What is the solution of [tex]3+\frac{x-2}{x-3} \leq 4[/tex]?

A. [tex]x \ \textless \ 3[/tex]
B. [tex]x \leq 3[/tex]
C. [tex]x \ \textgreater \ 3[/tex]
D. [tex]x \geq 3[/tex]



Answer :

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]