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.