Solving linear systems
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Solving linear systems

[From: ] [author: ] [Date: 11-10-08] [Hit: ]
this means u = 0 since its consistent in both equation 2 and 3. Sub u = 0 for any equation, lets say 2).Therefore, both u and v are = 0.Hope this helps.......
ive been doing practice questions all day and doing fine but I cannot seem to get this one. can anyone help?
Solve the given linear system
6u − v = 0
– 8u + 4v = 0
8u − 2v = 0

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Hi Chuckie.

Let me arrange the 3 into equations 1-3.

1) 6u − v = 0
2) 4v – 8u = 0
3) 8u − 2v = 0

Arrange 1) into v = 6u

Sub v=6u in both equation 2 and 3.

2) 4(6u) - 8u = 0
16u = 0

3) 8u - 2(6u) = 0
-4u = 0

Now hold on, this means u = 0 since it's consistent in both equation 2 and 3. Sub u = 0 for any equation, let's say 2).

4v -8(0) = 0
v = 0

Therefore, both u and v are = 0.

Hope this helps. :D

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u=v=0
1
keywords: Solving,linear,systems,Solving linear systems
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