Problem 61 - Precalculus Mathematics for Calculus

Question

A gas station sells regular gas for \(\$ 2.20\) per gallon and premium gas for \(\$ 3.00\) a gallon. At the end of a business day 280 gallons of gas were sold, and receipts totaled S680. How many gallons of each type of gas were sold?

Step-by-Step Solution

Answer

200 gallons of regular gas and 80 gallons of premium gas were sold.

Step 1: Identify Variables

Let \( x \) be the number of gallons of regular gas sold, and \( y \) be the number of gallons of premium gas sold. We need to find the values of \( x \) and \( y \).

Step 2: Set Up Equations

We know two things: the total gallons sold is 280, and the total receipts are $680. This gives us the two equations:1. \( x + y = 280 \) (total gallons)2. \( 2.20x + 3.00y = 680 \) (total receipts)

Step 3: Solve for One Variable

Solve the first equation for \( y \):\[ y = 280 - x \]

Step 4: Substitute to Solve

Substitute \( y = 280 - x \) into the second equation:\[ 2.20x + 3.00(280 - x) = 680 \]

Step 5: Simplify the Equation

Expand and simplify the equation:\[ 2.20x + 840 - 3.00x = 680 \]

Step 6: Combine Like Terms

Combine the terms in the equation:\[ -0.80x + 840 = 680 \]

Step 7: Isolate \( x \)

Subtract 840 from both sides:\[ -0.80x = -160 \]

Step 8: Solve for \( x \)

Divide both sides by -0.80:\[ x = 200 \]

Step 9: Solve for \( y \)

Use \( x = 200 \) in the first equation to find \( y \):\[ y = 280 - 200 = 80 \]

Step 10: Verify Solution

Check the solution:- Total gallons = 200 + 80 = 280, which matches.- Total receipts = \( 2.20(200) + 3.00(80) = 440 + 240 = 680 \), which matches.

Key Concepts

Linear Equations

Linear equations are foundational in mathematics and are often used to solve real-world problems. In the context of our exercise, we have two linear equations. One represents the total amount of gasoline sold, and the other represents the total revenue from the sales of these gases.
When you see a phrase like "total gallons sold," it usually translates to an equation. Here, the equation is:

• \( x + y = 280 \)

This equation states that the sum of regular and premium gas sold must equal 280 gallons.

The second equation involves the costs:

• \( 2.20x + 3.00y = 680 \)

It shows the relationship between the price per gallon and the total revenue of $680. These equations are termed 'linear' because they graphically represent straight lines. Understanding how these lines represent relationships between variables is key to solving the system of equations.

Problem Solving

Problem-solving in mathematics begins with understanding the problem statement and the available data. For our problem, the key is identifying what needs to be calculated - here, the amount of each type of gas sold.

Firstly, determine unknowns by defining them as variables. We defined \( x \) as the gallons of regular gas and \( y \) as the gallons of premium gas. This clear definition lets you translate words into algebraic equations.

Next, break down the problem into manageable steps. By creating equations from the word problem, and then systematically solving these equations, you create a logical path to find the solution. Finally, verify that the solution fits all conditions outlined in the problem, such as checking whether the total gallons sold and the receipts match. This thorough check ensures accuracy and solves the problem fully.

Algebraic Manipulation

Algebraic manipulation involves rearranging and simplifying equations to find the values of variables. We start by expressing one variable in terms of another. From the equation \( x + y = 280 \), solving for \( y \) gives \( y = 280 - x \).

Substituting this expression into the second equation \( 2.20x + 3.00y = 680 \) allows us to work with a single variable. This substitution turns the equation into \( 2.20x + 3.00(280-x) = 680 \), simplifying our work effectively.

• Expand it: \( 2.20x + 840 - 3.00x = 680 \).
• Combine like terms: \(-0.80x + 840 = 680 \).
• Isolate \( x \): Subtract 840 from both sides, resulting in \(-0.80x = -160 \).
• Solve for \( x \) by dividing both sides by \(-0.80\): \( x = 200 \).

Once \( x \) is known, back-substitute to find \( y \): \( y = 280 - 200 = 80 \). Through manipulating these equations, you showed that the number of gallons sold fits the conditions of the problem.