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JPHamlett
04-29-2011, 03:09 AM
1. Samantha’s student loans total $30,000. Part was a personal loan at 10% interest; the other was a Stafford loan at 5%. After one year the loans accumulated $2,300 in interest. What was the amount of each loan?
a) Develop an equation to represent the sum of the loans, use ‘x’ for the personal loan, use ‘y’ for the Stafford loan..

b) Develop an equation to represent the interest accumulated in the first year. Use ‘x’ for the personal loan, use ‘y’ for the Stafford loan.

c) Use the equations from part a and b of this exercise as a system of equations. Use substitution to determine the amount of each loan. Show all your work from problem statement through the final result.

d) What are the intercepts of the equation from part a of this problem?
Where would the lines intersect if you solved the system by graphing?\


I honestly have NO idea even where to begin, so any help is GREATLY appreciated.

Method
04-29-2011, 04:14 AM
J_Pizzle: Hey, are you good at math?
Method: what do you need help with?
J_Pizzle: Ill show you the problem one sec
J_Pizzle: http://pastebin.com/AmTh9a0E
Method: where are you stuck?
J_Pizzle: Part A, would it be .1x + .05y?
Method: my guess would be 'x + y = 30000'
Method: you know she has two loans, and they add up to 30k total
J_Pizzle: Yes, so its x + y = 30000?
Method: Method: my guess would be 'x + y = 30000'
Method: i can't really see any other possibility
J_Pizzle: Ok
J_Pizzle: Thn
J_Pizzle: Thnx*
Method: you're welcome
J_Pizzle: Would part B be my .1x + .05y = 2300?
Method: yes

Did I have this conversation with myself???

Sex
04-29-2011, 04:17 AM
4 days ago and you're still on a simple problem like this?

alach11
04-29-2011, 04:22 AM
a.
Let x = money in personal loan at the beginning of the year
Let y = money in Stafford loan at the beginning of the year
Total Loan = $30,000 = x + y

b. To get just the interest you take 10% of the money in personal loans and 5% of the Stafford loan.
.10x + .05y = $2300

c. Solve the equation in part a for x or y.
y = $30,000 - x
Now since the values of x and y (starting amount in each loan) are the same in both equations you can replace y in part b with 30,000-x
We get the equation:
.10x + .05(30,000-x) = $2300
Then you solve for x. Then plug in your x value into one of your other equations to find y.