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Which of the following are solutions to [tex]|x+4|=3x-5[/tex]? Check all that apply.

A. [tex]x=\frac{1}{4}[/tex]
B. [tex]x=-\frac{1}{4}[/tex]
C. [tex]x=\frac{9}{2}[/tex]
D. [tex]x=-\frac{9}{2}[/tex]


Sagot :

To solve the equation [tex]\(|x+4| = 3x - 5\)[/tex], we need to consider the definition of the absolute value function, which implies two possible cases:

### Case 1: [tex]\(x + 4 \geq 0\)[/tex]
If [tex]\(x + 4 \geq 0\)[/tex], then [tex]\(|x + 4| = x + 4\)[/tex]. So the equation becomes:
[tex]\[ x + 4 = 3x - 5 \][/tex]

Solve for [tex]\(x\)[/tex]:
[tex]\[ x + 4 = 3x - 5 \][/tex]
Subtract [tex]\(x\)[/tex] from both sides:
[tex]\[ 4 = 2x - 5 \][/tex]
Add 5 to both sides:
[tex]\[ 9 = 2x \][/tex]
Divide by 2:
[tex]\[ x = \frac{9}{2} \][/tex]

Check if [tex]\(\frac{9}{2} + 4 \geq 0\)[/tex]:
[tex]\[ \frac{9}{2} + 4 = \frac{9}{2} + \frac{8}{2} = \frac{17}{2} \][/tex]
Since [tex]\(\frac{17}{2} > 0\)[/tex], [tex]\(x = \frac{9}{2}\)[/tex] is a valid solution for this case.

### Case 2: [tex]\(x + 4 < 0\)[/tex]
If [tex]\(x + 4 < 0\)[/tex], then [tex]\(|x + 4| = -(x + 4)\)[/tex]. So the equation becomes:
[tex]\[ -(x + 4) = 3x - 5 \][/tex]

Solve for [tex]\(x\)[/tex]:
[tex]\[ -x - 4 = 3x - 5 \][/tex]
Add [tex]\(x\)[/tex] to both sides:
[tex]\[ -4 = 4x - 5 \][/tex]
Add 5 to both sides:
[tex]\[ 1 = 4x \][/tex]
Divide by 4:
[tex]\[ x = \frac{1}{4} \][/tex]

Check if [tex]\(\frac{1}{4} + 4 < 0\)[/tex]:
[tex]\[ \frac{1}{4} + 4 = \frac{1}{4} + \frac{16}{4} = \frac{17}{4} \][/tex]
Since [tex]\(\frac{17}{4} > 0\)[/tex], [tex]\(x = \frac{1}{4}\)[/tex] is not a valid solution for this case.

Therefore, the correct solution is:
- [tex]\(x = \frac{9}{2}\)[/tex]

Let's check the provided options again:

A. [tex]\(x = \frac{1}{4}\)[/tex] - Not a solution
B. [tex]\(x = -\frac{1}{4}\)[/tex] - Not a solution
C. [tex]\(x = \frac{9}{2}\)[/tex] - Solution
D. [tex]\(x = -\frac{9}{2}\)[/tex] - Not a solution

Hence, the correct answer is:
- C. [tex]\(x = \frac{9}{2}\)[/tex]