SOLUTION: You invested $7000 in two accounts paying 7% and 8% annual interest, respectively. If the total interest earned for the year was $536, how much was invested at each rate?

Question 651633: You invested $7000 in two accounts paying 7% and 8% annual interest, respectively. If the total interest earned for the year was $536, how much was invested at each rate?
Answer by Algebraic(50) (Show Source): You can put this solution on YOUR website!
Let's break this down into simpler terms so you can understand.
You invested $7000 in a 7% and 8% account (always remember: the 'x' variable is always given to the biggest percentage)
Total of interest: 536 (this always goes at the end of the equation, after the equal sign)
Step 1: Let's set up the equation:
Here are how percents are written: (7% = 0.07, 8% = 0.08, other examples: 50% = 0.50)
0.08x + 0.07(7,000-x)=536
You're adding both because you want to see how much each account paid together.
Step 2: Distribute the 0.07 to everything in the parenthesis
0.08x + 0.07(7,000-x)=536
0.08x + 490 - 0.07x =536
Step 3: Combine like terms
0.08x + 490 - 0.07x =536
0.01x + 490 =536
Step 4: Continue combining like terms
0.01x + 490 =536
0.01x = 46
Step 5: Divide both sides by 0.01 to get the 'x' by itself
0.01x = 46
x = 4,600
This answer, x = 4,600, says to us that $4,600 was invested in the 8% account.
To find out how much was invested in the 7% account, just replace the x = 4,600 back into this part of the original equation: 7,000-x
7,000-x
7,000-4,600
2,400
So.. $4,600 was invested in the 8% account and $2,400 was invested in the 7% account.
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