SOLUTION: I have a substitution problem that I would like help with, please (I have a test tomorrow). x + y + z = 3 3x + 3y + 3z = 10 x - 3y + 4z = 6 I know that I could start by sol

Question 159260: I have a substitution problem that I would like help with, please (I have a test tomorrow).
x + y + z = 3
3x + 3y + 3z = 10
x - 3y + 4z = 6
I know that I could start by solving for x in the third equation:
x = 3y - 4z +6
I can also solve for y in the first equation: y = 3 - x - z
I could then substitute y in the second and third equation:
x + 3 - x - z + z = 3 but this makes all variables 0
x - 3(3 - x - z) + 4z = 6 x - 9 + 3x + 3z + 4z = 3 4x + 7z = 12
Now I'm stuck. Thank you!
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
I have a substitution problem that I would like help with, please (I have a test tomorrow).
x + y + z = 3
3x + 3y + 3z = 10
x - 3y + 4z = 6
I know that I could start by solving for x in the third equation:
x = 3y - 4z +6
I can also solve for y in the first equation: y = 3 - x - z
I could then substitute y in the second and third equation:
x + 3 - x - z + z = 3 but this makes all variables 0
x - 3(3 - x - z) + 4z = 6 x - 9 + 3x + 3z + 4z = 3 4x + 7z = 12
Now I'm stuck. Thank you!
----------------
The best approach is to solve for one variable, then sub into the other 2 eqns. The result will be 2 eqns in 2 variables. Then sub for one of those.
------------------
x + y + z = 3
3x + 3y + 3z = 10
x - 3y + 4z = 6
I know that I could start by solving for x in the third equation:
x = 3y - 4z +6
--------------
Now sub for x in the other 2 eqns
x + y + z = 3
3x + 3y + 3z = 10
---------------
3y-4z+6 + y + z = 3
3(3y-4z+6) +3y +3z = 10
Collect terms
4y - 3z = -3
12y -9z = -8
Solve the 1st for y
y = (-3+3z)/4
Sub into the 2nd
12((-3+3z))/4 - 9z = -8
-9 + 9z - 9z = -8
-9 = -8 This means there's no solution.
If you look at the 1st and 2nd original eqns, all 3 variables in the 2nd are 3 times in the 1st, so there are actually 2 eqns (with an inconsistency) in 3 unknowns. There is no one solution, there are an infinite number of solutions.

RELATED QUESTIONS
2x-3y+4z=1 x+3y-2z=4 3x-y+z=2 So far i have tried this.. 2x-3y+4z=-1 x+3y-2x=4... (answered by Fombitz,earmstrong620)

If you could please help me I have four equations that i have to solve for w, x, y, and... (answered by stanbon)

2x-3y-z=0 x+y+z=3 4x-5y-z=2 I would like to know how to do this either by... (answered by scott8148)

I am have a problem simplifying (x^4y^3z)^3(-2x^3y^4) Please... (answered by mathchemprofessor)

how do you find the answer to x, y, and z if you have to use matrix row reduction? this (answered by scott8148)

solve the following when: x=4 and y=1 and z=4 11x+6-3y+2z= y+4z+2x= 2x+10-2z=... (answered by solver91311)

3x+3y+6z=9 2x+y+3z=7 x+2y-z=-10 I was provided work for this problem using the... (answered by Alan3354)

Hi,- I have a problem with substitution that does not allow the use of matrices: 2x + (answered by Alan3354)

I need to solve the following system of linear equations with three variables. Here is... (answered by ptaylor)