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

1) Let's solve in

ℝ

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

2) Let's add fractions:

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

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

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

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

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

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

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

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

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

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

We study the sign of the numerator

Let's study the sign of

2 x

+ 3

- 5 x \cdot (x - 2)

+ 2 \cdot (x - 2)

by solving in

ℝ

2 x

+ 3

- 5 x \cdot (x - 2)

+ 2 \cdot (x - 2) \leq 0

ⓘ Expand by multiplying each term inside the parentheses:

- 5 x \cdot (x - 2)

- (5 x^{2} - 10 x)

2 \cdot (x - 2)

2 x - 4

2 x

+ 3

- (5 x^{2} - 10 x)

+ 2 x

- 4 \leq 0

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

- (5 x^{2} - 10 x)

- 5 x^{2} + 10 x

■Start from

3

and move back

4

on the number line, going below zero:

3 - 4

-1

■Like terms are grouped by adding their coefficients:

2 x + 2 x

4 x

4 x

- 1

- 5 x^{2}

+ 10 x \leq 0

ⓘ Like terms are grouped by adding their coefficients:

4 x + 10 x

14 x

14 x - 1 - 5 x^{2} \leq 0

Given the quadratic form

a x^{2} + b x + c

, let's compute the discriminant:

Δ = b^{2} - 4 a c

. Here

a = -5

b = 14

and

c = -1

Δ = 14^{2} - 4 \cdot (-5) \cdot (-1)

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

14^{2}

14 \cdot 14

196

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

-4 \cdot (-5)

20

■Multiplying by -1 gives the opposite:

20 \cdot (-1)

-20

Δ = 196 - 20

ⓘ Start from

196

and move back

20

on the number line:

196 - 20

176

Δ = 176

Since

Δ > 0

, the equation has two roots:

The quadratic formula gives

x_{1}

\frac{- b - \sqrt{Δ}}{2 a}

\frac{- 14 - \sqrt{176}}{2 \cdot (-5)}

x_{1} = \frac{- 14 - \sqrt{176}}{2 \cdot (-5)}

ⓘ We look for the largest square dividing

176

16

4^{2}

, and take it out of the root:

\sqrt{176}

\sqrt{16 \cdot 11}

\sqrt{16} \cdot \sqrt{11}

4 \cdot \sqrt{11}

■The minus sign is factored out, changing the sign of each term:

- 14 - 4 \cdot \sqrt{11}

- (14 + 4 \cdot \sqrt{11})

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

2 \cdot (-5)

-10

x_{1} = \frac{- (14 + 4 \cdot \sqrt{11})}{-10}

ⓘ The minus signs of the numerator and the denominator cancel out:

\frac{- (14 + 4 \cdot \sqrt{11})}{-10}

\frac{14 + 4 \cdot \sqrt{11}}{10}

x_{1} = \frac{14 + 4 \cdot \sqrt{11}}{10}

ⓘ A fraction is simplified by dividing the numerator and the denominator by the same number:

\frac{14 + 4 \cdot \sqrt{11}}{10}

\frac{7 + 2 \cdot \sqrt{11}}{5}

x_{1} = \frac{7 + 2 \cdot \sqrt{11}}{5}

The quadratic formula gives

x_{2}

\frac{- b + \sqrt{Δ}}{2 a}

\frac{- 14 + \sqrt{176}}{2 \cdot (-5)}

x_{2} = \frac{- 14 + \sqrt{176}}{2 \cdot (-5)}

ⓘ We look for the largest square dividing

176

16

4^{2}

, and take it out of the root:

\sqrt{176}

\sqrt{16 \cdot 11}

\sqrt{16} \cdot \sqrt{11}

4 \cdot \sqrt{11}

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

2 \cdot (-5)

-10

x_{2} = \frac{- 14 + 4 \cdot \sqrt{11}}{-10}

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

\frac{- 14 + 4 \cdot \sqrt{11}}{-10}

- (\frac{- 14 + 4 \cdot \sqrt{11}}{10})

x_{2} = - (\frac{- 14 + 4 \cdot \sqrt{11}}{10})

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

- (\frac{- 14 + 4 \cdot \sqrt{11}}{10})

\frac{14 - 4 \cdot \sqrt{11}}{10}

x_{2} = \frac{14 - 4 \cdot \sqrt{11}}{10}

ⓘ A fraction is simplified by dividing the numerator and the denominator by the same number:

\frac{14 - 4 \cdot \sqrt{11}}{10}

\frac{7 - 2 \cdot \sqrt{11}}{5}

x_{2} = \frac{7 - 2 \cdot \sqrt{11}}{5}

The quadratic has the sign of a =

-5

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 (-∞, \frac{7 - 2 \cdot \sqrt{11}}{5}] \cup [\frac{7 + 2 \cdot \sqrt{11}}{5}, +∞)

⇥So the expression

2 x

+ 3

- 5 x \cdot (x - 2)

+ 2 \cdot (x - 2)

is negative or zero on

(-∞, \frac{7 - 2 \cdot \sqrt{11}}{5}] \cup [\frac{7 + 2 \cdot \sqrt{11}}{5}, +∞)

and positive otherwise

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

⇥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	$-∞$		$\frac{7 - 2 \cdot \sqrt{11}}{5}$		$2$		$\frac{7 + 2 \cdot \sqrt{11}}{5}$		$+∞$
$2 x$ $+ 3$ $- 5 x \cdot (x - 2)$ $+ 2 \cdot (x - 2)$		-	0	+		+	0	-
$x - 2$		-		-	0	+		+
$\frac{2 x + 3}{x - 2} - (5 x - 2)$		+	0	-	❌	+	0	-

Candidates, pending the domain check:

x \in [\frac{7 - 2 \cdot \sqrt{11}}{5}, 2) \cup [\frac{7 + 2 \cdot \sqrt{11}}{5}, +∞)

4) Let's find the domain of

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

For the expression

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

to exist, you must check:

x - 2 \neq 0

ⓘ We add

2

to both sides

Domain constraints:

x \neq 2

▶Solution:

x \in [\frac{7 - 2 \cdot \sqrt{11}}{5}, 2) \cup [\frac{7 + 2 \cdot \sqrt{11}}{5}, +∞)

5) The numeric value of:

x \in [\frac{7 - 2 \cdot \sqrt{11}}{5}, 2) \cup [\frac{7 + 2 \cdot \sqrt{11}}{5}, +∞)

equals:

x \in [\frac{7 - 2 \cdot \approx 3.3166}{5}, 2) \cup [\frac{7 + 2 \cdot \approx 3.3166}{5}, +∞)

x \in [\frac{7 - \approx 6.6332}{5}, 2) \cup [\frac{7 + \approx 6.6332}{5}, +∞)

x \in [\frac{\approx 0.3668}{5}, 2) \cup [\frac{\approx 13.6332}{5}, +∞)

▷

x \in [\approx 7.335 \times 10^{- 2}, 2) \cup [\approx 2.7266, +∞)