college algebra word problem study guide

Question

a pawnbroker buys a computer and a guitar for $500. He sells the computer at a markup of 40% but he has to take a loss of 10% on the guitar in order to sell it. If he makes a profit of $110 on the sale of the two items, what did he pay for the computer?

Answer 1

Let x = cost of computer and y = cost of guitar

 

The first equation is simple, cost of computer + cost of guitar = $500

 

x + y = 500

 

For the second equation, we look at the profit/loss on each item. If he sells the computer at a 40% mark up, we can express that as x(1+.4) or 1.4x. Since he lost 10% on the guitar, that would be y(1-.1) or .9y. He profited $110 which means his total money he received was $610 (500 + 110), so, 

 

1.4x + .9y = 610

 

We now have a system of equations that we can solve by graphing, linear combinations (elimination), or substitution. For the sake of this problem, I'll go with elimination and multiply the entire original equation by -.9. Then we can add the two equations together to eliminate the "y" value. Here's what that looks like now:

 

-.9x + -.9y = -500

1.4x +  .9y =  610

________________

0.5x           =  110

 

x = 220, he paid $220 for the computer.

 

Since x + y = 500, that means he paid $280 for the guitar.