)}

\frac{x \cdot (4 \cdot (x + 6) - (x - 1))}{(x - 1) \cdot (x + 6)}

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

\frac{x \cdot (4 \cdot (x + 6) - (x - 1))}{(x - 1) \cdot (x + 6)} - 4

\frac{x \cdot (4 \cdot (x + 6) - (x - 1)) - 4 \cdot (x - 1) \cdot (x + 6)}{(x - 1) \cdot (x + 6)}

\frac{x \cdot (4 \cdot (x + 6) - (x - 1)) - 4 \cdot (x - 1) \cdot (x + 6)}{(x - 1) \cdot (x + 6)} \leq 0

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

- (x - 1)

- x + 1

\frac{x \cdot (4 \cdot (x + 6) - x + 1) - 4 \cdot (x - 1) \cdot (x + 6)}{(x - 1) \cdot (x + 6)} \leq 0

We study the sign of the numerator

⇥2.1) Let's study the sign of

x \cdot (4 \cdot (x + 6) - x + 1) - 4 \cdot (x - 1) \cdot (x + 6)

by solving in

ℝ

x \cdot (4 \cdot (x + 6) - x + 1) - 4 \cdot (x - 1) \cdot (x + 6) \leq 0

⇥ⓘ Expand by multiplying each term inside the parentheses:

4 \cdot (x + 6)

4 x + 24

x \cdot ((4 x + 24) - x + 1)

x \cdot (4 x + 24) + x \cdot (- x) + x

4 \cdot (x - 1)

4 x - 4

⇥

x \cdot (4 x + 24)

+ x \cdot (- x)

+ x

- (4 x - 4) \cdot (x + 6) \leq 0

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

x \cdot (- x)

- x \cdot x

⇥

x \cdot (4 x + 24)

- x \cdot x

+ x

- (4 x - 4) \cdot (x + 6) \leq 0

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

x \cdot x

x^{2}

⇥

x \cdot (4 x + 24)

- x^{2}

+ x

- (4 x - 4) \cdot (x + 6) \leq 0

⇥ⓘ Expand by multiplying each term inside the parentheses:

x \cdot (4 x + 24)

4 x^{2} + x \cdot 24

- (4 x - 4) \cdot (x + 6)

- (4 \cdot x \cdot x

+ 4 x \cdot 6

- 4 x

- 4 \cdot 6)

⇥

4 x^{2}

+ x \cdot 24

- x^{2}

+ x

- (4 \cdot x \cdot x

+ 4 x \cdot 6

- 4 x

- 4 \cdot 6) \leq 0

⇥ⓘ Like terms are grouped by adding their coefficients:

4 x^{2} + x \cdot 24 - x^{2}

3 x^{2} + 24 x

24 x + x

25 x

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

- (4 \cdot x \cdot x

+ 4 x \cdot 6

- 4 x

- 4 \cdot 6)

- 4 \cdot x \cdot x

- 4 x \cdot 6

+ 4 x

+ 4 \cdot 6

⇥

3 x^{2}

+ 25 x

- 4 \cdot x \cdot x

- 4 x \cdot 6

+ 4 x

+ 4 \cdot 6 \leq 0

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

x \cdot x

x^{2}

■Multiplications are done before additions and subtractions:

4 \cdot 6

24

■Like terms are grouped by adding their coefficients:

25 x + 4 x

29 x

⇥

3 x^{2}

+ 29 x

- 4 x^{2}

- 24 x

+ 24 \leq 0

⇥ⓘ Like terms are grouped by adding their coefficients:

3 x^{2} - 4 x^{2}

- x^{2}

29 x - 24 x

5 x

⇥

- x^{2} + 5 x + 24 \leq 0

⇥Given the quadratic form

a x^{2} + b x + c

, you can calculate the discriminant:

Δ = b^{2} - 4 a c

. Here

a = -1

b = 5

and

c = 24

⇥

Δ = 5^{2} - 4 \cdot (-1) \cdot 24

⇥ⓘ Powers are computed before multiplications, additions and subtractions. An integer power is a repeated product:

5^{2}

5 \cdot 5

25

■Multiplications are done before additions and subtractions. Two minus signs cancel out, the product of two negative numbers is positive:

-4 \cdot (-1)

4

⇥

Δ = 25 + 4 \cdot 24

⇥ⓘ Multiplications are done before additions and subtractions:

4 \cdot 24

4 \cdot (20 + 4)

4 \cdot 20 + 4 \cdot 4

80 + 16

96

⇥

Δ = 25 + 96

⇥ⓘ Start from

25

and move forward

96

on the number line:

25 + 96

121

⇥

Δ = 121

⇥As

Δ > 0

, the equation has two roots:

⇥The quadratic formula gives

x_{1}

\frac{- b - \sqrt{Δ}}{2 a}

\frac{- 5 - \sqrt{121}}{2 \cdot (-1)}

⇥

x_{1} = \frac{- 5 - \sqrt{121}}{2 \cdot (-1)}

⇥ⓘ The square root of a number is the positive number whose square is that number:

\sqrt{121}

11

;

11^{2}

121

■A single minus sign does not cancel out, the product of a negative number by a positive number is negative:

2 \cdot (-1)

-2

⇥

x_{1} = \frac{- 5 - 11}{-2}

⇥ⓘ The sign of a fraction goes in front of the bar, whether it comes from the numerator or the denominator:

\frac{- 5 - 11}{-2}

- (\frac{- 5 - 11}{2})

⇥

x_{1} = - (\frac{- 5 - 11}{2})

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

- 5 - 11

-16

⇥

x_{1} = - (\frac{-16}{2})

⇥ⓘ Two minus signs cancel out:

- (\frac{-16}{2})

\frac{16}{2}

⇥

x_{1} = \frac{16}{2}

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

2

, their greatest common divisor:

\frac{16}{2}

8

⇥

x_{1} = 8

⇥The quadratic formula gives

x_{2}

\frac{- b + \sqrt{Δ}}{2 a}

\frac{- 5 + \sqrt{121}}{2 \cdot (-1)}

⇥

x_{2} = \frac{- 5 + \sqrt{121}}{2 \cdot (-1)}

⇥ⓘ The square root of a number is the positive number whose square is that number:

\sqrt{121}

11

;

11^{2}

121

■A single minus sign does not cancel out, the product of a negative number by a positive number is negative:

2 \cdot (-1)

-2

⇥

x_{2} = \frac{- 5 + 11}{-2}

⇥ⓘ The sign of a fraction goes in front of the bar, whether it comes from the numerator or the denominator:

\frac{- 5 + 11}{-2}

- (\frac{- 5 + 11}{2})

⇥

x_{2} = - (\frac{- 5 + 11}{2})

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

- 5 + 11

6

⇥

x_{2} = - \frac{6}{2}

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

2

, their greatest common divisor:

- \frac{6}{2}

-3

⇥

x_{2} = -3

⇥The quadratic has the sign of a =

-1

outside its roots, so it is indeed negative outside its roots; the roots, where it is zero, are included.

⇥Candidates, pending the domain check:

x \in (-∞, -3] \cup [8, +∞)

⇥So the expression

x \cdot (4 \cdot (x + 6) - x + 1) - 4 \cdot (x - 1) \cdot (x + 6)

is negative or zero on

(-∞, -3] \cup [8, +∞)

and positive otherwise

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

⇥2.2) Let's solve in

ℝ

(x - 1) \cdot (x + 6) < 0

⇥We study the sign of each factor of the product

⇥⇥The expression

x - 1

is zero at

x = 1

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

⇥⇥The expression

x + 6

is zero at

x = -6

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

3) Let's draw up the sign table:

x	$-∞$		$-6$		$-3$		$1$		$8$		$+∞$
$x \cdot (4 \cdot (x + 6) - x + 1) - 4 \cdot (x - 1) \cdot (x + 6)$		-		-	0	+		+	0	-
$x - 1$		-		-		-	0	+		+
$x + 6$		-	0	+		+		+		+
$\frac{4 x}{x - 1} - (\frac{x}{x + 6} + 4)$		-	❌	+	0	-	❌	+	0	-

Candidates, pending the domain check:

x \in (-∞, -6) \cup [-3, 1) \cup [8, +∞)

4) Let's find the domain of

\frac{4 x}{x - 1} \leq \frac{x}{x + 6} + 4

For the expression

\frac{4 x}{x - 1}

to exist, you must check:

x - 1 \neq 0

For the expression

\frac{x}{x + 6}

to exist, you must check:

x + 6 \neq 0

5) Let's solve in

ℝ

x - 1 \neq 0

ⓘ We add

1

to both sides

Solution found:

x \neq 1

6) Let's solve in

ℝ

x + 6 \neq 0

ⓘ We subtract

6

from both sides

Solution found:

x \neq -6

Let's create the table of the different solutions to find the overall solution:

x	$-∞$		$-6$		$1$		$+∞$
$x - 1 \neq 0$		✅	✅	✅	❌	✅
$x + 6 \neq 0$		✅	❌	✅	✅	✅
Domain of definition:		✅	❌	✅	❌	✅

Domain constraints:

x \in ℝ \ {-6, 1}

▶Solution:

x \in (-∞, -6) \cup [-3, 1) \cup [8, +∞)