SOLUTION: Hi..trying to help my son on his homework, and running into a bit of a wall on this one. I can intuit my way to the correct answer, but can't show how to get there algebraically.

Question 515509: Hi..trying to help my son on his homework, and running into a bit of a wall on this one. I can intuit my way to the correct answer, but can't show how to get there algebraically.
"Samantha invested her $10,000 savings at two different banks for one year. She put part of the money in Bank A which paid 8% interest. The rest of the money was invested at 7% at Bank B.At the end of the year she had accumulated $735 in interest. How much money did she invest at Bank A? Hint:7% interest on an unknown amount of money is .07n."
I can get the answer (6500$ at 7%, %3500 at 8%), but I can't see how to write out the equation. Any help you can provide would be greatly appreciated, as we've spent three hours on this assignment already! :) Thank you again, we both greatly appreciate any time you give us!!!
-greg and gabe scott
Found 2 solutions by stanbon, drcole:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Samantha invested her $10,000 savings at two different banks for one year. She put part of the money in Bank A which paid 8% interest. The rest of the money was invested at 7% at Bank B.At the end of the year she had accumulated $735 in interest. How much money did she invest at Bank A?
-----
Using one variable.
---
Interest Equation:
Interest + Interest = Interest
0.08x + 0.07(10,000-x) = 735
-------------------------------------------------
Multiply thru by 100 to get:
8x + 7*10,000 - 7x = 73500
----
x = 73500-70,000
x = $3500 (Amount invested at Bank A)
=================
Cheers,
Stan H.
===================

Answer by drcole(72) (Show Source): You can put this solution on YOUR website!
Let x be the amount of money Samantha invested in Bank A, and let y be the amount she invested in Bank B. The total amount of money she invested in the two banks was $10000, and we can represent this as an algebraic equation as:
x + y = 10000
Since Bank A paid 8% interest, she earned 0.08x in interest at Bank A. Likewise, since Bank B paid 7% interest, she earned 0.07y at Bank B. The total amount of interest she earned at the two banks was $735, and we can represent this fact as an algebraic equation as:
0.08x + 0.07y = 735
Now we have a system of two linear equations in two unknowns. We can solve this system using the substitution method. First, we will take the first equation and solve for x:
x + y = 10000
x = 10000 - y (subtracting y from both sides)
Now we substitute 10000 - y for x in the second equation, yielding an equation involving y only:
0.08(10000 - y) + 0.07y = 735
Now we solve for y:
800 - 0.08y + 0.07y = 735 (distributing 0.08 in the left side)
800 - 0.01y = 735 (combining like terms)
-0.01y = -65 (subtracting 800 from both sides)
y = 6500 (dividing both sides by -0.01)
So Samantha invested $6500 in Bank B, as you figured out intuitively. Substituting 6500 back in for y in our equation for x, we then get:
x = 10000 - y = 10000 - 6500 = 3500
so, as you also figured out, she invested $3500 in Bank A.
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