Solving a Systems of Linear Equations and Matrices Problem

In summary, the problem involves a small company that took out three loans with interest rates of 8%, 9%, and 10% to fund a planned expansion. They borrowed $1000 more at 9% than they did at 10%. The total annual interest on the loans was $2190. By setting up a system of linear equations, it can be determined that the company borrowed $12,000 at 8%, $7,000 at 9%, and $6,000 at 10%.
  • #1
bbrudi
4
0
Problem: To get the necessary funds for a planned expansion, a small company took out three loans totaling $25,000. Company owners were able to get interest rates of 8%, 9%, and 10%. They borrowed $1000 more at 9% than they borrowed at 10%. The total annual interest on the loans was $2190.

a) How much did they borrow at each rate?

b) Suppose we drop the condition that they borrowed $1000 more at 9% than at 10%. What can you say about the amount borrowed at 10%? What is the solution if the amount borrowed at 10% is $5000.

Could someone help me with this "Systems of Linear and Matrices problem...thank you.
 
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  • #2
Perhaps you should post some work or thoughts...

Look at the part of the problem that says:
They borrowed $1000 more at 9% than they borrowed at 10%.

So if the amount they borrowed at 9% is called [tex]x[/tex] and the amount they borrowed at 10% is called [tex]y[/tex]...then you would have the equation [tex]x = 1000 + y[/tex], now you need to do the same thing for the rest of the problem.
 
  • #3
So should the problem look something like this?

x + y + z = 25,000
.08x + .09y + .10z = 2,190
x = 1000 + y
 
  • #4
No one can answer that because you didn't say what "x", "y", and "z" mean!

I can guess that you mean "x is the amount borrowed at 8%, y is the amount borrowed at 9% and z is the amount borrowed at 10%" (not the assignments daveyinaz used) but you should say that. Assuming that, then your first two equations are correct but the last is not. What you have there says "they borrowed $1000 more at 8% than they did at 9%".
 
  • #5
That's where I'm stuck how do I express that he borrowed $1000 more at 9%? Would it be possibly y + 1000 = 0?
 
  • #6
[tex]
\begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & -1 \\ 8\% & 9\% & 10\% \end{bmatrix} \vec{x} = \begin{bmatrix} 25000 \\ 1000 \\ 2190 \end{bmatrix},\rightarrow \vec{x}=\begin{bmatrix} 12000 \\ 7000 \\ 6000 \end{bmatrix}
[/tex]
 
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  • #7
Thank you
 

Related to Solving a Systems of Linear Equations and Matrices Problem

1. How do I determine the number of solutions for a system of linear equations?

The number of solutions for a system of linear equations can be determined by looking at the number of variables and the number of equations in the system. If the number of variables is greater than the number of equations, the system will have infinitely many solutions. If the number of variables is equal to the number of equations, the system will have a unique solution. If the number of variables is less than the number of equations, the system will have no solution.

2. What is the Gauss-Jordan method for solving systems of linear equations?

The Gauss-Jordan method is a systematic approach for solving systems of linear equations. It involves using row operations to transform the system into an equivalent one with a simpler solution. The goal is to reduce the system to an upper triangular form, where the solution can easily be determined by back substitution.

3. Can matrices be used to solve systems of linear equations?

Yes, matrices can be used to solve systems of linear equations. The coefficients of the variables in the system can be represented as a matrix, and the constants can be represented as a column vector. By performing row operations on the augmented matrix, the system can be transformed into an equivalent one that is easier to solve.

4. What is the difference between consistent and inconsistent systems of linear equations?

A consistent system of linear equations is one that has at least one solution, while an inconsistent system has no solutions. Inconsistent systems have equations that are contradictory, making it impossible to find a solution that satisfies all the equations. Consistent systems, on the other hand, can have either a unique solution or infinitely many solutions.

5. Are there any real-world applications for solving systems of linear equations and matrices?

Yes, systems of linear equations and matrices are commonly used in various fields such as engineering, economics, and physics. For example, they can be used to model and solve problems involving electrical circuits, chemical reactions, and optimization of resources. They are also used in computer graphics and data analysis, among other applications.

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