More Algebra 2 problems.

__June 2017, Part III__

All Questions in Part III are worth up to 4 credits. Partial credit is possible.

*33. Solve the following system of equations algebraically for all values of x, y, and z:
*

*x + y + z = 1*

2x + 4y + 6z = 2

-x + 3y - 5z = 11

2x + 4y + 6z = 2

-x + 3y - 5z = 11

**Answer: **

Any system of three equations with three variables can be solved by eliminating one of the variables and then solving the remaining system of equations with two variables.

I was away from the PC for a while the other day, so I worked the solution out long hand, two different ways. First by using elimination, then by using substitution, to remove the x. After that, the work would be the same.

Note that the *2x + 2y + 2z = 2* came from multiplying the first equation by 2.

At this point, by either method, you should be left with the two equations:

4y - 4z = 12

Once you have y, you can find z.

Once you have y and z, you can find x.

*34. Jim is looking to buy a vacation home for $172,600 near his favorite southern beach. The formula to compute a mortgage payment, M, is M = P • ( r (1+r)*

^{N}) / ( (1 + r)^{N}- 1) where P is the principal amount of the loan, r is the monthly interest rate, and N is the number of monthly payments. Jim’s bank offers a monthly interest rate of 0.305% for a 15-year mortgage.With no down payment, determine Jim’s mortgage payment, rounded to the nearest dollar.

*Algebraically determine and state the down payment, rounded to the nearest dollar, that Jim needs
to make in order for his mortgage payment to be $1100.
*

**Answer: **

For the first part of the question, use the given formula with the following substitutions:

P = 172,600, r = 0.00305, N = 12 * 15 = 180 payments

M = P • ( r (1+r)^{N}) / ( (1 + r)^{N} - 1)

M = 172600 • ( 0.00305 (1+0.00305)^{180}) / ( (1 + 0.00305)^{180} - 1)

M = 1247.4933... = 1247 to the nearest dollar. (Round 49 cents down.)

The down payment reduces the Principal that needs to be borrowed. To answer the second part, we need to solve for P when M = 1100. Once we know that, we need to subtract 172600 - P to find the amount of the down payment.

M = P • ( r (1+r)^{N}) / ( (1 + r)^{N} - 1)

1100 = P • ( 0.00305 (1+0.00305)^{180}) / ( (1 + 0.00305)^{180} - 1)

P = 1100 / • ( 0.00305 (1+0.00305)^{180}) / ( (1 + 0.00305)^{180} - 1)

P = 152193.19...

The down payment is 172600 - 152193.19 = 20406.81 = 20407.

Comments and questions welcome.

More Algebra 2 problems.