Certainly! Let's break this problem down step-by-step.
### Part (a): Find the equation that relates [tex]\( y \)[/tex] and [tex]\( x \)[/tex].
Given that [tex]\( y \)[/tex] varies directly with [tex]\( x \)[/tex], we can express this relationship as:
[tex]\[ y = kx \][/tex]
where [tex]\( k \)[/tex] is the constant of variation. We need to determine the value of [tex]\( k \)[/tex].
We are provided with the information that [tex]\( y = 56 \)[/tex] when [tex]\( x = 14 \)[/tex]. Plugging these values into the direct variation equation, we get:
[tex]\[ 56 = k \cdot 14 \][/tex]
To find [tex]\( k \)[/tex], solve for [tex]\( k \)[/tex]:
[tex]\[ k = \frac{56}{14} = 4 \][/tex]
Therefore, the equation that relates [tex]\( y \)[/tex] and [tex]\( x \)[/tex] is:
[tex]\[ y = 4x \][/tex]
### Part (b): Find [tex]\( y \)[/tex] when [tex]\( x = 21 \)[/tex].
Now that we have the equation [tex]\( y = 4x \)[/tex], we can find [tex]\( y \)[/tex] for any value of [tex]\( x \)[/tex]. We are asked to determine [tex]\( y \)[/tex] when [tex]\( x = 21 \)[/tex].
Substitute [tex]\( x = 21 \)[/tex] into the equation:
[tex]\[ y = 4 \cdot 21 \][/tex]
Perform the multiplication:
[tex]\[ y = 84 \][/tex]
Hence, when [tex]\( x = 21 \)[/tex], [tex]\( y = 84 \)[/tex].
### Summary:
- The equation that relates [tex]\( y \)[/tex] and [tex]\( x \)[/tex] is [tex]\( y = 4x \)[/tex].
- When [tex]\( x = 21 \)[/tex], [tex]\( y = 84 \)[/tex].
