SOLUTION: Consider the following system of equations 1. 4x+3y+3z=-8 2. 2x+y+z=-4 3. 3x-2y+(m^2-6)z=m-4 Determine the value(s) of m for which this system of equations will have: a.

Algebra ->  Coordinate Systems and Linear Equations -> SOLUTION: Consider the following system of equations 1. 4x+3y+3z=-8 2. 2x+y+z=-4 3. 3x-2y+(m^2-6)z=m-4 Determine the value(s) of m for which this system of equations will have: a.       Log On


   

Question 775466: Consider the following system of equations
1. 4x+3y+3z=-8
2. 2x+y+z=-4
3. 3x-2y+(m^2-6)z=m-4
Determine the value(s) of m for which this system of equations will have:
a. no solution
b. one solution
c. an infinite number of solution
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
1. 4x%2B3y%2B3z=-8
2. 2x%2By%2Bz=-4
3. 3x-2y%2B%28m%5E2-6%29z=m-4+
__________________
1. 4x%2B3y%2B3z=-8
2. 2x%2By%2Bz=-4...multiply by 2
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1. 4x%2B3y%2B3z=-8
2. 4x%2B2y%2B2z=-8......and subtract from 1.
___________________
4x%2B3y%2B3z-%284x%2B2y%2B2z%29=-8-%28-8%29
4x%2B3y%2B3z-4x-2y-2z=-8%2B8
y%2Bz=0.......solve for y
y=-z........go to 2. .substitute y
2x-z%2Bz=-4
2x=-4
x=-4%2F2
highlight%28x=-2%29.......first solution
go to 3.,substitute x and y solve for z

3%28-2%29-2%28-z%29%2B%28m%5E2-6%29z=m-4+
-6%2B2z%2B%28m%5E2-6%29z=m-4+
2z%2B%28m%5E2-6%29z=m-4+%2B6
%28m%5E2-4%29z=m%2B2
z=%28m++%2B2%29%2F%28m%5E2-4%29
z=%28m++%2B2%29%2F%28%28m-2%29%28m%2B2%29%29
z=cross%28%28m++%2B2%29%291%2F%28%28m-2%29cross%28%28m%2B2%29%29%29
z=1%2F%28m-2%29 => y=-1%2F%28m-2%29 since y=-z
so, solution will exist for all real numbers m except m=2
if %28m-2%29+=0 => m=2=>z=1%2F%282-2%29=1%2F0........there will be no solutions
if m=1=>z=1%2F%281-2%29=-1 and y=1
or
if m=3=>z=1%2F%283-2%29=1 and y=-1
so, real integer solutions are:
x=-2, y=1,z=-1 for m=1
and
x=-2, y=-1,z=1 for m=3
and
x=-2, y=-1%2F2,z=1%2F2 for m=4
and so on......your answer is:
c. an infinite number of solution