Problem 42 - Precalculus with Limits

Question

MEASUREMENT When buying gasoline, you notice that 14 gallons of gasoline is approximately the same amount of gasoline as 53 liters. Use this information to find a linear model that relates liters \(y\) to gallons \(x\). Then use the model to find the numbers of liters in 5 gallons and 25 gallons.

Step-by-Step Solution

Answer

The linear model that relates liters \(y\) to gallons \(x\) is \(y = (53/14)x\). Using this model, 5 gallons and 25 gallons are approximately equal to 18.93 liters and 94.64 liters respectively.

Step 1: Establish known quantities

We start by identifying the data points we have. We know that 14 gallons of gas is equivalent to 53 liters. So, we have the points (0,0) and (14,53).

Step 2: Calculate the slope of the line

The slope of the line \(m\) that goes through the points (x1, y1) and (x2, y2) is given by the formula \(m = (y2 - y1)/(x2 - x1)\). Substituting our points into the formula gives \(m = (53 - 0)/(14 - 0) = 53/14\).

Step 3: Establish the linear model

A linear equation is generally of the form \(y = mx + b\), where \(m\) is the slope and \(b\) is the y-intercept. Because our graph passes through the origin (0,0), the y-intercept \(b\) is 0, producing the linear equation \(y = (53/14)x\). This equation represents the relationship between the amount of gasoline in liters \(y\) to gallons \(x\).

Step 4: Use the model to find the liters in 5 and 25 gallons

We can substitute \(x = 5\) and \(x = 25\) into our formula to find out how many liters they are each equivalent to. For \(x = 5\), \(y = (53/14) * 5\). And for \(x = 25\), \(y = (53/14) * 25\).

Key Concepts

Slope

The slope of a line in a graph is a measure of its steepness. It tells us how much the value of 'y' changes for a unit change in 'x'.
To find the slope, we use the formula: \(m = \frac{{y_2 - y_1}}{{x_2 - x_1}}\).
In this exercise, we identified two points: \((0,0)\) and \((14,53)\), representing gallons and liters, respectively.
By inserting these values into the slope formula, we get:

• \(y_1 = 0, y_2 = 53\)
• \(x_1 = 0, x_2 = 14\)

The slope comes out as \(\frac{53}{14}\).
This slope tells us that for each gallon of gasoline, the equivalent in liters increases by approximately 3.7857 (when calculated \(\frac{53}{14}\)).This provides the constant rate of change between gallons and liters.

Linear Equation

A linear equation creates a straight line on a graph.
The equation is typically represented as \(y = mx + b\), where:

• '\(m\)' is the slope.
• '\(b\)' is the y-intercept.

In our situation, 'b' is zero because our line goes through the origin point \((0,0)\).
Thus, our linear equation becomes \(y = \frac{53}{14}x\).
This equation describes how many liters, \(y\), can be obtained from a certain number of gallons, \(x\).
Every gallon you consider is simply multiplied by the slope, \(\frac{53}{14}\), to get the equivalent in liters.
For example:

• If \(x = 5\) gallons, \(y = \frac{53}{14} \times 5\) liters becomes the result.
• For \(x = 25\) gallons, \(y = \frac{53}{14} \times 25\) liters is calculated.

So, a linear equation like this one provides an easy way to convert and find values between two related quantities.

Unit Conversion

Unit conversion is the process of converting one measurement unit to another.
In our exercise, we're converting gallons of gasoline into liters.
The key to this conversion is our linear model, \(y = \frac{53}{14}x\).
This model directly connects gallons to liters by treating the conversion factor \(\frac{53}{14}\) as equivalent to 3.7857 liters per gallon.
Unit conversion steps:

• Identify the conversion factor (here, \(\frac{53}{14}\)) that links the two units.
• Multiply the quantity in the original unit (gallons) by this factor to convert to the desired unit (liters).

For example:

• To find out how many liters are in 5 gallons, calculate \(\frac{53}{14} \times 5\), resulting in approximately 18.93 liters.
• Similarly, for 25 gallons, use \(\frac{53}{14} \times 25\), which equals roughly 94.64 liters.

Learning to perform unit conversions is extremely useful in real-life scenarios, like cooking or filling a gas tank, where different measurement systems are often used.