# Solutions for linear equations

• Oct 11th 2009, 10:34 AM
Frostking
Solutions for linear equations
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
• Oct 11th 2009, 11:47 AM
skeeter
Quote:

Originally Posted by Frostking
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

note that the first equation = 2nd equation - 3rd equation

so the system really consists of just two equations w/ three variables, yielding an infinite number of solutions.
• Oct 11th 2009, 12:51 PM
jeremyparker10
Re: linear eqns with 3 variables
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