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Let's go through the problems step by step and solve them. 

### Test II: Find the value of the variables of the following equations.
The formula given is \( y = kx \), where \( y \) varies directly as \( x \). 
To find:
- \( y \), multiply \( k \) and \( x \).
- \( x \), divide \( y \) by \( k \).
- \( k \), divide \( y \) by \( x \).

#### 1. Find \( y \) if \( x = 6 \) when \( k = 5 \)
The formula is \( y = kx \).
\[
y = 5 \times 6 = 30
\]
So, \( y = 30 \).

#### 2. Find \( x \) if \( y = 27 \) when \( k = 3 \)
To find \( x \), use the formula \( x = \frac{y}{k} \).
\[
x = \frac{27}{3} = 9
\]
So, \( x = 9 \).

#### 3. Find \( k \) if \( y = 48 \) when \( x = 2 \)
To find \( k \), use the formula \( k = \frac{y}{x} \).
\[
k = \frac{48}{2} = 24
\]
So, \( k = 24 \).

#### 4. Find \( k \) if \( y = 6 \) when \( x = 24 \)
Again, use \( k = \frac{y}{x} \).
\[
k = \frac{6}{24} = \frac{1}{4} = 0.25
\]
So, \( k = 0.25 \).

#### 5. Find \( x \) if \( y = 5 \) when \( k = 25 \)
Use \( x = \frac{y}{k} \).
\[
x = \frac{5}{25} = \frac{1}{5} = 0.2
\]
So, \( x = 0.2 \).

---

### Test III: In each of the following, \( y \) varies directly as \( x \). Find the value as indicated.

#### 1. If \( y = 12 \) when \( x = 4 \), find \( y \) when \( x = 12 \)
First, find \( k \) using \( y = kx \). When \( y = 12 \) and \( x = 4 \):
\[
k = \frac{12}{4} = 3
\]
Now, use this value of \( k \) to find \( y \) when \( x = 12 \):
\[
y = 3 \times 12 = 36
\]
So, \( y = 36 \).

#### 2. If \( y = 25 \) when \( x = 5 \), find \( x \) when \( y = 100 \)
First, find \( k \) using \( y = kx \). When \( y = 25 \) and \( x = 5 \):
\[
k = \frac{25}{5} = 5
\]
Now, use \( k = 5 \) to find \( x \) when \( y = 100 \):
\[
x = \frac{100}{5} = 20
\]
So, \( x = 20 \).

#### 3. If \( y = -8 \) when \( x = 2 \), find \( x \) when \( y = 24 \)
First, find \( k \) using \( y = kx \). When \( y = -8 \) and \( x = 2 \):
\[
k = \frac{-8}{2} = -4
\]
Now, use \( k = -4 \) to find \( x \) when \( y = 24 \):
\[
x = \frac{24}{-4} = -6
\]
So, \( x = -6 \).

#### 4. If \( y = 14 \) when \( x = 7 \), find \( y \) when \( x = 21 \)
First, find \( k \) using \( y = kx \). When \( y = 14 \) and \( x = 7 \):
\[
k = \frac{14}{7} = 2
\]
Now, use \( k = 2 \) to find \( y \) when \( x = 21 \):
\[
y = 2 \times 21 = 42
\]
So, \( y = 42 \).

#### 5. If \( y = -18 \) when \( x = -6 \), find \( x \) when \( y = 90 \)
First, find \( k \) using \( y = kx \). When \( y = -18 \) and \( x = -6 \):
\[
k = \frac{-18}{-6} = 3
\]
Now, use \( k = 3 \) to find \( x \) when \( y = 90 \):
\[
x = \frac{90}{3} = 30
\]
So, \( x = 30 \).

---

### Do you have any questions or would you like further details on any solution?

Here are 5 related questions to explore further:
1. What happens if the value of \( k \) is negative in direct variation?
2. How does changing the value of \( k \) affect the relationship between \( y \) and \( x \)?
3. Can you derive a formula for inverse variation from direct variation concepts?
4. How does direct variation differ from joint variation?
5. If \( y \) varies directly as \( x \), what happens if both \( x \) and \( k \) double?

**Tip:** Direct variation means that the ratio \( \frac{y}{x} \) remains constant, so understanding this ratio helps in solving any related problems.

Find the value of the variables in direct variation problems involving y = kx.

Learn how to solve problems involving direct variation, where y = kx. This includes finding values of y, x, and the constant k based on proportional relationships. Suitable for Grades 6-8.

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