Problem 11 - Introductory Algebra for College Students

Question

In a discount clothing store, all sweaters are sold at one fixed price and all shirts are sold at another fixed price. If one sweater and three shirts cost 42 dollar while three sweaters and two shirts cost 56 dollar find the price of one sweater and the price of one shirt.

Step-by-Step Solution

Answer

The price of a sweater is $12 and a shirt is $10.

Step 1 - Setup Equations

First, set up the system of equations based on the problem: \n1) \( S + 3T = 42 \)\n2) \( 3S + 2T = 56 \)

Step 2 - Use Substitution or Elimination Method

You can use either substitution or elimination method to solve the system of equations. In this case, the elimination method is chosen: \n Multiply equation 1) by 3 and equation 2) by 1: \n3) \( 3S + 9T = 126 \)\n4) \( 3S + 2T = 56 \)\nThen subtract equation 4) from equation 3): \( 7T = 70 \)

Step 3 - Find Value of T

Divide both side by 7 to isolate T: \( T = 70 / 7 = 10 \)

Step 4 - Find Value of S

Plug T = 10 into the first equation: \( S + 3(10) = 42 \) gives \( S = 42 -30 = 12 \)

Key Concepts

Substitution Method

The substitution method is a technique used to solve systems of equations where you solve one equation for one variable, and then substitute that expression into the other equation. This approach is particularly useful when one equation is easily solvable for one variable, providing a straightforward substitution into the other equation.
Let's understand this with a simple example:

• Consider a system of equations: 1) \( x + y = 5 \)2) \( 2x - y = 3 \)
• First, solve equation (1) for \( y \): \( y = 5 - x \).
• Substitute \( y = 5 - x \) into equation (2): \( 2x - (5 - x) = 3 \).
• Simplify and solve for \( x \): \( 2x - 5 + x = 3 \), which simplifies to \( 3x = 8 \), giving \( x = \frac{8}{3} \).
• Substitute \( x = \frac{8}{3} \) back into the expression for \( y \): \( y = 5 - \frac{8}{3} \), which results in \( y = \frac{7}{3} \).

This method works well when substitution doesn't overly complicate computations, leading to a simpler solution process.

Elimination Method

The elimination method is often used to solve systems of equations by eliminating one variable, allowing for the direct solution of the other. The goal is to manipulate the equations so that adding or subtracting them cancels out one of the variables.
Here's how it works:

• Take two equations: 1) \( a_1x + b_1y = c_1 \)2) \( a_2x + b_2y = c_2 \)
• Multiply each equation, if needed, so that the coefficients of one of the variables are the same or opposites. For instance, get both equations to have the same or opposite coefficient for \( x \) or for \( y \).
• In the example from the exercise, by multiplying the first equation by 3 and the second as is, you facilitate canceling the \( S \) variable:
• Resultant equations: \( 3S + 9T = 126 \) and \( 3S + 2T = 56 \).
• By subtracting the second from the first: \( 7T = 70 \), directly find \( T = 10 \).

After finding the value of one variable, substitute it back into any original equation to find the other. The elimination method is particularly powerful when substitution complicates calculations.

System of Equations

A system of equations consists of two or more equations with the same set of variables. Solving a system means finding the values of each variable that satisfy all the equations at the same time.
There are several methods for solving systems of equations such as substitution and elimination. Each method has situations where it might be more preferable.

• Graphical Method: The graphical method involves plotting each equation on the same graph and identifying the intersection point, which represents the solution. This is often limited in precision and used mainly for visualization.
• Substitution Method: Explored earlier, best used when one equation is easily solvable for one variable.
• Elimination Method: Efficient and useful in many contexts, especially when working with systems involving more than two equations.

Choosing the right method often depends on the specific structure and constraints of the system being solved.
In the given problem, both methods can solve the system effectively, providing the same result: each variable's value that satisfies both financial equation scenarios.