SOLUTION: The problem is: Suppose that a lot costs $60,000 and a builder charges $84 per square foot of living area in a house. How big, to the nearestr square foot, is a house that can be

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Question 56273This question is from textbook Algebra 2
: The problem is:
Suppose that a lot costs $60,000 and a builder charges $84 per square foot of living area in a house. How big, to the nearestr square foot, is a house that can be purchased for $180,000?
What I've tryied using so far is 180000=-84x-60000, and have come up with three answers... 2143 square feet, 2857 square feet, and 1429 square feet. Which one (if any) is right? This question is from textbook Algebra 2

Answer by funmath(2933) About Me  (Show Source):
You can put this solution on YOUR website!
The problem is:
Suppose that a lot costs $60,000 and a builder charges $84 per square foot of living area in a house. How big, to the nearest square foot, is a house that can be purchased for $180,000?
Your model should be:
180000=84x%2B60000
180000-60000=84x
120000=84x
120000%2F84=84x%2F84
1428.57=x That was one of your choices if you round off to the nearest whole number: x=1429 sq ft.
Happy Calculating!!!