SOLUTION: A man has $295,000 invested in three properties. One earns 12%, one 10% and one 8%. His annual income from the properties is $28,100 and the amount invested at 8% is twice that inv

Algebra ->  Coordinate Systems and Linear Equations  -> Linear Equations and Systems Word Problems -> SOLUTION: A man has $295,000 invested in three properties. One earns 12%, one 10% and one 8%. His annual income from the properties is $28,100 and the amount invested at 8% is twice that inv      Log On


   

Question 1140876: A man has $295,000 invested in three properties. One earns 12%, one 10% and one 8%. His annual income from the properties is $28,100 and the amount invested at 8% is twice that invested at 12%.
Answer by ikleyn(52794) About Me  (Show Source):
You can put this solution on YOUR website!
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The question (which you forgot to formulate) is to find each investment.


Let x be the amount invested at 12% (in dollars).


Then the amount invested at 8% is (2x) dollars, according to the condition;

the amount invested at 10% is the rest  (295000-x - 2x) = (295000-3x)  dollars.


The money equation is the total income from three properties


    0.08*(2x) + 0.1*(295000-3x) + 0.12*x = 28100.


It is the equation for one single unknown, which you can easily solve.


When you solve it and will find x, do not forget to calculate all other participating amounts.