Answer :
To solve the problem where [tex]\( y \)[/tex] varies inversely with [tex]\( x \)[/tex], we follow these steps:
1. Understanding inverse variation:
When [tex]\( y \)[/tex] varies inversely with [tex]\( x \)[/tex], it means that the product of [tex]\( y \)[/tex] and [tex]\( x \)[/tex] is a constant. Mathematically, this can be expressed as:
[tex]\[ y \cdot x = k \][/tex]
where [tex]\( k \)[/tex] is a constant.
2. Find the constant [tex]\( k \)[/tex]:
We are given that [tex]\( y \)[/tex] is 5 when [tex]\( x \)[/tex] is 14. Using the inverse variation relation:
[tex]\[ y_1 \cdot x_1 = k \][/tex]
Plugging in the given values [tex]\( y_1 = 5 \)[/tex] and [tex]\( x_1 = 14 \)[/tex]:
[tex]\[ 5 \cdot 14 = k \][/tex]
[tex]\[ k = 70 \][/tex]
3. Determine [tex]\( y \)[/tex] for a different [tex]\( x \)[/tex]:
We need to find [tex]\( y \)[/tex] when [tex]\( x \)[/tex] is 7. Using the same inverse variation relation:
[tex]\[ y \cdot x = k \][/tex]
Let the new [tex]\( y \)[/tex] be [tex]\( y_2 \)[/tex] and the new [tex]\( x \)[/tex] be [tex]\( x_2 \)[/tex]:
[tex]\[ y_2 \cdot 7 = 70 \][/tex]
Solving for [tex]\( y_2 \)[/tex]:
[tex]\[ y_2 = \frac{70}{7} \][/tex]
[tex]\[ y_2 = 10 \][/tex]
Hence, the value of [tex]\( y \)[/tex] when [tex]\( x \)[/tex] is 7 is:
[tex]\[ \boxed{10} \][/tex]
1. Understanding inverse variation:
When [tex]\( y \)[/tex] varies inversely with [tex]\( x \)[/tex], it means that the product of [tex]\( y \)[/tex] and [tex]\( x \)[/tex] is a constant. Mathematically, this can be expressed as:
[tex]\[ y \cdot x = k \][/tex]
where [tex]\( k \)[/tex] is a constant.
2. Find the constant [tex]\( k \)[/tex]:
We are given that [tex]\( y \)[/tex] is 5 when [tex]\( x \)[/tex] is 14. Using the inverse variation relation:
[tex]\[ y_1 \cdot x_1 = k \][/tex]
Plugging in the given values [tex]\( y_1 = 5 \)[/tex] and [tex]\( x_1 = 14 \)[/tex]:
[tex]\[ 5 \cdot 14 = k \][/tex]
[tex]\[ k = 70 \][/tex]
3. Determine [tex]\( y \)[/tex] for a different [tex]\( x \)[/tex]:
We need to find [tex]\( y \)[/tex] when [tex]\( x \)[/tex] is 7. Using the same inverse variation relation:
[tex]\[ y \cdot x = k \][/tex]
Let the new [tex]\( y \)[/tex] be [tex]\( y_2 \)[/tex] and the new [tex]\( x \)[/tex] be [tex]\( x_2 \)[/tex]:
[tex]\[ y_2 \cdot 7 = 70 \][/tex]
Solving for [tex]\( y_2 \)[/tex]:
[tex]\[ y_2 = \frac{70}{7} \][/tex]
[tex]\[ y_2 = 10 \][/tex]
Hence, the value of [tex]\( y \)[/tex] when [tex]\( x \)[/tex] is 7 is:
[tex]\[ \boxed{10} \][/tex]
