|10x-6|<= |2x+2| — step-by-step solution

1) Let's solve in

ℝ

| 10 x - 6 | \leq | 2 x + 2 |

Both sides are nonnegative. Comparing them is equivalent to comparing their squares, with the same comparison sign

2) Let's solve in

ℝ

{(10 x - 6)}^{2} \leq {(2 x + 2)}^{2}

{(10 x - 6)}^{2} - {(2 x + 2)}^{2} \leq 0

For

{(10 x - 6)}^{2} - {(2 x + 2)}^{2}

, we recognize a difference of two squares:

{(10 x - 6)}^{2} - {(2 x + 2)}^{2}

. Apply

a^{2} - b^{2} = (a - b) \cdot (a + b)

, with

a = 10 x - 6

and

b = 2 x + 2

You can factor:

{(10 x - 6)}^{2} - {(2 x + 2)}^{2}

into:

((10 x - 6) - (2 x + 2)) \cdot ((10 x - 6) + (2 x + 2))

3) Let's solve in

ℝ

((10 x - 6) - (2 x + 2)) \cdot ((10 x - 6) + (2 x + 2)) \leq 0

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

- (2 x + 2)

- 2 x - 2

■Parentheses are computed first:

- 6 + 2

-4

■Like terms are grouped by adding their coefficients:

10 x + 2 x

12 x

(10 x

- 6

- 2 x

- 2) \cdot (12 x - 4) \leq 0

ⓘ Parentheses are computed first:

- 6 - 2

-8

■Like terms are grouped by adding their coefficients:

10 x - 2 x

8 x

(8 x - 8) \cdot (12 x - 4) \leq 0

We study the sign of each factor of the product

⇥The expression

8 x - 8

is zero at:

⇥

x = \frac{8}{8}

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

8

, their greatest common divisor:

\frac{8}{8}

1

⇥

x = 1

⇥Its coefficient of

x

8

, being positive, it is negative before and positive after

1

⇥The expression

12 x - 4

is zero at:

⇥

x = \frac{4}{12}

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

4

, their greatest common divisor:

\frac{4}{12}

\frac{1 \cdot 4}{3 \cdot 4}

\frac{1}{3}

⇥

x = \frac{1}{3}

⇥Its coefficient of

x

12

, being positive, it is negative before and positive after

\frac{1}{3}

4) Let's draw up the sign table:

▶Solution obtained:

x \in [\frac{1}{3}, 1]

5) The numeric value of:

x \in [\frac{1}{3}, 1]

is:

▷

x \in [\approx 0.3333, 1]