I am trying to understand how to determine the difference between no solutions for a system and infinitely many solutions.
I have the set of equations:
x + 2y - z = 1
2x + y + z = 1
x - y + 2z = 1
To start, I make sure the determinants do not equal zero.
They do indeed equal zero so there are no solutions.
My next system is :
u + 2v -w = 2
2u + v + w = 1
u - v + 2w = - 1
Again, I check and the determinants equal zero. So, I assume that there are no solutions.
However, my textbook states that for this second system there are infinitely many solutions where u = - w, v = w + 1 and w can have any value. Please explain the difference to me!!!! Thanks Frostking
See this sample question,,,,it will really help u
Sample: Solve for x, y and z.
2x - y + 6z = 1 (equation 1)
x - y + z = 2 (equation 2)
x + y + z = 1 (equation 3)
Steps:
1) Solve equation 1 for y.
After applying algebra, equation 1 becomes y = 2x + 6z - 1.
2) Plug the value of y or (2x + 6z -1) into equations 2 and 3.
Equation 2:
x - y + z = 2
x -(2x + 6z -1) + z = 2
x -2x - 6z + 1 = 2
-x - 6z = 2-1
-x -6z = 1
Equation 3:
x + y + z = 1
x + 2x + 6z -1 + z = 1
3x + 7z -1 = 1
3x + 7z = -1+1
3x + 7z = 0
We now have come down to a system of two equations and two variables.
Here are the two equations and two variables:
-x - 6z = 1 (equation A)
3x + 7z = 0 (equation B)
3) Solve equation A for x.
-x - 6z = 1
-x = 6z + 1
x = -6z - 1
4) Plug the value for x into equation B to solve for z.
3x + 7z = 0
3(-6z - 1) + 7z = 0
-18z - 3 + 7z = 0
-11z - 3 = 0
-11z = 3
z = - 11/3
5) Now plug the value for z into equation A to find x.
-x - 6z = 1
-x -6(-11/3) = 1
-x + 22 = 1
-x = -22 + 1
-x = -21
x = 21
Lastly or step 6, go back to ANY of the original 3 equations to find the value of y by substituting what you found for x and z. I will select equation 3.
Equation 3:
x + y + z = 1
21 + y -11/3 = 1
52/3 + y = 1
y = -52/3 + 1
y = -49/3
Final answer: x = 21, y = -49/3 and z = -11/3