SOLUTION: Ms. Brown invested $29,000 in two accounts. One yielding 6% interetst and the other yielding 10%. If she recieved a total of $2,500 in interest at the end of the year, how much did

Question 389440: Ms. Brown invested $29,000 in two accounts. One yielding 6% interetst and the other yielding 10%. If she recieved a total of $2,500 in interest at the end of the year, how much did she invest in each account?
Answer by gwendolyn(128) (Show Source): You can put this solution on YOUR website!
Let A be the amount invested in the 6% account.
Let B be the amount invested in the 10% account.
Mr Brown invested a total of $29,000 so,
A + B = 29,000
We subtract B from both sides to solve to A (picked at random):
A + B - B = 29,000 - B
A = 29,000 - B
We also know that the interest totaled $2,500. So, let's form an equation from the interest contributed to that total by each account:
.06 * A + .10 * B = 2,500
I like working with whole numbers, so I will multiply this equation by 100:
100(.06*A + .10*B) = 100(2,500)
Distribute the 100:
100*.06*A + 100*.10*B = 100*2,500
6*A + 10*B = 250,000
Substitute our value for A from the first equation:
6*(29,000 - B) + 10*B = 250,000
Distribute the 6 on the right:
6*29,000 - 6*B + 10*B = 250,000
174,000 + 4*B = 250,000
Subtract 174,000 from both sides to isolate the terms with the variable:
174,000 + 4*B - 174,000 = 250,000 - 174,000
4*B = 76,000
Divide both sides by 4 to solve for B:

B = 19,000
Find A by substituting B back into the first equation:
A = 29,000 - B
A = 29,000 - 19,000
A = 10,000

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