Solve the following system of equations. 2x + 2y + z = 10 3x - y + 3z = 10 2x + 3y - 2z = 6 I'm not good at systems of three equations. Anyone willing to help me start?
From what I attempted: 2x+2y+z=10 3x-y+3x=10 I multiplied the bottom equation by negative three and I got the equation: 8x+7z=30 after adding them. Was I supposed to do this first?
Can you tell me what you did to get those two equations?
Also, with those two equations do I use them together to solve for one of the variables?
2x + 2y + z = 10 3x - y + 3z = 10 -->(2) -------------- 2x + 2y + z = 10 6x - 2y + 6z = 20 (result of multiplying by 2) --------------add 8x + 7z = 30 3x - y + 3z = 10 -->(3) 2x + 3y -2z = 6 ------------- 9x - 3y + 9z = 30 (result of multiplying by 3) 2x + 3y - 2z = 6 ------------add 11x + 7z = 36 8x + 7z = 30 --->(-1) 11x + 7z = 36 ------------ -8x -7z = -30 (result of multiplying by -1) 11x + 7z = 36 -----------add 3x = 6 x = 2 8x + 7z = 30 8(2) + 7z = 30 16 + 7z = 30 7z = 30 - 16 7z = 14 z = 2 2x + 2y + z = 10 2(2) + 2y + 2 = 10 4 + 2y + 2 = 10 2y + 6 = 10 2y = 10 - 6 2y = 4 y = 2 check... 3x - y + 3z = 10 3(2) - 2 + 3(2) = 10 6 - 2 + 6 = 10 12 - 2 = 10 10 = 10 (correct) x,y,and z, are all 2
Thanks to both of you! I'll use Kellie's work for reference on any more problems, and thanks DSS for starting me off
you have to be really careful with elimination because if you mess up at the beginning, that mistake carries all the way through to the end
Thanks for the advice, it'll help a lot since Elimination is the main form I use to solve these
just remember, whatever variable you eliminate in the first two equation you use, you have to eliminate the same variable in the second set of equations you use.
I would suggest checking your answers at the end to make sure they are correct
okay. By the way, do you know a link to any good Logarithm videos?
no sorry...not really
good luck finding some :) and I am glad I could help with the elimination
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