ThisIsMyName Posted September 22, 2006 Report Share Posted September 22, 2006 I'm in pre-calc...The question is:x + y - z = -1 x + y + z = 33x - 2y - z = -4 I got this far on that question: 2( x + y - z = -1 ) 2x + 2y - 2z = -2 + 3x - 2y - z = -4 5x - 3z = -6 I remember getting stuck the same way when I was in Algebra 2 and finding out I was making the dumbest mistake... I'm doing the elimination process because I think it's easier... Do you think I should try the same problem with substitution? If someone can point me in the right direction or tell me where I went wrong, I'd be a-mighty appriciative! Thanks! [PS- No I don't expect you to give me the answer... All I want is to [a] make sure I'm doing this right & find out what I'm doing wrong and any pointers someone can give me] Quote Link to comment Share on other sites More sharing options...

CraigD Posted September 22, 2006 Report Share Posted September 22, 2006 I'm in pre-calc...The question is:x + y - z = -1 x + y + z = 33x - 2y - z = -4... All I want is to [a] make sure I'm doing this right & find out what I'm doing wrong and any pointers someone can give me]What you should do is attempt to multiply then add each equation to make the coefficient of each term zero in all but one of the equations. For example, your first step could be to writex + y - z = -1 (x + y + z) – (x + y – z) = 3 + 1(3x - 2y – z) -3(x + y - z) = -4 +3then expand and simplify it tox + y - z = -1 0 +2y + 0 = 40 -5y +2z = -1 In the next step, make the coefficient of y zero in every equation but one (but not the one where x has a non zero coefficient), and in the final step, make the coefficient of z zero all but 1 equation. This is known as Gaussian elimination. A nice thing about it is it will also tell you that the system of equations has no solution, if any step creates a line with all coefficients zero. Post your answer when you get it, or let us know if you need more hints… IDMclean 1 Quote Link to comment Share on other sites More sharing options...

IDMclean Posted September 22, 2006 Report Share Posted September 22, 2006 Wouldn't a matrix be good here... I sort of remember they were excellent for solving multi-variable equations. Quote Link to comment Share on other sites More sharing options...

Turtle Posted September 22, 2006 Report Share Posted September 22, 2006 Wouldn't a matrice be good here... I sort of remember they were excellent for solving multi-variable equations. A matrix (plural matices) does have such an application in linear programming, however This Is My Name makes clear in the context of the query that it isn't the method under study. Craigs reply lays the proper foundation.:cocktail: Quote Link to comment Share on other sites More sharing options...

IDMclean Posted September 22, 2006 Report Share Posted September 22, 2006 Just contributing what I know. I can't say that I really know calculus at all. As for the mistake in grammar, I point you to my reason for edit. I's ain't sucha gud speller, heer-yuk. Quote Link to comment Share on other sites More sharing options...

Turtle Posted September 22, 2006 Report Share Posted September 22, 2006 Just contributing what I know. I can't say that I really know calculus at all. As for the mistake in grammar, I point you to my reason for edit. While arguably OT, I will reply in a manner consistant with the thread that is on topic. First, this isn't a Calculus problem; it is an issue of linear programming.Second, the grammar mistake is in regard to number and inasmuch as the subject at hand is math, mistakes in number are non-trivial.Last, I appreciate your input and acknowledged in it the correctness as well as the error. Again, this is math and without rigor it is worthless. In all probability I know these terms from having mistaken and misused them in error myself. I strive to make new mistakes, rather than repeat those of the past.:cocktail: Quote Link to comment Share on other sites More sharing options...

Racoon Posted September 22, 2006 Report Share Posted September 22, 2006 I'm in pre-calc...The question is:x + y - z = -1 x + y + z = 33x - 2y - z = -4 I got this far on that question: 2( x + y - z = -1 ) 2x + 2y - 2z = -2 + 3x - 2y - z = -4 5x - 3z = -6 I remember getting stuck the same way when I was in Algebra 2 and finding out I was making the dumbest mistake... I'm doing the elimination process because I think it's easier... I'm trying to forget how to do those. :) Thanks for reminding me I need to do the next level math class. Listen to Craig and Turtle. I don't have any advice beyond whats already been given, but just wanted to say good thread... They get much easier to make a mistake with when larger numbers, fractions and decimals become involved. :) Quote Link to comment Share on other sites More sharing options...

cwes99_03 Posted September 22, 2006 Report Share Posted September 22, 2006 Linear Programming AKA linear algebra. Simplest way is to do the top {(x + y + z) – (x + y – z) = 3 + 1} and then solve for y. Using y solve the second part for z, then using y and z solve for x.Always look for the simplest two equations to work with first.You have to view these things as three different lines going through the same point. (x,y,z). Thus you need at least three lines because you are in 3-dimensional space and none of the equations specifies one of the values (such as x=2).As a general rule1) Solve the two simplest equations for a particular value z= ?x +#y . 2) then substitute that equation into one of the other equations (thereby eliminating the term z) and repeat. 3) The second repetition will set x=&y (or y=&x) and now you can substitute that into the final equation to get x=? or y=?. This is the very same thing you do when you put all the numbers into a matrix, however, once you learn how to use a matrix you can do all that math in much less time, and you can do much more powerful things with a matrix. Quote Link to comment Share on other sites More sharing options...

Southtown Posted September 22, 2006 Report Share Posted September 22, 2006 This is known as Gaussian elimination. A nice thing about it is it will also tell you that the system of equations has no solution, if any step creates a line with all coefficients zero.Nice link dude! I actually solved it. (dropout) Neat little solution too. Someone rep that mofo for me. Quote Link to comment Share on other sites More sharing options...

CraigD Posted September 24, 2006 Report Share Posted September 24, 2006 Wouldn't a matrix be good here... I sort of remember they were excellent for solving multi-variable equations.Indeed they are. Matrix notation can be thought of as a compact way to write systems of equations. Here’s the solution to 3-equation, 3-variable equation, using algebraic notation:a +2b +3c = 14a +4b -c = 63a -2b +6c = 17 a +2b +3c = 14(a +4b -c) -(a +2b +3c) = 6 -14(3a -2b +6c) -3(a +2b +3c) = 17 -3*14 a +2b +3c = 142b -4c = -8-8b -3c = -25 (a +2b +3c) -2(b -2c) = 14 -2*-4b -2c = -4(-8b -3c) -8(b -2c) = -25 -8*-4 (a +2b +3c) -2(b -2c) = 14 -2*-4b -2c = -4(-8b -3c) +8(b -2c) = -25 +8*-4 a +7c = 22b -2c = -419c = 57 (a +7c) -7c = 22 -7*3(b -2c) +2c = -4 +2*3c = 3 a = 1b = 2c = 3 Here it is using matrix notation (with notes in the right margin): [math]\begin{bmatrix}1 & 2 & 3 &14 \\ 1 & 4 & -1 & 6 \\ 3 & -2 & 6 & 17 \end{bmatrix}[/math] [math]\begin{bmatrix}1 & 2 & 3 & 14 \\ 0 & 1 & -2 & -4 \\ 0 & -8 & -3 & -25 \end{bmatrix} \begin{matrix} \\ R2=(R2-R1)/2 \\ R3=R3-3R1 \end{matrix}[/math] [math]\begin{bmatrix}1 & 0 & 7 & 22 \\ 0 & 1 & -2 & -4 \\ 0 & 0 & 1 & 3 \end{bmatrix} \begin{matrix}R1=R1-2R2 \\ \\ R3=(R3+8R2)/19 \end{matrix}[/math] [math]\begin{bmatrix}1 & 0 & 0 & 1 \\ 0 & 1 & 0 & 2 \\ 0 & 0 & 1 & 3 \end{bmatrix} \begin{matrix}R1=R1-7R3 \\ R2=R2+2R3 \end{matrix}[/math] Both notations require the same amount of arithmetic, but the matrices are easier on the eye (expecially with hypography’s nice [math]\LaTeX{}[/math] support!) :) Quote Link to comment Share on other sites More sharing options...

eric l Posted September 24, 2006 Report Share Posted September 24, 2006 The question is:x + y - z = -1 x + y + z = 33x - 2y - z = -4 Actually, this one is rather easy if we work by elimination. Equations 1 and 2 botht have (1x + 1y) (I added the coëfficient 1 to make things clearer). We can substract them which gives :[x + y + z] - [x + y - z] = 3 - [-1]or : z - [-z] = 3 + 1 = 4or : 2 z = 4 or z = 2put this in equation 2 and we have : x + y + 2 = 3or x + y = 1 (equation 4)putting the same in equation 3 gives :3x - 2y - 2 = - 4or 3 x - 2y = - 2 (equation 5)Now we eliminate y by adding up 2 times equation 4 and 1 time equation 5.This gives :2 [x + y] + [3 x - 2 y] = 2 [1] +[- 2]or 2x + 2 y + 3x - 2y = 2 - 2or 5x = 0or x = 0In the last stage we put this value for x in either equation 4 or equation 5. Equation 4 is easier so we have0 + y = 1or : y = 1Result : x = 0; y = 1 and z = 2This we check in the original equations. If you have equations with simple coëfficients (like these), elimination is the easiest way to proceed : you find your variables one by one, though not necessarily in alphabetical order. Turtle 1 Quote Link to comment Share on other sites More sharing options...

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