Sure, let's solve the equation [tex]\( |x - 3| = 3x + 1 \)[/tex] step-by-step.
The equation involves an absolute value, which means we need to consider two cases: one where the expression inside the absolute value is non-negative ([tex]\( x - 3 \geq 0 \)[/tex]) and one where it is negative ([tex]\( x - 3 < 0 \)[/tex]).
### Case 1: [tex]\( x - 3 \geq 0 \)[/tex]
If [tex]\( x - 3 \geq 0 \)[/tex], then [tex]\( |x - 3| = x - 3 \)[/tex]. So our equation becomes:
[tex]\[ x - 3 = 3x + 1 \][/tex]
Now, solve for [tex]\( x \)[/tex]:
[tex]\[ x - 3 = 3x + 1 \][/tex]
Subtract [tex]\( x \)[/tex] from both sides:
[tex]\[ -3 = 2x + 1 \][/tex]
Subtract 1 from both sides:
[tex]\[ -4 = 2x \][/tex]
Divide by 2:
[tex]\[ x = -2 \][/tex]
Now check if this [tex]\( x \)[/tex] satisfies the condition [tex]\( x - 3 \geq 0 \)[/tex]:
[tex]\[ -2 - 3 \geq 0 \Rightarrow -5 \geq 0 \quad (\text{False}) \][/tex]
Since this contradicts our assumption for this case, [tex]\( x = -2 \)[/tex] is not a solution in this scenario.
### Case 2: [tex]\( x - 3 < 0 \)[/tex]
If [tex]\( x - 3 < 0 \)[/tex], then [tex]\( |x - 3| = -(x - 3) = -x + 3 \)[/tex]. So our equation becomes:
[tex]\[ -x + 3 = 3x + 1 \][/tex]
Now, solve for [tex]\( x \)[/tex]:
[tex]\[ -x + 3 = 3x + 1 \][/tex]
Add [tex]\( x \)[/tex] to both sides:
[tex]\[ 3 = 4x + 1 \][/tex]
Subtract 1 from both sides:
[tex]\[ 2 = 4x \][/tex]
Divide by 4:
[tex]\[ x = \frac{1}{2} \][/tex]
Now check if this [tex]\( x \)[/tex] satisfies the condition [tex]\( x - 3 < 0 \)[/tex]:
[tex]\[ \frac{1}{2} - 3 < 0 \Rightarrow -\frac{5}{2} < 0 \quad (\text{True}) \][/tex]
This does satisfy our assumption for this case, so [tex]\( x = \frac{1}{2} \)[/tex] is a valid solution.
### Summary
The only solution to the equation [tex]\( |x - 3| = 3x + 1 \)[/tex] is:
[tex]\[ x = \frac{1}{2} \][/tex]
