SOLUTION: Chelsea sold 2 cars for $6,000 each, making 20% on car#1 & losing 20% on car#2. Did he make money, lose money or break even?

Algebra ->  Algebra  -> Coordinate Systems and Linear Equations  -> Linear Equations and Systems Word Problems -> SOLUTION: Chelsea sold 2 cars for $6,000 each, making 20% on car#1 & losing 20% on car#2. Did he make money, lose money or break even?      Log On

Ad: You enter your algebra equation or inequality - Algebrator solves it step-by-step while providing clear explanations. Free on-line demo .
Ad: Algebrator™ solves your algebra problems and provides step-by-step explanations!
Ad: Algebra Solved!™: algebra software solves algebra homework problems with step-by-step help!

   

Question 13241: Chelsea sold 2 cars for $6,000 each, making 20% on car#1 & losing 20% on car#2. Did he make money, lose money or break even?
Answer by bam878s(77) About Me  (Show Source):
You can put this solution on YOUR website!
Chelsea made a total of 12,000 dollars in sales (6,000 + 6,000)
For car 1, note he made 20% profit. So the price, call it x, was the price Chelsea paid for the car in the beginning. We have 6000 = x + x(.20). We can factor out an x on the right hand side of this equation to get 6000 = x(1 + .20). So to solve for x (the price Chelsea paid for car 1) divide 6000 by 1.20 to get x = 5,000.
For car 2, note he lost 20% on the car. Let y be the price Chelsea paid for car 2. Then 6,000 = y - y(.20). Factoring out a y on each side yields 6000 = y(1-.20) or 6,000 = y(.80). To solve for y we divide 6,000 by .80 to get $7,500.
So the total price Chelsea paid for the cars was x + y = $5,000 + $7,500 = $12,500. However he only sold the cars for $6,000 each and gained a total of $12,000. So, in the long run, Chelsea actually lost $500.