Respuesta :
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:
[tex]x + y = 60[/tex]
In all, the partnership raised $348,000. So:
[tex]4000x + 8000y = 348000[/tex]
I am going to simplify by 4000. So:
[tex]x + 2y = 87[/tex]
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:
[tex]1)x + y = 60[/tex]
[tex]2)x + 2y = 87[/tex]
I am going to multiply 1) by -1. So we have
[tex]1)-x - y = -60[/tex]
[tex]2)x + 2y = 87[/tex]
By addition, the x are going to cancel each other
[tex]-x + x - y + 2y = -60 + 87[/tex]
[tex]y = 27[/tex]
For x:
[tex]x + y = 60[/tex]
[tex]x = 60-y = 60-27 = 33[/tex]
There were 33 $4,000 investors and 27 $8,000 investors.
2) Solve the system by row-reducing the corresponding augmented matrix.
[tex]2x + y = 17[/tex]
[tex]x + y = 13[/tex]
This system has the following augmented matrix:
[tex]\left[\begin{array}{ccc}2&1&17\\1&1&13\end{array}\right][/tex]
To help the row reducing, i am going to swap the first with the second line:
[tex]L1 <-> L2[/tex]
So we have:
[tex]\left[\begin{array}{ccc}1&1&13\\2&1&17\end{array}\right][/tex]
Now, reducing the first column.
[tex]L2 = L2 - 2L1[/tex]
So we have:
[tex]\left[\begin{array}{ccc}1&1&13\\0&-1&-9\end{array}\right][/tex]
Now we do:
[tex]L2 = -L2[/tex]
And the matrix is:
[tex]\left[\begin{array}{ccc}1&1&13\\0&1&9\end{array}\right][/tex]
Now to reduce the second column, we do:
[tex]L1 = L1 - L2[/tex]
[tex]\left[\begin{array}{ccc}1&0&4\\0&1&9\end{array}\right][/tex].
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:
[tex]x + y = 80[/tex]
They are worth $6.80. So:
[tex]0.05x + 0.10y = 6.80[/tex]
Solving the system:
[tex]1)x + y = 80[/tex]
[tex]2)0.05x + 0.10y = 6.80[/tex]
I am going to divide 1) by -10, so we can cancel y. So:
[tex]1)-0.10x - 0.10y = -8[/tex]
[tex]2)0.05x + 0.10y = 6.80[/tex]
Adding:
[tex]-0.10x + 0.05x - 0.10y + 0.10y = -8 + 6.80[/tex]
[tex]-0.05x = -1.2[/tex] *(-100)
[tex]5x = 120[/tex]
[tex]x = \frac{120}{5}[/tex]
[tex]x = 24[/tex]
Also
[tex]x + y = 80[/tex]
[tex]y = 80-x = 80-24 = 56[/tex]
There were 24 nickels and 56 dimes.
