To solve the inequality
[tex]\[
\frac{-x + 3}{x - 7} > 0,
\][/tex]
we need to determine the values of [tex]\( x \)[/tex] that make this expression positive. Here are the steps to solve the inequality:
### Step 1: Identify the critical points
First, we need to find the critical points where the numerator and denominator change signs.
1. Numerator: [tex]\(-x + 3 = 0\)[/tex]
Solve for [tex]\( x \)[/tex]:
[tex]\[
-x + 3 = 0 \implies x = 3
\][/tex]
2. Denominator: [tex]\( x - 7 = 0 \)[/tex]
Solve for [tex]\( x \)[/tex]:
[tex]\[
x - 7 = 0 \implies x = 7
\][/tex]
### Step 2: Determine the intervals
These critical points divide the number line into three intervals. We will test the sign of the expression in each interval:
1. [tex]\( (-\infty, 3) \)[/tex]
2. [tex]\( (3, 7) \)[/tex]
3. [tex]\( (7, \infty) \)[/tex]
### Step 3: Test each interval
To determine whether the expression is positive or negative in each interval, choose a test point in each interval and plug it into the inequality:
1. Interval [tex]\( (-\infty, 3) \)[/tex]: Test with [tex]\( x = 0 \)[/tex]
[tex]\[
\frac{-(0) + 3}{0 - 7} = \frac{3}{-7} = -\frac{3}{7} \quad (\text{Negative})
\][/tex]
2. Interval [tex]\( (3, 7) \)[/tex]: Test with [tex]\( x = 4 \)[/tex]
[tex]\[
\frac{-(4) + 3}{4 - 7} = \frac{-1}{-3} = \frac{1}{3} \quad (\text{Positive})
\][/tex]
3. Interval [tex]\( (7, \infty) \)[/tex]: Test with [tex]\( x = 8 \)[/tex]
[tex]\[
\frac{-(8) + 3}{8 - 7} = \frac{-5}{1} = -5 \quad (\text{Negative})
\][/tex]
### Step 4: Combine the intervals
The expression [tex]\(\frac{-x + 3}{x - 7}\)[/tex] is positive in the interval [tex]\( (3, 7) \)[/tex].
Finally, in interval notation, the solution to the inequality is:
[tex]\[
(3, 7)
\][/tex]
Thus, the values of [tex]\( x \)[/tex] that satisfy the inequality [tex]\(\frac{-x + 3}{x - 7} > 0\)[/tex] are in the interval:
[tex]\[
(3, 7)
\][/tex]
