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Chapter 6: Problem 13

Solve each of the following verbal problems algebraically. You may use either a one or a two-variable approach. Pat and Carlos both belong to the same book club. Pat orders two regular selections and three specially discounted ones for a total of \(\$ 56.90 .\) Carlos orders three regular selections and four specially discounted ones for a total of \(\$ 80.85\) What are the prices of a regular and a specially discounted selection?

Short Answer

Expert verified
The price of a regular selection is \( \$ 14.95 \) and the price of a specially discounted selection is \( \$ 9.00 \).

Step by step solution

01

- Define the Variables

Let \( r \) be the price of a regular selection, and \( d \) be the price of a specially discounted selection.
02

- Set Up the Equations

From the problem, we have two pieces of information which can be translated into equations. For Pat: \( 2r + 3d = 56.90 \). For Carlos: \( 3r + 4d = 80.85 \).
03

- Solve the System of Equations Using Substitution or Elimination

We can use either substitution or elimination. Here, we will use the elimination method. First, multiply the first equation by 3 and the second equation by 2 to align the coefficients of \( r \). Thus, we get: \( 6r + 9d = 170.70 \) and \( 6r + 8d = 161.70 \).
04

- Eliminate One Variable

Subtract the second equation from the first: \( (6r + 9d) - (6r + 8d) = 170.70 - 161.70 \). This simplifies to: \( d = 9.00 \).
05

- Solve for the Other Variable

Substitute \( d = 9.00 \) back into one of the original equations, let's use \( 2r + 3d = 56.90 \). Substituting, we get \( 2r + 3(9.00) = 56.90 \). This simplifies to \( 2r + 27.00 = 56.90 \) and further to \( 2r = 29.90 \). Finally, \( r = 14.95 \).
06

- Verify the Solution

Substitute \( r = 14.95 \) and \( d = 9.00 \) into the other equation to verify: \( 3(14.95) + 4(9.00) = 44.85 + 36.00 = 80.85 \). The solution is verified.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Algebraic Methods
When solving systems of equations, algebraic methods are crucial. These methods allow us to find the values of unknown variables that satisfy multiple equations simultaneously.
Algebraic methods typically involve:
  • Substitution Method
  • Elimination Method
Both methods transform the system of equations into a simpler form, leading us to the solution. Choosing an appropriate method depends on the structure of the equations and personal preference. Let's explore these in greater detail as we continue.
Word Problems
Solving word problems with algebra involves translating real-world scenarios into mathematical equations. Here's how to approach them:
  • Identify what you need to find. These are typically unknown quantities, which will become your variables.
  • Define variables clearly. In our exercise, let \( r \) represent the price of a regular selection and \( d \) the price of a specially discounted selection.
  • Set up equations based on the problem's statements. For instance, Pat's and Carlos's purchases translated into two separate equations.
Word problems can seem tricky at first, but breaking them down step-by-step and translating them into algebraic form makes them manageable.
Substitution Method
The substitution method involves solving one equation for one variable and then substituting that expression into the other equation. Here's a general approach:
1. Solve one equation for one of the variables. For example, from \(2r + 3d = 56.90\), solve for \(r\): \( r = (56.90 - 3d)/2 \).
2. Substitute this expression for \(r\) in the other equation: \( 3((56.90 - 3d)/2) + 4d = 80.85 \).
3. Simplify and solve for the remaining variable. This step eventually results in a single variable equation that's easy to solve.
The substitution method is particularly useful when one equation is easily rearranged to isolate one of the variables.
Elimination Method
In the elimination method, we eliminate one variable by adding or subtracting two equations. Here鈥檚 how it worked in our example:
  • Align coefficients by multiplying each equation as necessary. We multiplied the first equation by 3 and the second by 2.
  • Subtract one equation from the other to eliminate \(r\): \( (6r + 9d) - (6r + 8d) = 170.70 - 161.70 \).
  • Solve for the remaining variable. This resulted in \(d = 9.00\).
  • Substitute the value of \(d\) back into one of the original equations to find \(r\).
The elimination method is robust and often preferred when both equations are in standard form, as it can eliminate variables efficiently and quickly.

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Most popular questions from this chapter

In Exercises \(29-62,\) solve the system of equations using any method you choose. $$\left\\{\begin{aligned} 5 x+2 y &=4 y+9 \\ y &=x-3 \end{aligned}\right.$$

An orchestral society put on a concert. The members sold 200 tickets in advance and 75 tickets at the door. They charged \(\$ 1.50\) more for tickets at the door than for advance-purchase tickets. If they collected a total of \(\$ 1075,\) how much did they charge for tickets at the door?

Cellmate Communications offers two monthly cellular phone plans. The Standard plan costs \(\$ 15\) per month plus \(\$ 0.22\) per minute of air time. The Deluxe plan costs \(\$ 35\) per month plus \(\$ 0.14\) per minute of air time. (A) Write an equation for the monthly cost \(C\) of the Standard plan and the Deluxe plan for a month in which you use \(m\) minutes. (B) Sketch the graphs of the two equations obtained in part (a). Label the horizontal axis \(m\) and the vertical axis \(C\). (C) Using the graphs obtained in part (b), determine how many air time minutes per month make it more economical to buy the Standard plan.

Solve each of the following verbal problems algebraically. You may use either a one or a two-variable approach. Two cars start at the same place and time, and travel in opposite directions. One car is traveling 15 kph faster than the other. After 5 hours the two cars are \(275 \mathrm{km}\) apart. Find the speed of each car.

Sketch a graph that represents the scenario described in the exercise. Be sure to clearly label any variables and the coordinate axes. Keep in mind that various graphs may be drawn to represent each situation. The price of a certain stock starts the day at \(\$ 15\) per share. Over the first 2 hours of trading, the price of the stock steadily declines to \(\$ 13\) per share. It remains at that price for 3 hours and then declines to \(\$ 11.50\) per share over the next hour.
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