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(B) Let the natural length of the string be $L$ and the spring constant be $k$. According to Hooke's Law,the tension $F$ is proportional to the extens

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$5y - 4x$

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(B) Let the natural length of the string be $L$ and the spring constant be $k$. According to Hooke's Law,the tension $F$ is proportional to the extension $(l - L)$,where $l$ is the stretched length.$F = k(l - L)$For the first case: $8 = k(x - L) \quad ...(1)$For the second case: $10 = k(y - L) \quad ...(2)$Let the length be $z$ when the tension is $18 \, N$: $18 = k(z - L) \quad ...(3)$Subtracting $(1)$ from $(2)$: $10 - 8 = k(y - L) - k(x - L) \Rightarrow 2 = k(y - x) \Rightarrow k = \frac{2}{y - x}$.Substitute $k$ into $(2)$: $10 = \frac{2}{y - x}(y - L) \Rightarrow 5(y - x) = y - L \Rightarrow L = y - 5y + 5x = 5x - 4y$.Now,substitute $k$ and $L$ into $(3)$: $18 = \frac{2}{y - x}(z - (5x - 4y))$$9(y - x) = z - 5x + 4y$$9y - 9x = z - 5x + 4y$$z = 9y - 4y - 9x + 5x = 5y - 4x$.

The length of an elastic string is $x \, m$ when the tension is $8 \, N$ and $y \, m$ when the tension is $10 \, N$. The length in meters when the tension is $18 \, N$ is:

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