x+3y+z = a^2?

Question

x+3y+z = a^2
2x+5y+2az= 0
x+y+a^2z =-9

solve the system, for what values of a is the system inconsistant.

Ive been working on this for 3 days now, its my last problem, class ends tomorrow, no matter what

Thanks for the help, the one I chose as the best answer was absolutely right, and I came up with the same, finally. and based on the value of z. a=1 and a=3 made the system inconsistant. Appreciate the help, and glad to be done with College Algebra.

Answer 1

You need to do your own homework.

Answer 2

First express x,y and z in terms of a. Then work on the inconsistency part. If you can solve a system of 3 equations with 3 variables, then this shouldn't be too much of a problem.

x+3y+z = a^2 (equation1)
2x+5y+2az= 0 (equation2)
x+y+a^2z =-9 (equation3)

So from 1:
x = a^2 - 3y - z (equation4)
Put that into equation2
2(a^2 - 3y - z) + 5y + 2az = 0
2a^2 - 6y - 2z + 5y + 2az = 0
2a^2 - y - 2z + 2az = 0
So, y = 2a^2 - 2z + 2az

Equation 4 becomes:
x = a^2 - 3y - z
x = a^2 - 3(2a^2 - 2z + 2az) - z
x = a^2 - 6a^2 + 6z - 6az - z
x = -5a^2 + 5z - 6az

Substitute x and y into the final equation (equation3) to solve for z in terms of 'a'
x+y+a^2z =-9 (equation3)
(-5a^2 + 5z - 6az) + (2a^2 - 2z + 2az) + a^2z =-9
-3a^2 +3z - 4az + a^2z = -9
Separate z terms on one side
3z - 4az + a^2z = -9 + 3a^2
Factor out z:
z(3-4a+a^2) = -9 + 3a^2
Divide by (3-4a+a^2):
z = (-9 + 3a^2)/(3-4a+a^2)
Observe that (3-4a+a^2) can be factored as follows:
3-4a+a^2 = a^2 - 4a + 3 = (a-1)(a-3)

So z= (-9 + 3a^2)/(a-1)(a-3)
Substitute this expression for z into x and y:

Remember the expressions are:

x = -5a^2 + 5z - 6az
y = 2a^2 - 2z + 2az

When you are done, that will be the solution. I guess the system will be inconsistent when x=1,3, although the final solution will depend on what the expressions for x and y look like.