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!
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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.
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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
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Now sub for x in the other 2 eqns
x + y + z = 3
3x + 3y + 3z = 10
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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.
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