Problem 19 - Understanding Elementary Algebra with Geometry

Question

Solve each of the following verbal problems algebraically. You may use either a one or a two-variable approach. Two retailers are ordering from the same source. One retailer orders eight stereo receivers and four turntables at a total cost of \(\$ 2060 .\) A second retailer orders five of the same receivers and six of the same turntables at a total cost of \(\$ 1690\). What are the costs of an individual receiver and an individual turntable?

Step-by-Step Solution

Answer

The cost of one stereo receiver is \$200\ and one turntable is \$115\.

Step 1 - Define Variables

Let the cost of one stereo receiver be denoted by \( R \) dollars and the cost of one turntable be denoted by \( T \) dollars.

Step 2 - Set Up Equations

Based on the information provided, set up the following system of linear equations: \[ 8R + 4T = 2060 \] \[ 5R + 6T = 1690 \]

Step 3 - Solve One Equation for a Variable

Solve the first equation for \(R\) in terms of \(T\). From \[ 8R + 4T = 2060, \] we get: \[ R = \frac{2060 - 4T}{8} \]

Step 4 - Substitute into the Second Equation

Substitute \( R = \frac{2060 - 4T}{8} \) into the second equation: \[ 5\left(\frac{2060 - 4T}{8}\right) + 6T = 1690 \]

Step 5 - Solve for T

Solve the equation \[ 5\left(\frac{2060 - 4T}{8}\right) + 6T = 1690 \] to find \(T\).

Step 5a - Simplify Equation

First, simplify \[ 5\left(\frac{2060 - 4T}{8}\right): \] \[ \frac{5(2060 - 4T)}{8} = \frac{10300 - 20T}{8} = 1287.5 - 2.5T \]

Step 5b - Solve for T

Combine like terms and solve for T: \[ 1287.5 - 2.5T + 6T = 1690 \] \[ 1287.5 + 3.5T = 1690 \] \[ 3.5T = 402.5 \] \[ T = \frac{402.5}{3.5} \approx 115 \]

Step 6 - Substitute T to Find R

Substitute \( T = 115 \) back into \ 8R + 4T = 2060. \

Step 6a - Solve for R

Substituting \ T = 115 \: \[ 8R + 4(115) = 2060 \] \[ 8R + 460 = 2060 \] \[ 8R = 1600 \] \[ R = \frac{1600}{8} = 200 \]

Step 7 - Verify Solution

Verify the solutions by substituting \ R = 200 \ and \ T = 115 \ into the second retailer's equation: \[ 5R + 6T = 1690 \] \[ 5(200) + 6(115) = 1690 \] \[ 1000 + 690 = 1690 \] \ which confirms our solution. \

Key Concepts

linear equations

A linear equation is a mathematical statement where the highest power of the variable is one. They are called 'linear' because they represent straight lines when graphed.
In the exercise, we have two linear equations:

• 8R + 4T = 2060
• 5R + 6T = 1690

Here, the variables are 'R' and 'T' representing the cost of receivers and turntables respectively. Notice there are no variables raised to any power other than one. Understanding this is crucial, as it tells us our relationships will graph as straight lines and can be solved using methods such as substitution and elimination.

solving equations

Solving equations can be done by finding the value of the variables that make the equation true. For the given problem, let's take one step to explore this:
First, solve one of the linear equations for one of the variables. For example, transform 8R + 4T = 2060 to isolate R:
R = (2060 - 4T) / 8
Next, substitute this into the other equation. This technique simplifies the problem to one equation with one variable,
making it straightforward to solve for the final variable ‘T’ in this case.
Finally, use this solved value to determine the other variable from our initial solved equation. This ensures all equations balance out correctly.

system of linear equations

A system of linear equations consists of two or more linear equations involving the same set of variables. In this exercise, we have:

• 8R + 4T = 2060
• 5R + 6T = 1690

The goal is to find a common solution for both equations, meaning a set of values for 'R' and 'T' that satisfy both equations simultaneously. Various methods can solve these systems, including graphing, substitution, and elimination.
Systems of linear equations can have one solution, infinitely many solutions, or no solution at all. Here, our unique solution represents the specific costs for the receiver and turntable.

substitution method

The substitution method is a technique for solving systems of linear equations. Here's a simplified guide through our exercise:

• Step 1: Solve one of the equations for one variable. Here, from 8R + 4T = 2060, we get R = (2060 - 4T) / 8.
  
• Step 2: Substitute this expression in the other equation (5R + 6T = 1690):
• 5[(2060 - 4T) / 8] + 6T = 1690 simplifies to find T.
• Step 3: Solve for T, then plug back into the expression for R to find the cost.
After solving, we get R = \(200 and T = \)115. This method is useful when you can easily solve for one variable in terms of the other, making the system easier to handle step-by-step.