(2x)/3-((x-1))/4 <((x-1))/8 — step-by-step solution

1) Let's solve in

ℝ

\frac{2 x}{3} - \frac{x - 1}{4} < \frac{x - 1}{8}

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

- \frac{x - 1}{4}

\frac{- x + 1}{4}

\frac{2 x}{3} + \frac{- x + 1}{4} < \frac{x - 1}{8}

2) Let's add fractions:

\frac{2 x}{3} + \frac{- x + 1}{4} - \frac{x - 1}{8} < 0

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

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

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

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

- \frac{x - 1}{8}

\frac{- x + 1}{8}

\frac{2 x \cdot 4 + (- x + 1) \cdot 3}{3 \cdot 4} + \frac{- x + 1}{8} < 0

ⓘ Multiplications are done before additions and subtractions:

2 \cdot 4

8

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

\frac{8 x + (- x + 1) \cdot 3}{12} + \frac{- x + 1}{8}

\frac{(8 x + (- x + 1) \cdot 3) \cdot 2 + (- x + 1) \cdot 3}{12 \cdot 2}

ⅰ

3 \cdot 4

12

\frac{(8 x + (- x + 1) \cdot 3) \cdot 2 + (- x + 1) \cdot 3}{12 \cdot 2} < 0

ⅰ

12 \cdot 2

(10 + 2) \cdot 2

10 \cdot 2 + 2 \cdot 2

20 + 4

24

\frac{(8 x + (- x + 1) \cdot 3) \cdot 2 + (- x + 1) \cdot 3}{24} < 0

Let's study the sign of

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

by solving in

ℝ

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

ⓘ Expand by multiplying each term inside the parentheses:

(- x + 1) \cdot 3

- x \cdot 3 + 3

(8 x + (- x \cdot 3 + 3)) \cdot 2

16 x + (- x \cdot 3 + 3) \cdot 2

16 x

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

- x \cdot 3

+ 3 < 0

ⓘ Like terms are grouped by adding their coefficients:

16 x + (- x \cdot 3 + 3) \cdot 2 - x \cdot 3

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

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

ⓘ Expand by multiplying each term inside the parentheses:

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

6 \cdot (- x) + 6

13 x

+ 6 \cdot (- x)

+ 6

+ 3 < 0

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

6 \cdot (- x)

- 6 x

■Start from

6

and move forward

3

on the number line:

6 + 3

9

13 x - 6 x + 9 < 0

ⓘ Like terms are grouped by adding their coefficients:

13 x - 6 x

7 x

7 x + 9 < 0

7 x < -9

ⓘ Both sides are divided by

7

Solution obtained:

x < - \frac{9}{7}

⇥So the expression

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

is negative on

(-∞, - \frac{9}{7})

and positive or zero otherwise

▶Solution obtained:

x < - \frac{9}{7}

3) The numeric value of:

x < - \frac{9}{7}

equals:

▷

x < -1.2857 (approximate value)