A building contractor is planning to build an apartment complex with one, two or three bedroom apartments. Let x,y,z respectively denote the number of apartments of each type to be built. Suppose that the builder will spend a total of $8,277,000, and that the costs for the three types of apartments are $25,000, $35,000, and $54,000 respectively. Then the 'cost equation' for the builder is:

?x + ?y + 54 z =?

Suppose he plans to build a total of 236 apartments, then complete the equation in x, y, and z which describes this:

x +(?)y +(?)z = 0.

Finally, find the solution to these equations: x =?, y =?, and z =?

The two equations, based on your fill-in-the-blanks above are,

25x + 35y + 54z = 8277
x + y + z = 236

Are you sure this is the exact problem as written?

Are far as I know, this problem cannot be solved to get an exact number for x, y and z .

To solve simultaneous equations in three unknown requires three equations (I've been taught).

If this can be solved with only two equations it is beyond my scope of my knowledge regarding simultaneous equations.

To find the cost equation for the builder, we need to multiply the cost per apartment by the number of apartments of each type and then sum them up.

Let's represent the number of one-bedroom apartments as x, the number of two-bedroom apartments as y, and the number of three-bedroom apartments as z.

The cost equation for the builder is:

25,000x + 35,000y + 54,000z = ?

Since the builder plans to build a total of 236 apartments, the equation in x, y, and z describing this would be:

x + y + z = 236

To find the solution to these equations, we need one more equation. This can be obtained from the fact that the builder will spend a total of $8,277,000. Using the cost equation, we can set it equal to this amount:

25,000x + 35,000y + 54,000z = 8,277,000

Now we have a system of three equations:

25,000x + 35,000y + 54,000z = 8,277,000
x + y + z = 236

We can solve this system to find the values of x, y, and z.

To solve this system of equations, we can use various methods such as substitution, elimination, or matrix methods. Let's use the elimination method.

First, let's multiply the second equation by 25,000 to eliminate x:

25,000(x + y + z) = 25,000 * 236
25,000x + 25,000y + 25,000z = 5,900,000

Now, subtract this equation from the first equation:

25,000x + 35,000y + 54,000z - (25,000x + 25,000y + 25,000z) = 8,277,000 - 5,900,000

Cancel out common terms:

10,000y + 29,000z = 2,377,000

We now have a system of two equations:

10,000y + 29,000z = 2,377,000 (Equation 1)
x + y + z = 236 (Equation 2)

To solve this system, we can either solve one equation for one variable and substitute it into the other equation or use matrix methods. Let's solve Equation 2 for x:

x = 236 - y - z

Substituting this into Equation 1, we get:

10,000y + 29,000z = 2,377,000

From here, we can solve for y and z. Once we have the values for y and z, we can substitute them back into Equation 2 to calculate the value of x.