A house and a lot together cost more than $86,000. The house costs $2000 more than six times the costs of the lot. How much does the lot cost by itself?
If house costs x and lot costs y then x+y=86000, and x=6y+2000 or x-6y=2000.
Subtracting second eq. from first we get
6y+y=86000-2000 or 7y=84000 or y=12000.
Thus lot costs $12000
House costs 86000-12000=$74000.
To find the cost of the lot, we'll first set up a system of equations based on the given information.
Let's assume the cost of the lot is "x" dollars.
According to the problem, the cost of the house is $2000 more than six times the cost of the lot. Therefore, the cost of the house can be expressed as 6x + $2000.
We also know that the combined cost of the house and lot is more than $86,000. So, we can set up the following equation:
x + (6x + $2000) > $86,000
Now, let's simplify and solve the equation:
7x + $2000 > $86,000
Subtracting $2000 from both sides:
7x > $84,000
Dividing both sides by 7:
x > $12,000
Therefore, the cost of the lot by itself is more than $12,000.