Solving a Systems of Linear Equations and Matrices Problem

[bbrudi]

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.

 

[daveyinaz]

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 x and the amount they borrowed at 10% is called y...then you would have the equation x = 1000 + y, now you need to do the same thing for the rest of the problem.

 

[bbrudi]

So should the problem look something like this?

x + y + z = 25,000
.08x + .09y + .10z = 2,190
x = 1000 + y

 

[HallsofIvy]

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%".

 

[bbrudi]

That's where I'm stuck how do I express that he borrowed $1000 more at 9%? Would it be possibly y + 1000 = 0?

 

[Physics2009]

<br /> \begin{bmatrix} 1 &amp; 1 &amp; 1 \\ 0 &amp; 1 &amp; -1 \\ 8\% &amp; 9\% &amp; 10\% \end{bmatrix} \vec{x} = \begin{bmatrix} 25000 \\ 1000 \\ 2190 \end{bmatrix},\rightarrow \vec{x}=\begin{bmatrix} 12000 \\ 7000 \\ 6000 \end{bmatrix}<br />

 

[bbrudi]

Thank you