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OpenStudy (anonymous):

Solve the system by elimination.. HELP :(

OpenStudy (anonymous):

ur still having trouble

OpenStudy (anonymous):

its a different problem an older one

OpenStudy (anonymous):

anyone there???????

OpenStudy (anonymous):

yeah im trying

OpenStudy (anonymous):

i cant help sorry

OpenStudy (anonymous):

:P

OpenStudy (anonymous):

okay thanks i really appreciate it. can u show ur steps so i understand more

OpenStudy (anonymous):

who are u talking to

OpenStudy (anonymous):

-2x -y +z = -3 2x + 3y +3z = 5 --------------- 0x +2y +4z = 2 <-added together

OpenStudy (anonymous):

huh

OpenStudy (anonymous):

I'm not sure if the above helped...

OpenStudy (anonymous):

okay now what @phanta_seea

OpenStudy (anonymous):

dont i have to take the third equation and do something with that

OpenStudy (anonymous):

I'm really just trying..... -2x + 2y +3z = 0 2x +3y +3z = 5 -------------- 0 + 5 y + 6z = 5

OpenStudy (anonymous):

so I did two additions of original lines now, I can add the results together as well

OpenStudy (anonymous):

0x +2y +4z = 2 0 + 5 y + 6z = 5 want to eliminate y or z, I will eliminate y: 0 x + 10y + 20z = 10 first line * 5 -0 x - 10y - 12 z = -10 second line * -2 -------------------------------- 8z = 0

OpenStudy (anonymous):

i dont get what you did to eliminate y

OpenStudy (anonymous):

is 8z=0 the answeR???

OpenStudy (anonymous):

what I did was the following: I did add row 2 and 3 --- and got a "result row 1" (R1) then, added row 1 and 3 --- and got a "result row 2" (R2) here are the result rows I obtained: R1: +2y +4z = 2 R2: +5y +6z = 5 what do the result rows have that the original rows don't? they don't have any x values to worry about.

OpenStudy (anonymous):

@blahkatherine yep, z=0 is part of the answer, and you can use "backsubstitution" to find all the other values :)

OpenStudy (anonymous):

so where do i plug in 0 ... which equation

OpenStudy (anonymous):

an equation that has only one additional variable. that way you can also determine the value of the additional variable, use one of the result rows: R1: +2y +4z = 2 R2: +5y +6z = 5

OpenStudy (anonymous):

ill use R1 so 2y + 4(0) = 2 so 2y + 0 =2 ... 2y = 2 ... y = 1 right?

OpenStudy (anonymous):

true :)

OpenStudy (anonymous):

now we know 2 variables so we can use ANY of the original rows to get x

OpenStudy (anonymous):

So now what do i do :o

OpenStudy (anonymous):

well you take any of the original equations and plug the z, y values

OpenStudy (anonymous):

to find x, which is the only variable still missing

OpenStudy (anonymous):

so my answer is y=1 and z= 0

OpenStudy (anonymous):

oh pellet okay

OpenStudy (anonymous):

yep, we already have y=1 and z=0 now plug those in any of the 3 original equations to get x

OpenStudy (anonymous):

that one is right :)

OpenStudy (anonymous):

I checked the values x=1, y=1, z=0 in all the three equations and they're correct

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