If y varies directly as x and inversely as z, given y = 36 when x = 10 and z = 5, find y if x = 3 and z =15

Since \( y \) varies directly as \( x \) and inversely as \( z \), we can express the relationship as:

\[
y = k \frac{x}{z}
\]

where \( k \) is the constant of proportionality.

### Step 1: Find \( k \) using the given values
We're given:
- \( y = 36 \)
- \( x = 10 \)
- \( z = 5 \)

Plugging these values into the equation:

\[
36 = k \frac{10}{5}
\]

Simplify the fraction:

\[
36 = k \cdot 2
\]

Now, solve for \( k \):

\[
k = \frac{36}{2} = 18
\]

So, \( k = 18 \).

### Step 2: Find \( y \) when \( x = 3 \) and \( z = 15 \)
Now we use the equation with the new values for \( x \) and \( z \):

\[
y = 18 \frac{3}{15}
\]

Simplify \( \frac{3}{15} \) to \( \frac{1}{5} \):

\[
y = 18 \cdot \frac{1}{5} = \frac{18}{5} = 3.6
\]

### Answer
When \( x = 3 \) and \( z = 15 \), \( y = 3.6 \).

Would you like further clarification on any part? 

#### Related Questions
1. What would \( y \) be if \( x = 6 \) and \( z = 10 \)?
2. How would \( y \) change if \( z \) were doubled, keeping \( x \) constant?
3. What if \( y \) varies directly as \( x \) and directly as \( z \) as well?
4. How would the solution change if \( y \) were inversely proportional to both \( x \) and \( z \)?
5. If \( y \) varies directly as \( x \) squared, how would that affect the formula?

#### Tip:
For direct and inverse variations, always isolate the constant \( k \) first using initial conditions; then apply it to find unknown values.

If y varies directly as x and inversely as z, given y = 36 when x = 10 and z = 5, find y if x = 3 and z = 15

Learn how to solve a direct and inverse variation problem where y varies directly as x and inversely as z. Given y = 36 when x = 10 and z = 5, find y when x = 3 and z = 15. Understand key concepts like direct and inverse proportionality with a step-by-step solution. Suitable for students in Grades 8-10.

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