paulo1913 Posted October 26, 2007 Posted October 26, 2007 how do you solve pairs of simultaneous linear equations?
NeonBlack Posted October 26, 2007 Posted October 26, 2007 There are many ways. My personal favorite is Cramer's rule, which I don't have time to explain right now, maybe you can look it up on wiki. The simplest is probably just to write your equations in the matrix form [math]Ax=B[/math] Then find the inverse matrix of A and then [math]x=A^{-1}B[/math]
paulo1913 Posted October 26, 2007 Author Posted October 26, 2007 Is there a way to do it which is simpler? I just don't understand that...
Cap'n Refsmmat Posted October 26, 2007 Posted October 26, 2007 There's the substitution method. [math]2x + 4y = 10[/math] [math]2x = y[/math] Your goal is to find x and y. You know that y is the same as 2x because of the second equation, so you can change the first to: [math]2x + (4 \times 2x) = 10[/math] or just [math]2x + 8x = 10[/math] Solve it and you find that x = 1. Now you just take the second equation to find y. It can be more complicated than that -- you may have to solve one equation for y or x to get started -- but substitution works that way in general.
Royston Posted October 26, 2007 Posted October 26, 2007 Alternatively there's the elimination method... [math]3 + 5x = 2y[/math] [math] 5 - x = y[/math] You need to find a way to make the multiple of either variable (x or y) the same in both equations, so the obvious would be to multiply by 5... [math] 3 + 5x = 2y[/math] [math] 25 - 5x = 5y[/math] Add the two values on the left, the x terms cancel, so you get... [math] 28 = 7y[/math] or [math]y = 4[/math] Now just substitute [math]y = 4[/math] into the original equation, and you get [math]x = 1[/math] I'll show you another alternative, but I'm just about to head out. Depending on the equation, some methods are more appropriate than others.
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