We are dealing with two linear temperature scales, $X$ and $Y$, where the freezing and boiling points of water are given. The problem asks to convert a specific temperature from the $Y$ scale to the $X$ scale.

Step 1: Set up the linear relationship between the two scales

Since the temperature scales are linear, the relationship between $X$ and $Y$ can be expressed in the form: $X = mY + b$ where $m$ is the slope and $b$ is the intercept.

Step 2: Determine the slope $m$

We know the freezing and boiling points on both scales:

• Freezing point: $-119.0^\circ X$ corresponds to $-84.00^\circ Y$,
• Boiling point: $339.0^\circ X$ corresponds to $-39.00^\circ Y$.

Using these two points, we can calculate the slope $m$ as: $m = \frac{\Delta X}{\Delta Y} = \frac{339.0 - (-119.0)}{-39.00 - (-84.00)} = \frac{458.0}{45.00} = 10.18$

Step 3: Determine the intercept $b$

We can use one of the known points to solve for $b$. Using the freezing point $(-119.0^\circ X, -84.00^\circ Y)$: $-119.0 = 10.18(-84.00) + b$ $-119.0 = -855.12 + b$ $b = 736.12$

Step 4: Find the temperature on the $X$ scale corresponding to $42.000^\circ Y$

Now that we have the equation $X = 10.18Y + 736.12$, we can substitute $Y = 42.000$ into this equation: $X = 10.18(42.000) + 736.12$ $X = 427.56 + 736.12 = 1163.68^\circ X$

Conclusion

A temperature of $42.000^\circ Y$ corresponds to $1163.68^\circ X$.

Would you like more details on any step?

---

Here are 5 related questions:

1. How can you derive the conversion formulas between other temperature scales?
2. What are other temperature scales similar to $X$ and $Y$?
3. How do you find the slope of a line given two points?
4. Why do we use a linear equation for temperature conversions?
5. How do temperature scales like Celsius and Fahrenheit compare to linear scales like $X$ and $Y$?

Tip: When dealing with temperature scales, always check if they are linear or logarithmic, as this affects the conversion method.