To solve this problem, we need to follow the concept of direct variation, which states that if [tex]\( y \)[/tex] varies directly with [tex]\( x \)[/tex], then [tex]\( y \)[/tex] can be expressed as:
[tex]\[ y = kx \][/tex]
where [tex]\( k \)[/tex] is the constant of variation.
### Step-by-Step Solution:
1. Find the constant of variation ([tex]\( k \)[/tex]):
- We are given that [tex]\( y \)[/tex] is 30 when [tex]\( x \)[/tex] is 6. Substitute these values into the direct variation equation to find [tex]\( k \)[/tex]:
[tex]\[
30 = k \cdot 6
\][/tex]
- To solve for [tex]\( k \)[/tex], divide both sides of the equation by 6:
[tex]\[
k = \frac{30}{6}
\][/tex]
- Simplify the fraction:
[tex]\[
k = 5
\][/tex]
2. Use the constant of variation [tex]\( k \)[/tex] to find the new [tex]\( y \)[/tex]:
- We need to find [tex]\( y \)[/tex] when [tex]\( x \)[/tex] is 11. Substitute [tex]\( x = 11 \)[/tex] and [tex]\( k = 5 \)[/tex] into the direct variation equation:
[tex]\[
y = kx
\][/tex]
- Substitute the values of [tex]\( k \)[/tex] and [tex]\( x \)[/tex]:
[tex]\[
y = 5 \cdot 11
\][/tex]
- Perform the multiplication:
[tex]\[
y = 55
\][/tex]
Therefore, [tex]\( y \)[/tex] is 55 when [tex]\( x \)[/tex] is 11.
