Question

2 | 3x + 1 | = 4 + x

Answer 1

2 | 3x + 1 | = 4 + x

ограничения x >= -4

раскрываем модуль

1. x >= -1/3

2 ( 3x + 1 ) = 4 + x

6x + 2 = 4 + x

5x = 2

x = 2/5

2. x < -1/3

2 (- 3x - 1 ) = 4 + x

-6x - 2 = 4 + x

-7x = 6

x = - 6/7 < -1/3

ответ -6/7, 2/5

Answer 2

Ответ:

Объяснение:

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Answer 3

To solve the equation $2|3x + 1| = 4 + x$, we'll need to consider two cases: one for when $3x + 1$ is positive (or zero) and one for when it's negative.

Case 1: $3x + 1$ is non-negative (i.e., $3x + 1 \geq 0$).

In this case, we don't need to do anything with the absolute value since it's already non-negative. So the equation becomes:

$2(3x + 1) = 4 + x$

Now, let's solve for $x$:

$6x + 2 = 4 + x$

Subtract $x$ from both sides:

$5x + 2 = 4$

Subtract 2 from both sides:

$5x = 2$

Divide both sides by 5:

$x = \frac{2}{5}$

So, for this case, $x = \frac{2}{5}$.

Case 2: $3x + 1$ is negative (i.e., $3x + 1 < 0$).

In this case, we need to negate the absolute value to make it positive, so the equation becomes:

$2(-(3x + 1)) = 4 + x$

Now, let's solve for $x$:

$-2(3x + 1) = 4 + x$

Distribute the -2 on the left side:

$-6x - 2 = 4 + x$

Add $6x$ to both sides:

$-2 = 4 + 7x$

Subtract 4 from both sides:

$-6 = 7x$

Divide both sides by 7:

$x = -\frac{6}{7}$

So, for this case, $x = -\frac{6}{7}$.

Therefore, the solution to the equation $2|3x + 1| = 4 + x$ consists of two values for $x$: $x = \frac{2}{5}$ and $x = -\frac{6}{7}$.