Mr. Brownwood invests a certain amount of money at 9% interest and $1,800 more than that amount in another account at 11% interest. At the end of one year, he earned a total of $818 in interest. How much money was invested in each account?
$3,500 at 9%; $4,300 at 11%
$3,400 at 9%; $3,200 at 11%
$3,100 at 9%; $4,900 at 11%

Question

Mr. Brownwood invests a certain amount of money at 9% interest and $1,800 more than that amount in another account at 11% interest. At the end of one year, he earned a total of $818 in interest. How much money was invested in each account?
$3,500 at 9%; $4,300 at 11%
$3,400 at 9%; $3,200 at 11%
$3,100 at 9%; $4,900 at 11%

mertsj · Answer

Let x = amount at 9%
Then x+1800 = amount at 11%
.09x+.11(x+1800)=818

mertsj · Answer

Solve it.

anonymous · Answer

Interest = x*.09 + (x + 1800)*.11 = 818
.09x + .11x + 198 = 818

anonymous · Answer

Even without solving the equation, the answer has to be 3100 at 9% and 4900 at 11% because the difference between 4900 and 3100 is 1800.

anonymous · Answer

1.01x+198=818

anonymous · Answer

then subtract 198from both sides right

anonymous · Answer

X is your amount in the first account. You will make 0.09*X. In your second account you have X+1800. You will make 0.11*(X+1800). To find X you simply solve the equation:
0.09X + 0.11(X+1800) = 818
0.2X + 198 = 818
X = 3100.

anonymous · Answer

ok thank you