# SOLUTION: Sam earned 1155 in interest on a total investment 15000 this past year, if part of the money was invested at 6% and the remainder at 9% how much was invested at 6%

Algebra ->  Algebra  -> Signed-numbers -> SOLUTION: Sam earned 1155 in interest on a total investment 15000 this past year, if part of the money was invested at 6% and the remainder at 9% how much was invested at 6%      Log On

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 Question 56970: Sam earned 1155 in interest on a total investment 15000 this past year, if part of the money was invested at 6% and the remainder at 9% how much was invested at 6%Answer by Earlsdon(6287)   (Show Source): You can put this solution on YOUR website!Let \$x = the amount invested at 6%, then the remainder of (\$15000 - \$x) = the amount invested at 9%. Changing the percents to their decimal equivalents (6% = 0.06 and 9% = 0.09), you can write the equation for this situation: 0.06x + 0.09(15000-x) = 1155 Simplify and solve for x. 0.06x + 1350 - 0.09x = 1155 -0.03x + 1350 = 1155 Subtract 1350 from both sides of the equation. -0.03x = -195 Divide both sides by -0.03 x = \$6500 The amount invested at 6% was \$6500