6/(x-1)-6/x >= 1 — step-by-step solution

1) Let's solve in

ℝ

\frac{6}{x - 1} - \frac{6}{x} \geq 1

2) Let's sum the fractions:

\frac{6}{x - 1} - \frac{6}{x} - 1 \geq 0

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

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

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

\frac{6 x - 6 \cdot (x - 1)}{(x - 1) \cdot x} - 1 \geq 0

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

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

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

\frac{(6 x - 6 \cdot (x - 1)) - 1 \cdot (x - 1) \cdot x}{(x - 1) \cdot x} \geq 0

ⓘ Multiplying by 1 changes nothing:

1 \cdot (x - 1) \cdot x

(x - 1) \cdot x

\frac{6 x - 6 \cdot (x - 1) - (x - 1) \cdot x}{(x - 1) \cdot x} \geq 0

We study the sign of the numerator

⇥2.1) Let's study the sign of

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

by solving in

ℝ

6 x - 6 \cdot (x - 1) - (x - 1) \cdot x \geq 0

⇥ⓘ Expand by multiplying each term inside the parentheses:

- 6 \cdot (x - 1)

- (6 x - 6)

- (x - 1) \cdot x

- (x^{2} - x)

⇥

6 x - (6 x - 6) - (x^{2} - x) \geq 0

⇥

6 x

- 6 x

+ 6

- x^{2}

+ x \geq 0

⇥ⓘ Like terms are grouped by adding their coefficients:

6 x

- 6 x

+ 6

- x^{2}

+ x

x + 6 - x^{2}

⇥

x + 6 - x^{2} \geq 0

⇥Given the quadratic form

a x^{2} + b x + c

, let's compute the discriminant:

Δ = b^{2} - 4 a c

. Here

a = -1

b = 1

and

c = 6

⇥

Δ = 1 - 4 \cdot (-1) \cdot 6

⇥ⓘ 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 = 1 + 4 \cdot 6

⇥ⓘ Multiplications are done before additions and subtractions:

4 \cdot 6

24

⇥

Δ = 1 + 24

⇥ⓘ Start from

1

and move forward

24

on the number line:

1 + 24

25

⇥

Δ = 25

⇥Since

Δ > 0

, the equation has two roots:

⇥The quadratic formula gives

x_{1}

\frac{- b - \sqrt{Δ}}{2 a}

\frac{- 1 - \sqrt{25}}{2 \cdot (-1)}

⇥

x_{1} = \frac{- 1 - \sqrt{25}}{2 \cdot (-1)}

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

\sqrt{25}

5

;

5^{2}

25

■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{- 1 - 5}{-2}

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

\frac{- 1 - 5}{-2}

- (\frac{- 1 - 5}{2})

⇥

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

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

- 1 - 5

-6

⇥

x_{1} = - (\frac{-6}{2})

⇥ⓘ Two minus signs cancel out:

- (\frac{-6}{2})

\frac{6}{2}

⇥

x_{1} = \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_{1} = 3

⇥The quadratic formula gives

x_{2}

\frac{- b + \sqrt{Δ}}{2 a}

\frac{- 1 + \sqrt{25}}{2 \cdot (-1)}

⇥

x_{2} = \frac{- 1 + \sqrt{25}}{2 \cdot (-1)}

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

\sqrt{25}

5

;

5^{2}

25

■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{- 1 + 5}{-2}

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

\frac{- 1 + 5}{-2}

- (\frac{- 1 + 5}{2})

⇥

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

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

- 1 + 5

4

⇥

x_{2} = - \frac{4}{2}

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

2

, their greatest common divisor:

- \frac{4}{2}

-2

⇥

x_{2} = -2

⇥The quadratic has the opposite sign to a =

-1

between its roots, so it is indeed positive between its roots; the roots, where it is zero, are included.

⇥Candidates, pending the domain check:

x \in [-2, 3]

⇥So the expression

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

is positive or zero on

[-2, 3]

and negative otherwise

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

⇥2.2) Let's solve in

ℝ

(x - 1) \cdot x > 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

is zero at

x = 0

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

3) Let's draw up the sign table:

x	$-∞$		$-2$		$0$		$1$		$3$		$+∞$
$6 x - 6 \cdot (x - 1) - (x - 1) \cdot x$		-	0	+		+		+	0	-
$x - 1$		-		-		-	0	+		+
$x$		-		-	0	+		+		+
$(\frac{6}{x - 1} - \frac{6}{x}) - 1$		-	0	+	❌	-	❌	+	0	-

Candidates, pending the domain check:

x \in [-2, 0) \cup (1, 3]

4) Let's evaluate the domain of

\frac{6}{x - 1} - \frac{6}{x} \geq 1

For the expression

\frac{6}{x - 1}

to exist, you must check:

x - 1 \neq 0

For the expression

- \frac{6}{x}

to exist, you must check:

x \neq 0

5) Let's solve in

ℝ

x - 1 \neq 0

ⓘ We add

1

to both sides

Solution obtained:

x \neq 1

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

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

Domain constraints:

x \in ℝ \ {0, 1}

▶Solution:

x \in [-2, 0) \cup (1, 3]