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Mathematics 15 Online
OpenStudy (anonymous):

What is the solution of the system of equations? 3x+2y+z=7 5x+5y+4z=3 3x+2y+3z=1

OpenStudy (anonymous):

are you familiar with the elimination method ?

OpenStudy (anonymous):

idk? what is it

OpenStudy (anonymous):

well...it is a little tedious...what you have to do is pick two of the three equations and eliminate a variable. Let me show you... 3x + 2y + z = 7 -->(-5) 5x + 5y + 4z = 3 -->(2) ------------------ -15x - 10y - 5z = -35 (result of multiplying by -5) 10x + 10y + 8z = 6 (result of multiplying by 2) ------------------add -5x + 3z = - 29 You see how I eliminated the y variable. Sometimes you have to multiply one or both of the equations by a number to be able to eliminate that variable. Do you understand so far ?

OpenStudy (anonymous):

yes:)

OpenStudy (anonymous):

now you take the equation you have not used yet and any one of the equations you have used and eliminate the y in those equations. Remember that what you eliminate in the first set of equations you have to eliminate in the second set. 3x + 2y + 3z = 1 -->(-5) 5x + 5y + 4z = 3 -->(2) ------------------ -15x - 10y - 15z = -5 (result of multiplying by -5) 10x + 10y + 8z = 6 (result of multiplying by 2) ------------------add -5x - 7z = 1 still with me ?

OpenStudy (anonymous):

yep soorrryyy

OpenStudy (e.mccormick):

@cutegirl Can you do the same sort of thing with the new equations? -5x + 3z = - 29 -5x - 7z = 1

OpenStudy (e.mccormick):

In those, I would eliminate the x because it is easy.

OpenStudy (anonymous):

do you know how to eliminate the x ? You will have to multiply one of the equations by a certain number, do you know what that number is cutegirl ?

OpenStudy (anonymous):

You want to make one of the x's positive and one negative.

OpenStudy (anonymous):

are you here ??

OpenStudy (anonymous):

sorry i had to help my bro

OpenStudy (anonymous):

-5x + 3z = -29 -->(-1) -5x - 7z = 1 -------------- 5x - 3z = 29 (result of multiplying by -1) -5x - 7z = 1 --------------add -10z = 30 z = -3 still with me ?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

so 3x+2y+-3=7 5x+5y+1=-12 3x+2y+0=1

OpenStudy (anonymous):

now sub -3 in for z in one of the equations that you eliminated y in -5x + 3z = -29 -5x + 3(-3) = -29 -5x - 9 = -29 -5x = -29 + 9 -5x = -20 x = -20/-5 x = 4 understand what I have done ?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

now we just have to find y. Sub in 4 for x and -3 for z in one of the original equations. 3x + 2y + 3z = 1 3(4) + 2y + 3(-3) = 1 12 + 2y - 9 = 1 2y + 3 = 1 2y = 1 - 3 2y = -2 y = -1 now we can check our answers.... 3x + 2y + 3z = 1 3(4) + 2(-1) + 3(-3) = 1 12 - 2 - 9 = 1 12 - 11 = 1 11 = 11 (correct) x = 4, y = -1 and z = -3 any questions ?

OpenStudy (anonymous):

thanks i get it know thanks again your the best:)

OpenStudy (anonymous):

Just be careful, because if you mess up on this problem in the beginning, it carries all the way to the end and you will get the wrong answer. e.mccormick was showing me another way and he explained it pretty good. This is the way he showed me...matrices.. http://openstudy.com/study#/updates/5275c700e4b00c1700b26a41

OpenStudy (anonymous):

thanks

OpenStudy (anonymous):

no problem :)

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