Directions: Write a system of equations to solve the following word problem graphically and algebraically (using Gauss and using substitution).
AMC Homes, Inc. is planning to build three- and four-bedroom homes in a housing development called Chestnut Hills. Consumer demand indicates a need for three times as many four-bedroom homes as for three-bedroom homes. The net profit for each three bedroom home is $16,000 and from each four-bedroom home, $17,000. If AMC Homes must net a total profit of $13.2 million from this development, how many homes of each type should they build?
Let T and F be the respective numbers of three and four bedroom houses built.
Then, 16,000T + 17,000F = 19,200,000
We also know that the four bedroom houses are to be three times the number of 3 bedroom houses.
Therefore, F = 3T.
Substitute F = 3T into the first expression and solve for T and then F.
let T be the number of 3 bedroom homes
let F be the number of 4 bedroom homes
16000T+17000F=13200000
16000T+17000(3T)=13200000
16000T+51000T=13200000
67000T=13200000
67000T/67000=13200000/67000
T=197
Therefore, there were 197 3 bedroom homes and 591 4 bedroom homes that were built.
Wow, looks like AMC Homes really has their numbers calculated! With 197 three bedroom homes and 591 four bedroom homes, they'll be on their way to netting that $13.2 million profit. You could say they're building a chestnut-tacular development!
To solve the word problem graphically, you can plot the equations on a graph and find the point of intersection.
The first equation is 16,000T + 17,000F = 19,200,000.
The second equation is F = 3T.
Let's plot these equations on a graph:
For the first equation, we have:
- When T = 0, F = 19,200,000/17,000 ≈ 1,129.4.
- When F = 0, T = 19,200,000/16,000 = 1,200.
For the second equation, we know that F = 3T.
Now, let's plot these points on a graph and find the intersection:
(T, F) = (0, 1,129.4) and (1,200, 0).
By graphing these equations, we find that the point of intersection is (197, 591).
Therefore, there were 197 three-bedroom homes and 591 four-bedroom homes that were built.
Using Gauss's method:
We'll start with the system of equations:
16,000T + 17,000F = 19,200,000
F = 3T
Substitute F = 3T into the first equation:
16,000T + 17,000(3T) = 19,200,000
Simplify the equation:
16,000T + 51,000T = 19,200,000
67,000T = 19,200,000
T = 19,200,000/67,000 ≈ 286.9
Since the number of three-bedroom homes cannot be a decimal, we can round it down to 286.
Now, substitute T = 286 back into the second equation to solve for F:
F = 3(286)
F = 858
Therefore, there were 286 three-bedroom homes and 858 four-bedroom homes that were built.
To solve this word problem, we can set up a system of equations. Let T represent the number of three-bedroom homes and F represent the number of four-bedroom homes to be built in the housing development.
According to the problem, the net profit for each three-bedroom home is $16,000 and for each four-bedroom home is $17,000. The total net profit must be $13.2 million.
The first equation can be written as: 16,000T + 17,000F = 13,200,000.
The problem also states that the number of four-bedroom homes should be three times the number of three-bedroom homes. This can be written as: F = 3T.
To solve graphically, we can plot the equations on a graph and find the point of intersection. The x-axis represents the number of three-bedroom homes (T), and the y-axis represents the number of four-bedroom homes (F).
To solve algebraically using Gaussian elimination, we can substitute F = 3T into the first equation:
16,000T + 17,000(3T) = 13,200,000
16,000T + 51,000T = 13,200,000
67,000T = 13,200,000
T = 13,200,000 / 67,000
T ≈ 197
Plugging this value back into F = 3T, we can find:
F = 3(197)
F ≈ 591
Therefore, to meet the net profit of $13.2 million, AMC Homes should build 197 three-bedroom homes and 591 four-bedroom homes.