SOLUTION: I need help with a word problem please. I paid $30.00 for 2 lawn chairs of different quality. At the same prices, 3 of the cheaper chairs would have cost $3.00 less than 2 of the

Algebra ->  Algebra  -> Coordinate Systems and Linear Equations  -> Linear Equations and Systems Word Problems -> SOLUTION: I need help with a word problem please. I paid $30.00 for 2 lawn chairs of different quality. At the same prices, 3 of the cheaper chairs would have cost $3.00 less than 2 of the      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 7672: I need help with a word problem please.
I paid $30.00 for 2 lawn chairs of different quality. At the same prices, 3 of the cheaper chairs would have cost $3.00 less than 2 of the better ones. Find the cost of each chair.
Thank you,
M.D.

Answer by prince_abubu(198) About Me  (Show Source):
You can put this solution on YOUR website!
So the situation says that the cheap chairs cost the same and the better chairs cost the same, too. So let's call price of one cheap chair C and the price of one better chair B.

You also said that you bought 1 chair of each type, so that would be two different chairs and it totalled $30.00. So, one of your equations has to be +B+%2B+C+=+30.00+.

The second sentence is quite tricky. They're giving you a situation here, which happens to be the relationship of the chairs' prices. "Three chairs cost" translates to 3C, and "2 of the better ones" would be 2B. Now, the trick here is with the "3.00 less". One way to think about this is to decide which of 3C and 2B costs less. The 3C costs less, and so it's NOT the one that needs to be lessened by $3.00. The more expensive deal, which is the 2B would need to be knocked down by $3.00 to equal the cost of three cheaper chairs. So you've got a second equation: +3C+=+2B+-+3.00+.

You have a system of 2 linear equations with 2 variables. Luckily one of your equations has variables that are already by themselves (with no coefficients unlike the 2 in the 2C and the 3 in the 3B). You can take the +B+%2B+C+=+30.00+ and solve for whichever variable you want and plug it into the other equation. Let's solve for C, the cheaper chair. +C+=+30.00+-+B+.

Now we can plug in +30.00+-+B+ in place of C into +3C+=+2B+-+3.00+ so that the resulting equation will only be in one variable:

+3%2830.00+-+B%29=+2B+-+3.00+ <----- substituted
+90.00+-+3B=+2B+-+3.00+ <----- used distributive property
+93.00+-+3B+=+2B+ <---- added 3.00 to both sides
+93.00+=+5B+ <----- combined like terms across the equals sign. (AKA, add 3B to both sides.)
+B+=+18.60+ <---- divide. Now you know that the better chair is 18.60. If you said you spend $30.00 and the better chair was 18.60, then the cheaper chair would have been $11.40.