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My question is how do you solve 4x + 7y = 10
2x + 3y = 3
thankyou
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.
Last edited by JaneFairfax (2007-04-21 01:00:49)
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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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For a 2×2 matrix?
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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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