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Sagot :
Sure, let's solve the given inequality step-by-step. The inequality we need to solve is:
[tex]\[ \frac{-6t}{t-2} < \frac{-30}{t-2} \][/tex]
### Step 1: Simplify by eliminating the denominators
Since both sides of the inequality have the same denominator, and [tex]\( t \neq 2 \)[/tex] (to avoid division by zero), we can safely multiply both sides of the inequality by [tex]\( t-2 \)[/tex] without changing the direction of the inequality (as long as [tex]\( t-2 \)[/tex] is either always positive or always negative within the range considered). It gives us:
[tex]\[ -6t < -30 \][/tex]
### Step 2: Solve the simplified inequality
Now, we'll isolate [tex]\( t \)[/tex]:
[tex]\[ -6t < -30 \][/tex]
To solve for [tex]\( t \)[/tex], divide both sides by [tex]\(-6\)[/tex]. Remember, dividing by a negative number reverses the inequality:
[tex]\[ t > 5 \][/tex]
### Step 3: Consider the critical point and restrictions
The original inequality's denominator, [tex]\( t-2 \)[/tex], cannot be zero. Thus, [tex]\( t \neq 2 \)[/tex] is an important restriction.
### Step 4: Combine the intervals
To find the solution for [tex]\( t \)[/tex]:
1. [tex]\( t > 5 \)[/tex] comes from solving [tex]\( -6t < -30 \)[/tex].
2. [tex]\( t \neq 2 \)[/tex] because the original inequality is undefined at [tex]\( t = 2 \)[/tex].
However, we must also consider the direction of the inequality and the number line's behavior around the critical point.
### Step 5: Test intervals around the critical points
When [tex]\( t < 2 \)[/tex], for a value like [tex]\( t = 1 \)[/tex]:
[tex]\[ \frac{-6 \cdot 1}{1-2} = \frac{-6}{-1} = 6 \][/tex]
And,
[tex]\[ \frac{-30}{1-2} = \frac{-30}{-1} = 30 \][/tex]
So, [tex]\( 6 < 30 \)[/tex] is true in this interval, meaning [tex]\( t < 2 \)[/tex] is also part of the solution.
When [tex]\( 2 < t < 5 \)[/tex], the inequality doesn’t hold:
For a value like [tex]\( t = 3 \)[/tex]:
[tex]\[ \frac{-6 \cdot 3}{3-2} = -18 \][/tex]
And,
[tex]\[ \frac{-30}{3-2} = -30 \][/tex]
So, [tex]\( -18 < -30 \)[/tex] is false.
So, this region [tex]\( 2 < t < 5 \)[/tex] doesn't work.
### Step 6: Conclude the final solution
Combining the intervals and considering the domain where both the inequality and denominator restrictions are satisfied, we have:
[tex]\[ (-\infty, 2) \cup (5, \infty) \][/tex]
Thus, the solution to the inequality [tex]\(\frac{-6t}{t-2} < \frac{-30}{t-2}\)[/tex] is:
[tex]\[ t \in (-\infty, 2) \cup (5, \infty) \][/tex]
[tex]\[ \frac{-6t}{t-2} < \frac{-30}{t-2} \][/tex]
### Step 1: Simplify by eliminating the denominators
Since both sides of the inequality have the same denominator, and [tex]\( t \neq 2 \)[/tex] (to avoid division by zero), we can safely multiply both sides of the inequality by [tex]\( t-2 \)[/tex] without changing the direction of the inequality (as long as [tex]\( t-2 \)[/tex] is either always positive or always negative within the range considered). It gives us:
[tex]\[ -6t < -30 \][/tex]
### Step 2: Solve the simplified inequality
Now, we'll isolate [tex]\( t \)[/tex]:
[tex]\[ -6t < -30 \][/tex]
To solve for [tex]\( t \)[/tex], divide both sides by [tex]\(-6\)[/tex]. Remember, dividing by a negative number reverses the inequality:
[tex]\[ t > 5 \][/tex]
### Step 3: Consider the critical point and restrictions
The original inequality's denominator, [tex]\( t-2 \)[/tex], cannot be zero. Thus, [tex]\( t \neq 2 \)[/tex] is an important restriction.
### Step 4: Combine the intervals
To find the solution for [tex]\( t \)[/tex]:
1. [tex]\( t > 5 \)[/tex] comes from solving [tex]\( -6t < -30 \)[/tex].
2. [tex]\( t \neq 2 \)[/tex] because the original inequality is undefined at [tex]\( t = 2 \)[/tex].
However, we must also consider the direction of the inequality and the number line's behavior around the critical point.
### Step 5: Test intervals around the critical points
When [tex]\( t < 2 \)[/tex], for a value like [tex]\( t = 1 \)[/tex]:
[tex]\[ \frac{-6 \cdot 1}{1-2} = \frac{-6}{-1} = 6 \][/tex]
And,
[tex]\[ \frac{-30}{1-2} = \frac{-30}{-1} = 30 \][/tex]
So, [tex]\( 6 < 30 \)[/tex] is true in this interval, meaning [tex]\( t < 2 \)[/tex] is also part of the solution.
When [tex]\( 2 < t < 5 \)[/tex], the inequality doesn’t hold:
For a value like [tex]\( t = 3 \)[/tex]:
[tex]\[ \frac{-6 \cdot 3}{3-2} = -18 \][/tex]
And,
[tex]\[ \frac{-30}{3-2} = -30 \][/tex]
So, [tex]\( -18 < -30 \)[/tex] is false.
So, this region [tex]\( 2 < t < 5 \)[/tex] doesn't work.
### Step 6: Conclude the final solution
Combining the intervals and considering the domain where both the inequality and denominator restrictions are satisfied, we have:
[tex]\[ (-\infty, 2) \cup (5, \infty) \][/tex]
Thus, the solution to the inequality [tex]\(\frac{-6t}{t-2} < \frac{-30}{t-2}\)[/tex] is:
[tex]\[ t \in (-\infty, 2) \cup (5, \infty) \][/tex]
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