what is the solution of $3+\frac{x - 2}{x - 3}leq4$?\n$x<3$\n$xleq3$\n$x>3$\n$xgeq3$

Answer:

A. $x < 3$

Explanation:

Step1: Isolate the fraction

Subtract 3 from both sides of $3+\frac{x - 2}{x - 3}\leq4$. $\frac{x - 2}{x - 3}\leq4 - 3$, so $\frac{x - 2}{x - 3}\leq1$.

Step2: Move 1 to the left - hand side

$\frac{x - 2}{x - 3}-1\leq0$. Rewrite 1 as $\frac{x - 3}{x - 3}$, then $\frac{x - 2-(x - 3)}{x - 3}\leq0$.

Step3: Simplify the numerator

Expand $x - 2-(x - 3)$: $x - 2 - x+3 = 1$. So we have $\frac{1}{x - 3}\leq0$.

Step4: Find the solution

For $\frac{1}{x - 3}\leq0$, since the numerator 1>0, the fraction is non - positive when the denominator $x - 3<0$. Solving $x - 3<0$ gives $x<3$.