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Question 159233: Hi,-
I have a problem with substitution that does not allow the use of matrices:
2x + y = -2
-x + 3y - 4z = -26
5x - 6y + z = 17
I started by solving for y in the first equation: y = -2 - 2x
I then solved for x in the second equation: 3y -4z +26 = x
I plugged the x and y values into the third equation
5(3(-2-2x) - 6(-2 - 2x) + z = 17
5(-6 -6x) + 12 + 12x + z = 17
-30 + 30x + 12 + 12x + z = 17
42 x + z -18 = 17
42x + z = 35
z = 35 - 42x
Now I'm stuck.
Thank you.
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! Hi, I have a problem with substitution that does not allow the use of matrices:
2x + y = -2
-x + 3y - 4z = -26
5x - 6y + z = 17
I started by solving for y in the first equation: y = -2 - 2x
I then solved for x in the second equation: 3y -4z +26 = x
I plugged the x and y values into the third equation
5(3(-2-2x) - 6(-2 - 2x) + z = 17
5(-6 -6x) + 12 + 12x + z = 17
-30 + 30x + 12 + 12x + z = 17 (should be -30x)
42 x + z -18 = 17
-18x + z = 35
z = 35 + 18x
Now I'm stuck.
Thank you.
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Since there's no z term in the 1st eqn, sub for y in eqns 2 & 3. Then you'll have 2 eqns in 2 unknowns.
I started by solving for y in the first equation: y = -2 - 2x
-x + 3y - 4z = -26
5x - 6y + z = 17
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-x + 3(-2-2x) -4z = -26
5x -6(-2-2x) +z = 17
collect terms
-x -6-6x -4z = -26
5x +12 +12x +z = 17
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-7x -4z = -20
17x + z = 5
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Take the easy one, z = 5-17x
Sub into -7x -4z = -20
-7x -4(5-17x) = -20
-7x -20 + 68x = -20
61x = 0
x = 0
So y = -2
z = 5
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Check each original eqn:
2x + y = -2
-x + 3y - 4z = -26
5x - 6y + z = 17
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The 1st is good.
0 + 3(-2) - 4*5 = -26 is good.
0 -6(-2) + 5 = 17 is good.
It all works.
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I think doing eliminating 1 variable at a time works better. Other methods might work, too, as you tried, but it's more confusing.
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