Sure, let's solve the inequality step-by-step:
We start with the inequality:
[tex]\[ 3 + \frac{x-2}{x-3} \leq 4 \][/tex]
1. Subtract 3 from both sides to isolate the fraction:
[tex]\[ \frac{x-2}{x-3} \leq 4 - 3 \][/tex]
This simplifies to:
[tex]\[ \frac{x-2}{x-3} \leq 1 \][/tex]
2. Subtract 1 from both sides to have a common denominator:
[tex]\[ \frac{x-2}{x-3} - 1 \leq 0 \][/tex]
3. Rewrite 1 as a fraction with the denominator [tex]\(x-3\)[/tex]:
[tex]\[ \frac{x-2}{x-3} - \frac{x-3}{x-3} \leq 0 \][/tex]
Now, combine the fractions:
[tex]\[ \frac{(x-2) - (x-3)}{x-3} \leq 0 \][/tex]
Simplify the numerator:
[tex]\[ \frac{x-2 - x + 3}{x-3} \leq 0 \][/tex]
[tex]\[ \frac{1}{x-3} \leq 0 \][/tex]
4. The expression [tex]\(\frac{1}{x-3}\)[/tex] is undefined for [tex]\(x = 3\)[/tex], so [tex]\(x \neq 3\)[/tex].
5. Now consider where [tex]\(\frac{1}{x-3}\)[/tex] is less than or equal to 0. Since the numerator (1) is positive, the fraction is non-positive only when the denominator is negative.
[tex]\[ x - 3 < 0 \][/tex]
This inequality implies:
[tex]\[ x < 3 \][/tex]
So, the solution set for this inequality is:
[tex]\[ x > 3 \][/tex]
