Eliminatory Excess?Date: 08/31/2015 at 11:03:26 From: cortney Subject: solving stytems of equations three vaitables (elemination) I'm supposed to eliminate one variable from this system of equations: -x - 5y - 5z = 2 -4x - 5y - z = -20 x + 5y - 2z = -20 I'm trying to get rid of x in the first and second equations, but the y is not letting me do that. The x's and y's cancel each other out as soon as you put the problem together! I know how to eliminate variables. I already know that you have to multiply the whole equation. But I find this system confusing because in the first and last equations, the x and y cancel each other out. I have seen this type of problem several times before and I never know what to do about it. Date: 08/31/2015 at 11:44:44 From: Doctor Peterson Subject: Re: solving stytems of equations three vaitables (elemination) Hi, Cortney. When something unexpected happens, you can either change your strategy to avoid it (e.g., eliminate a different variable or use a different pair of equations), or think of it as enabling you to skip a step. It's sort of like when you are going upstairs and think there is another step, so if you are not paying attention, you can stumble when your foot touches nothing. If you ARE paying attention, you just conclude, "I got there sooner than I expected!" So let's take what you were doing, and see how it is not really a problem. You have this system: -x - 5y - 5z = 2 [A] -4x - 5y - z = -20 [B] x + 5y - 2z = -20 [C] You chose to add equations [A] and [C], intending to eliminate x. But you found that you accidentally eliminated both x and y: [A] -x - 5y - 5z = 2 [C] x + 5y - 2z = -20 ------------------- -7z = -18 All that happened is that you prematurely solved for a variable: z = 18/7 That's a GOOD thing! Now you have to find x and y, so you'll have to do something a little different -- but not that different: just continue your plan and eliminate x from a different pair of equations, say [B] and [C]: [B] -4x - 5y - z = -20 4[C] 4x + 20y - 8z = -80 --------------------- 15y - 9z = -100 Your original plan at this point was to have a pair of equations in the two variables y and z. It's actually a little better than you planned: you have already solved for z, and you have another equation that you can solve for y by substituting 18/7 for z. Then you can carry on with the standard plan by substituting for both y and z in any of the original equations, and solving for x. (By the way, the fractions we're getting suggest an error in what you wrote. It may be that the solution really does involve fractions, but double-check the original statement of the problem to make sure.) The standard plan is to combine pairs of equations to get two equations in two unknowns, combine those to solve for one variable, then go back and pick up the variables you'd eliminated. I think of this procedure as like climbing a mountain, then going back down and picking up the equipment you discarded on the way. [F] one variable / \ [D] [E] two variables / \ / \ [A] [C] [B] three variables Your equation [D] happens to have only one variable, but you can still use it just as planned. If finding something unexpected throws you for a loop, you can also change your plan, as I said earlier, and choose a different strategy. You might avoid eliminating x or y first, and instead eliminate z by combining [A] and [B], then [B] and [C]. This will work fine. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ Date: 08/31/2015 at 12:01:21 From: cortney Subject: Thank you (solving stytems of equations three vaitables (elemination)) OMG thank you so much. I never realized that this could even happen! |
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