Matrices question.

[jesus_freak112]

Investor Company loaned out a total of $36,000, part at 6% interest and part at 9% interest. They reported that the annual earnings from both investments were the same amount that would have been earned by the total loan if it had been invested at 8%. Find the amount loaned at each rate.

I really don't have any clue what to do first. Please help. :?

 

[wjm11]

Investor Company loaned out a total of $36,000, part at 6% interest and part at 9% interest. They reported that the annual earnings from both investments were the same amount that would have been earned by the total loan if it had been invested at 8%. Find the amount loaned at each rate.

First, define some variables:

Let “x” be the amount invested at 6%.

Therefore, “36,000 – x” (the rest of the money) is the amount invested at 9%. Make sense?

If the entire amount were invested at 8%, the earnings would have been (.08)(36,000). Since the combined 6% and 9% earnings were equal to that amount, we can write:

(.06)(x) + (.09)(36,000 – x) = (.08)(36,000)

Now, just solve for x.

 

[mmm4444bot]

?

Why is this exercise a question about matrices? :?

?

 

[soroban]

Hello, jesus_freak112!

Does this really have to be solved by matrices?
It's much neater with normal Algebra . . . as wjm11 pointed out.

Investor Company loaned out a total of $36,000, part at 6% interest and part at 9% interest.
They reported that the annual earnings from both investments were the same amount
that would have been earned by the total loan if it had been invested at 8%.
Find the amount loaned at each rate.

Well, if we must use matrices . . .


\(\displaystyle \text{Let: }\:\begin{Bmatrix} x &=& \text{amt. at 6\%} \\ y &=& \text{amt. at 9\%} \end{Bmatrix}\)


\(\displaystyle \text{Then we have: }\;\begin{pmatrix} 1 & 1 \\ 0.06 & 0.09\end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} \;=\;\begin{pmatrix}36,\!000 \\ 0.08(x + y) \end{pmatrix}\)


. . \(\displaystyle \text{which gives us: }\;\begin{array}{ccc} x + y &=& 36,\!000 \\ 0.06x + 0.09y &=& 0.08x + 0.08y \end{array}\)

. . \(\displaystyle \text{and so on . . .}\)