# SOLUTION: Part of 8,000 was invested at 10% and the rest at 12%. If the annual income from these investments was 860.00 how much was invested at each rate. I need help with this one.Each ra

Algebra ->  Algebra  -> Customizable Word Problem Solvers  -> Misc -> SOLUTION: Part of 8,000 was invested at 10% and the rest at 12%. If the annual income from these investments was 860.00 how much was invested at each rate. I need help with this one.Each ra      Log On

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 Click here to see ALL problems on Miscellaneous Word Problems Question 260828: Part of 8,000 was invested at 10% and the rest at 12%. If the annual income from these investments was 860.00 how much was invested at each rate. I need help with this one.Each rate meaning 10% and 12 %Answer by Theo(3464)   (Show Source): You can put this solution on YOUR website!total invested is 8,000 let x be one part. let y be the other part. you have x + y = 8000 (equation 1) total annual income is 860.000 x part is invested at 10% y part is invested at 12% you have .10*x + .12*y = 860 (equation 2) you need to solve these 2 equations simultaneously. solve for y in equation 1 to get y = 8000 - x substitute for y in equation 2 to get .10*x + .12*(8000-x) = 860 solve for x in equation 2. equation is: .10*x + .12*(8000-x) = 860 simplify to get: .10*x + .12*8000 - .12*x = 860 simplify to get: -.02*x + 960 = 860 subtract 960 from both sides of this equation to get: -.02*x = -100 divide both sides of this equation by -.02 to get: x = 5000 that means y = 3000 because x + y = 8000. you have x = 5000 and y = 3000. substitute in equation 2 to get: .10*5000 + .12*3000 = 860 simplify to get: 500 + 360 = 860 which is true confirming the values for x and y are good. your answer is: \$5000 was invested at 10% and \$3000 was invested at 12%.