(5x+5)/6-1<(19x+1)/3-(8x+1)/4 — step-by-step solution

1) Let's solve in

ℝ

\frac{5 x + 5}{6} - 1 < \frac{19 x + 1}{3} - \frac{8 x + 1}{4}

ⓘ A fraction whose numerator is a sum splits term by term, (a + b)/c = a/c + b/c:

- \frac{8 x + 1}{4}

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

ⅰ

\frac{5 x + 5}{6} - 1

\frac{(5 x + 5) + 6 \cdot (-1)}{6}

\frac{(5 x + 5) + 6 \cdot (-1)}{6} < \frac{19 x + 1}{3} - (2 x + \frac{1}{4})

ⓘ Multiplications are done before additions and subtractions. A single minus sign does not cancel out, the product of a negative number by a positive number is negative:

6 \cdot (-1)

-6

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

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

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

\frac{5 x + 5 - 6}{6} < \frac{19 x + 1}{3} - 2 x - \frac{1}{4}

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

5 - 6

-1

ⅰ

\frac{19 x + 1}{3} - 2 x

\frac{(19 x + 1) + 3 \cdot (- 2 x)}{3}

\frac{5 x - 1}{6} < \frac{(19 x + 1) + 3 \cdot (- 2 x)}{3} - \frac{1}{4}

ⓘ Multiplications are done before additions and subtractions. A single minus sign does not cancel out, the product of a negative number by a positive number is negative:

3 \cdot (-2)

-6

\frac{5 x - 1}{6} < \frac{19 x + 1 - 6 x}{3} - \frac{1}{4}

ⓘ Like terms are grouped by adding their coefficients:

19 x - 6 x

13 x

\frac{5 x - 1}{6} < \frac{13 x + 1}{3} - \frac{1}{4}

2) Let's add fractions:

\frac{5 x - 1}{6} - \frac{13 x + 1}{3} + \frac{1}{4} < 0

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

\frac{5 x - 1}{6} - \frac{13 x + 1}{3}

\frac{(5 x - 1) - (13 x + 1) \cdot 2}{6}

\frac{(5 x - 1) - (13 x + 1) \cdot 2}{6} + \frac{1}{4} < 0

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

\frac{5 x - 1 - (13 x + 1) \cdot 2}{6} + \frac{1}{4}

\frac{(5 x - 1 - (13 x + 1) \cdot 2) \cdot 2 + 3}{6 \cdot 2}

\frac{(5 x - 1 - (13 x + 1) \cdot 2) \cdot 2 + 3}{6 \cdot 2} < 0

ⅰ

6 \cdot 2

12

\frac{(5 x - 1 - (13 x + 1) \cdot 2) \cdot 2 + 3}{12} < 0

Let's study the sign of

(5 x - 1 - (13 x + 1) \cdot 2) \cdot 2 + 3

by solving in

ℝ

(5 x - 1 - (13 x + 1) \cdot 2) \cdot 2 + 3 < 0

ⓘ Expand by multiplying each term inside the parentheses:

- (13 x + 1) \cdot 2

- (26 x + 2)

(5 x - 1 - (26 x + 2)) \cdot 2

10 x - 2 - (26 x + 2) \cdot 2

10 x

- 2

- (26 x + 2) \cdot 2

+ 3 < 0

ⓘ Start from

-2

and move forward

3

on the number line, going past zero:

- 2 + 3

1

10 x + 1 - (26 x + 2) \cdot 2 < 0

ⓘ Expand by multiplying each term inside the parentheses:

- (26 x + 2) \cdot 2

- (52 x + 4)

10 x + 1 - (52 x + 4) < 0

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

- (52 x + 4)

- 52 x - 4

10 x

+ 1

- 52 x

- 4 < 0

ⓘ Start from

1

and move back

4

on the number line, going below zero:

1 - 4

-3

■Like terms are grouped by adding their coefficients:

10 x - 52 x

-42 x

- 42 x - 3 < 0

ⓘ We add

3

to both sides

- 42 x < 3

We divide both sides by

-42

; since this number is negative, we reverse the inequality sign

x > \frac{3}{-42}

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

\frac{3}{-42}

- \frac{3}{42}

x > - \frac{3}{42}

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

3

, their greatest common divisor:

- \frac{3}{42}

- \frac{1 \cdot 3}{14 \cdot 3}

- \frac{1}{14}

Solution found:

x > - \frac{1}{14}

⇥So the expression

(5 x - 1 - (13 x + 1) \cdot 2) \cdot 2 + 3

is negative on

(- \frac{1}{14}, +∞)

and positive or zero otherwise

▶Solution found:

x > - \frac{1}{14}

3) The numeric value of:

x > - \frac{1}{14}

is:

▷

x > -7.1429 \times 10^{- 2} (approximate value)