Answer:
1) There were 33 $4,000 investors and 27 $8,000 investors.
2) The solution in x = 4, y = 9.
3) There were 24 nickels and 56 dimes.
Step-by-step explanation:
1) A lawyer has found 60 investors for a limited partnership to purchase an inner-city apartment building, with each contributing either $4,000 or $8,000. If the partnership raised $348,000, then how many investors contributed $4,000 and how many contributed $8,000?
I am going to say that:
x is the number of investors that contributed 4,000.
y is the number of investors that contributed 8,000.
Building the system:
There are 60 investors. So:

In all, the partnership raised $348,000. So:

I am going to simplify by 4000. So:

Solving the system:
The elimination method is a method in which we can transform the system such that one variable can be canceled by addition. So:


I am going to multiply 1) by -1. So we have


By addition, the x are going to cancel each other


For x:


There were 33 $4,000 investors and 27 $8,000 investors.
2) Solve the system by row-reducing the corresponding augmented matrix.


This system has the following augmented matrix:
![\left[\begin{array}{ccc}2&1&17\\1&1&13\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%261%2617%5C%5C1%261%2613%5Cend%7Barray%7D%5Cright%5D)
To help the row reducing, i am going to swap the first with the second line:

So we have:
![\left[\begin{array}{ccc}1&1&13\\2&1&17\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%261%2613%5C%5C2%261%2617%5Cend%7Barray%7D%5Cright%5D)
Now, reducing the first column.

So we have:
![\left[\begin{array}{ccc}1&1&13\\0&-1&-9\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%261%2613%5C%5C0%26-1%26-9%5Cend%7Barray%7D%5Cright%5D)
Now we do:

And the matrix is:
![\left[\begin{array}{ccc}1&1&13\\0&1&9\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%261%2613%5C%5C0%261%269%5Cend%7Barray%7D%5Cright%5D)
Now to reduce the second column, we do:

.
So the solution is:
x = 4, y = 9.
3) A jar contains 80 nickels and dimes worth $6.80. How many of each kind of coin are in the jar?
I am going to say that x is the number of nickels and y is the number of dimes.
Each nickel is worth 5 cents and each dime is worth 10 cents.
Building the system:
There are 80 coins in all:

They are worth $6.80. So:

Solving the system:


I am going to divide 1) by -10, so we can cancel y. So:


Adding:

*(-100)



Also


There were 24 nickels and 56 dimes.