Card Name (APR %) Existing Balance Credit Limit MarK2 (6.5%) $475.00 $3,000.00
Bee4 (10.1%) $1,311.48 $2,500.00

You have $400.00 each month to pay off these two credit cards. You decide to pay only the interest on the lower-interest card and the remaining amount to the higher interest card. Complete the following two tables to help you answer questions 1–3.

Higher-Interest Card (Payoff Option)
Month 1 2 3 4 5 6 7 8 9 10
Principal 1,311.48 925.09 535.45 142.53
Interest accrued 11.04 7.79 4.51 120
Payment (on due date) 397.43 397.43 397.43 143.73
End-of-month balance 925.09 545.35 142.53 0

Lower-Interest Card
Month 1 2 3 4 5 6 7 8 9 10
Principal 475 475 475 475 221.30
Interest accrued 2.57 2.57 2.57 2.57 1.20
Payment (on due date) 2.57 2.57 2.57 256.27 222.50
End-of-month balance 475 475 475 221.30 0

1) How long does it take to pay off the higher-interest card?
2) What is the amount of the last payment on the higher-interest card? Why?
3) At the end of the month that you pay off the higher-interest card, after you have started to pay down your debt on the lower-interest card, what is the balance of the lower-interest card? Why?

Complete the following two tables to help you answer questions 4–5.
4) Rework the problem so that you pay off the lower-interest card first.
5) How much money do you save by paying off the higher-interest card first?

Lower Interest Card (Payoff Option)
Month 1 2 3 4 5 6 7 8 9 10
Principal
Interest accrued
Payment (on due date)
End-of-month balance

Higher Interest Card
Month 1 2 3 4 5 6 7 8 9 10
Principal
Interest accrued
Payment (on due date)
End-of-month balance

Be sure to include in your response:
• All parts of the tables are complete and calculations correct
• The answers to the additional questions

(0,-3) see attached graph

step-by-step explanation:

a. to solve the system, graph the two equations.

x = y+3 becomes y=x-3 has m=1/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move up 1 unit and 1 unit to the right. plot a point and connect.

y=-4x+3 has m =-4/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move down four units and to the right 1 unit. plot a point and connect.

the solution is the point (x,y) where the two lines intersect. by the graph below it is (0,-3).

b. to solve by substitution, substitute y = -4x-3 into x= y+3

x= (-4x-3) +3

x=-4x

5x=0

x=0

now substitute x = 0 back in to find y.

y= -4(0)-3

y=0-3

y=-3

solution is (0,-3)

c. use elimination by adding the equations together.

x-y=3

4x+y=-3

5x=0

x=0

then substitute into one equation to find y.

y=-4(0)-3

y=-3

2 & 7

step-by-step explanation:

he climbs 300 ft/hr at a constant rate. the peak is at 1,500 ft.

when going up, he reaches 600 ft in 2 hours. remember, that he is going in a constant rate. divide 600 with 300:

600/300 = 2

assuming that he takes the same route back (meaning that it's a 1500 ft route back, you can solve.

note also that you are solving when he gets back to 600 ft after he reaches the peak.

first, find the amount of hours it takes to get up the peak. divide the total amount (1500), with the ft/her (300) to get the amount of hours (x):

1500/300 = x

x = 5

it takes jamie 5 hours to get up. now, he is returning down, and reaches 600 (his current destination). remember that he climbs at a constant 300 ft/hr. divide 600 with 300:

600/300 = x

x = 2

now add the two numbers together.

x = 5 + 2 = 7

jamie takes 7 hours to go up the mountain and reach his current position when descending.

(total amount/constant rate) = (time).

remember that it is a constant rate, so in total, when he gets back down the mountain, it will take the same amount of time as when he goes up. the total amount for the round trip is: 5 + 5 = 10.

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