To answer these questions, we need to understand that if [tex]\( y \)[/tex] varies directly as [tex]\( x \)[/tex], it means there is a constant [tex]\( k \)[/tex] such that [tex]\( y = kx \)[/tex].
### 9. Find [tex]\( x \)[/tex] when [tex]\( y = 24 \)[/tex] if [tex]\( y = 18 \)[/tex] when [tex]\( x = 6 \)[/tex].
First, we need to find the constant of proportionality [tex]\( k \)[/tex]. Given that [tex]\( y = 18 \)[/tex] when [tex]\( x = 6 \)[/tex],
[tex]\[
k = \frac{y}{x} = \frac{18}{6} = 3
\][/tex]
Now, we use this constant [tex]\( k = 3 \)[/tex] to find [tex]\( x \)[/tex] when [tex]\( y = 24 \)[/tex].
[tex]\[
y = kx \implies 24 = 3x \implies x = \frac{24}{3} = 8
\][/tex]
So, when [tex]\( y = 24 \)[/tex], [tex]\( x = 8 \)[/tex].
### 10. Find [tex]\( x \)[/tex] when [tex]\( y = 6 \)[/tex] if [tex]\( y = 18 \)[/tex] when [tex]\( x = 6 \)[/tex].
Again, we use the same constant [tex]\( k \)[/tex] from earlier, which is 3.
[tex]\[
y = kx \implies 6 = 3x \implies x = \frac{6}{3} = 2
\][/tex]
So, when [tex]\( y = 6 \)[/tex], [tex]\( x = 2 \)[/tex].
### Verification
Let's confirm our constant remains consistent:
For verification using [tex]\( y = -8 \)[/tex] when [tex]\( x = 4 \)[/tex],
[tex]\[
k = \frac{y}{x} = \frac{-8}{4} = -2
\][/tex]
This is a different scenario indicating either a typo or mistake in values because the constant from above should remain the same [tex]\( k = 3 \)[/tex]. Given the direct proportionality with the consistent [tex]\( k \)[/tex]:
Thus, results:
1. When [tex]\( y = 24 \)[/tex], [tex]\( x = 8 \)[/tex].
2. When [tex]\( y = 6 \)[/tex], [tex]\( x = 2 \)[/tex].
This solution aligns with consistent direct variation rules.
