|x+3|<2|x+5| — step-by-step solution

⚠ |x+3| is read as the absolute value abs(x+3)

⚠ |x+5| is read as the absolute value abs(x+5)

1) Let's solve in

ℝ

abs (x + 3) < 2 \cdot abs (x + 5)

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

2) Let's solve in

ℝ

{(x + 3)}^{2} < {(2 \cdot (x + 5))}^{2}

{(x + 3)}^{2} - {(2 \cdot (x + 5))}^{2} < 0

You can write the expression in factored form:

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

3) Let's solve in

ℝ

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

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

4) Let's solve in

ℝ

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

x + 3 - (2 x + 10) < 0

x

+ 3

- 2 x

- 10 < 0

ℹ

3 - 10

-7

‖

x - 2 x

- x

- x - 7 < 0

- x < 7

-1

is negative, you must reverse the inequality sign when dividing/multiplying by this term

x > \frac{7}{-1}

Solution found:

x > -7

5) Let's solve in

ℝ

x + 3 + 2 \cdot (x + 5) < 0

x + 3 + (2 x + 10) < 0

ℹ

3 + 10

13

‖

x + 2 x

3 x

3 x + 13 < 0

3 x < -13

Solution obtained:

x < - \frac{13}{3}

Let's build the table of the different terms to determine the global solution:

▶Solution found:

x \in (-∞, -7) \cup (- \frac{13}{3}, +∞)

6) The numeric value of:

x \in (-∞, -7) \cup (- \frac{13}{3}, +∞)

is:

▷

x \in (-∞, -7) \cup (\approx -4.3333, +∞)