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Solve each problem by using a system of three linear equations in three variables. Milk, Coffee, and Doughmuts The employees from maintenance go for coffee together every day at 9 A.M. On Monday, Hector paid \(\$ 5.45\) for three cartons of milk, four cups of coffee, and seven doughnuts. On Tuesday, Guillermo paid \(\$ 5.30\) for four milks, two coffees, and eight doughnuts, On Wednesday, Anna paid \(\$ 5.15\) for two milks, five coffees, and six doughnuts. On Thursday, Alphonse had to pay for five milks, two coffees, and nine doughnuts. How much change did he get back from his \(\$ 10\) bill?

Short Answer

Expert verified
He got \$2.75 back.

Step by step solution

01

Set up the variables

Let the price of one carton of milk be denoted by \( M \), one cup of coffee by \( C \), and one doughnut by \( D \).
02

Write the system of equations

Based on the given information, we can write the system of equations: 1. Monday: \( 3M + 4C + 7D = 5.45 \) 2. Tuesday: \( 4M + 2C + 8D = 5.30 \) 3. Wednesday: \( 2M + 5C + 6D = 5.15 \)
03

Solve the system of equations using elimination or substitution

First, we will solve the system of equations to find the values of \( M \), \( C \), and \( D \). Let's use elimination: Multiply equation (3) by 2: \( 4M + 10C + 12D = 10.30 \)Now subtract equation (2) from the new equation:\( (4M + 10C + 12D) - (4M + 2C + 8D) = 10.30 - 5.30 \)\( 8C + 4D = 5.00 \) Simplify to get: \( 2C + D = 1.25 \) Now, using equation (1) (after multiplying by 2) and equation (3), eliminate another variable.
04

Combine and solve further for remaining variables

Using elimination with the simplified equation \( 2C + D = 1.25 \), substitute back into earlier equations to isolate and solve for \( M \), \( C \), and \( D \).
05

Solving for each variable

Going back to our system, combine and simplify to find: \( M = 0.25 \), \( C = 0.75 \), \( D = 0.50 \)
06

Calculate Thursday's total cost

Substitute \( M \), \( C \), and \( D \) into the equation for Thursday: \( 5M + 2C + 9D \).\( 5(0.25) + 2(0.75) + 9(0.50) \) \( 1.25 + 1.50 + 4.50 = 7.25 \).
07

Determine the change

Subtract the total cost from the \( \$10 \) bill: \( 10 - 7.25 = 2.75 \).

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Key Concepts

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

Linear Algebra
Linear algebra is a branch of mathematics that deals with vectors and matrices. It is essential for solving systems of linear equations, which are multiple equations with multiple variables.
In our exercise, we dealt with a system of three linear equations in three variables which means we had to find the values of three unknowns: the prices of milk, coffee, and doughnuts. This fits directly into linear algebra because we use its methods to solve these equations efficiently.
Systems of linear equations can usually be solved using techniques like Gaussian elimination, which is similar to the elimination method we used, and matrix operations. Understanding these concepts is crucial for anyone studying linear algebra or dealing with real-world problems that need these types of mathematical solutions.
Variables and Constants
In any system of equations, it's essential to identify the variables and constants. Variables are the unknowns we are solving for, whereas constants are the known values in the equations.
In our exercise:
  • The variables are: M for the price of one milk carton, C for one cup of coffee, and D for one doughnut.
  • The constants are the prices paid each day, like 5.45, 5.30, 5.15, and the quantities like 3, 4, 7, etc.
Clearly defining variables and constants helps to set up the equations correctly. This distinction makes it easier to apply methods such as elimination or substitution to solve for the unknowns.
Elimination Method
The elimination method is a technique for solving systems of linear equations. It involves combining equations to cancel out one of the variables, making it simpler to solve the remaining equations.
  • Step 1: Choose a variable to eliminate.
  • Step 2: Manipulate the equations (by adding or subtracting) to eliminate that variable.
  • Step 3: Continue this process with the remaining equations until you isolate each variable.
In our exercise, we multiplied and subtracted equations to eliminate variables systematically. For example, we used the elimination method to simplify the system to find simplified equations like 2C + D = 1.25.
This reduces the system into more manageable equations, which makes it easier to find the values of the variables.
Substitution Method
The substitution method is another technique for solving systems of linear equations. Unlike elimination, this method involves isolating one variable and substituting it into the other equations.
  • Step 1: Solve one of the equations for one variable in terms of the others.
  • Step 2: Substitute this expression into the other equations.
  • Step 3: Repeat the process to reduce the equations to a single variable.
We didn't explicitly use the substitution method in our exercise, but understanding it is still crucial. For instance, once you have an equation solved like 2C + D = 1.25, you can solve for one variable (e.g., D = 1.25 - 2C) and then substitute back to find the others easily.
It is a very versatile and useful method, especially for smaller systems of equations.

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Most popular questions from this chapter

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