||2х+3|-4|=х
Let's solve the equation:
\[ |\,|2x + 3| - 4| = x \]
Step 1: Consider the inner absolute value \( |2x + 3| \)
Define: \[ A = |2x + 3| \]
The equation becomes: \[ |A - 4| = x \]
Since \(A \geq 0\), the possible cases are:
Case 1: \(A - 4 \geq 0 \Rightarrow A \geq 4 \)
In this case: \[ |A - 4| = A - 4 \] So: \[ A - 4 = x \]
Recall that \(A = |2x + 3|\), so: \[ |2x + 3| = A \]
But since \(A = x + 4\), we have: \[ |2x + 3| = x + 4 \]
Now, consider two subcases:
Subcase 1.1: \(2x + 3 \geq 0\)
\[ 2x + 3 \geq 0 \Rightarrow x \geq -\frac{3}{2} \] Then: \[ 2x + 3 = x + 4 \] \[ 2x + 3 = x + 4 \] \[ 2x - x = 4 - 3 \] \[ x = 1 \]
Check if \(A = |2x + 3| \geq 4\):
\[ |2(1) + 3| = |2 + 3| = 5 \geq 4 \]
Yes, so \(x = 1\) is a valid solution.
Subcase 1.2: \(2x + 3 < 0\)
\[ 2x + 3 < 0 \Rightarrow x < -\frac{3}{2} \] In this case: \[ |2x + 3| = -(2x + 3) = -2x - 3 \] and the equation becomes: \[ A = x + 4 \] \[ -2x - 3 = x + 4 \] \[ -2x - 3 = x + 4 \] \[ -2x - x = 4 + 3 \] \[ -3x = 7 \] \[ x = -\frac{7}{3} \]
Check whether this satisfies the condition \(x < -\frac{3}{2}\):
\[ -\frac{7}{3} \approx -2.3333 < -1.5 \] Yes, so:
\[ x = -\frac{7}{3} \]
Now verify \(A = |2x + 3|\):
\[ 2 \times \left(-\frac{7}{3}\right) + 3 = -\frac{14}{3} + 3 = -\frac{14}{3} + \frac{9}{3} = -\frac{5}{3} \] \[ |2x + 3| = \frac{5}{3} \] and check if \(A \geq 4\):
\[ \frac{5}{3} \approx 1.6667 < 4 \] But this does not satisfy the earlier assumption \(A \geq 4\), so discard this solution.
Case 2: \(A - 4 < 0 \Rightarrow A < 4\)
In this case: \[ |A - 4| = 4 - A \] The equation becomes: \[ 4 - A = x \] But \(A = |2x + 3|\): \[ 4 - |2x + 3| = x \] which implies: \[ |2x + 3| = 4 - x \]
Since absolute value is non-negative, then: \[ 4 - x \geq 0 \Rightarrow x \leq 4 \]
Similarly, the cases within this are:
Subcase 2.1: \(2x + 3 \geq 0\)
\[ 2x + 3 \geq 0 \Rightarrow x \geq -\frac{3}{2} \]
The equation becomes: \[ 2x + 3 = 4 - x \]
Solve: \[ 2x + 3 = 4 - x \] \[ 2x + x = 4 - 3 \] \[ 3x = 1 \] \[ x = \frac{1}{3} \]
Check whether \(x \geq -\frac{3}{2}\):
\[ \frac{1}{3} \approx 0.333 \geq -1.5 \] Yes, so the condition holds.
Verify if \(A = |2x + 3|\):
\[ |2 \times \frac{1}{3} + 3| = |\frac{2}{3} + 3| = |\frac{2}{3} + \frac{9}{3}| = |\frac{11}{3}| = \frac{11}{3} \] Check if \(A < 4\):
\[ \frac{11}{3} \approx 3.6667 < 4 \] Yes. Therefore, \(x = \frac{1}{3}\) is a valid solution.
Subcase 2.2: \(2x + 3 < 0\)
\[ 2x + 3 < 0 \Rightarrow x < -\frac{3}{2} \]
Now, the equation:
\[ |2x + 3| = -(2x + 3) = -2x - 3 \]
And:
\[ 4 - |2x + 3| = x \]
Thus:
\[ 4 - (-2x -3) = x \] \[ 4 + 2x + 3 = x \] \[ 2x + 7 = x \] \[ 2x - x = -7 \] \[ x = -7 \]
Check whether \(x < -\frac{3}{2}\):
\[ -7 < -1.5 \] Yes, so this is valid.
Now verify \(A = |2x+3|\):
\[ 2 \times (-7) + 3 = -14 + 3 = -11 \] \[ |2x + 3| = 11 \]
Check if \(A < 4\):
\[ 11 < 4 \quad \text{? No} \] It does not satisfy the earlier assumption \(A < 4\), so discard.
Final solutions:
\[ \boxed{ x = 1, \quad x = \frac{1}{3} } \]
Summary:
The solutions to the equation \( |\,|2x + 3| - 4| = x \) are:
\[ \boxed{ x = 1 \quad \text{and} \quad x = \frac{1}{3} } \]
If you have further questions or need clarification, feel free to ask!
