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#1 2007-04-21 00:11:25

mumble3
Guest

Algebra equation problem

My question is how do you solve      4x + 7y = 10
                                                    2x + 3y = 3
thankyou

#2 2007-04-21 00:58:56

JaneFairfax
Member
Registered: 2007-02-23
Posts: 6,868

Re: Algebra equation problem

Use matrices.

However, I just noticed that if you multiplied the second equation by −2 and add to the first equation, you get y = 4 straightaway. tongue

Last edited by JaneFairfax (2007-04-21 01:00:49)

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#3 2007-04-21 04:34:46

Ricky
Moderator
Registered: 2005-12-04
Posts: 3,791

Re: Algebra equation problem

Isn't finding the inverse of a matrix computationally harder than reducing a matrix to row echelon form?


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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#4 2007-04-21 04:45:39

JaneFairfax
Member
Registered: 2007-02-23
Posts: 6,868

Re: Algebra equation problem

For a 2×2 matrix?

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#5 2007-04-21 07:50:04

Zhylliolom
Real Member
Registered: 2005-09-05
Posts: 412

Re: Algebra equation problem

First of all, I think the poster is in an algebra class, not a linear algebra class. Here is how you would solve this problem in an algebra class:

We have the equations 4x + 7y = 10, 2x + 3y = 3. If we multiply the second equation by -2 (which makes it -4x -6y = -6) and add it to the first, look what happens!

     4x + 7y = 10
+(-4x - 6y = -6)
             y = 4.

The x's cancelled, and now we know what y is. We can then put y = 4 into one of our equations to find x. Let's put it in the second equation:

2x + 3y = 3
2x + 3(4) = 3
2x + 12 = 3
2x = -9
x = -9/2.

So the solution is x = -9/2, y = 4.

And yes, it will usually take you n times as long to invert an n×n matrix as opposed to using row reduction. Even in the case of a 2×2 matrix, row reducing will only take one or two simple steps (scaling a row and adding basically), while using inverses requires you to calculate the inverse (sure it's a simple formula) and then multiply this inverse on the right by the b vector. It's a little more work than just adding rows. It's common to run into singular matrices in homework assignments as well.

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