Please help with these 2 problems! Will give medal to best answer! Please show steps. 1. Solve the system by elimination. -2x+2y+3z=0 -2x-y+z=-3 2x-3y+3z=5 2. Solve the system by substitution. -x-y-z=-8 -4x+4y+5z=7 2x+2z=4
@jim_thompson5910 can you help?
@hartnn @thomaster
how far did you get?
I tried on the first one and ended up erasing it because I didn't think it was right. It didn't make sense. I got 2x-5y+6z=-2 I think.
do you know how elimination works? don't worry about equations with 3 variables I'm asking just for 2 variable systems right now
Not really.
ok then go ahead and read this page to get more familiar with the concept of elimination http://www.purplemath.com/modules/systlin5.htm I recommend you practice the equations with 2 variables first before you move up to the 3 variable equations
Okay thanks.
np, let me know if that page helps or not
That is the page I looked at before I made this post. I feel like all her equations are put together really well with the same or similar terms right on top of each other. It doesn't really help me much
@jonnymiller
hmm alright, I was afraid of that, let's see how far we get with this
let's focus on the first two equations -2x+2y+3z=0 -2x-y+z=-3 and let's say we want to get rid (or eliminate) the y terms we could add up +2y and -y to get 2y + (-y) = 1y = y but that's NOT 0y or 0 like we want
to fix this we can turn that -y into -2y notice now that 2y + (-2y) = 0y = 0 so the y terms have gone away
so basically you double everything in equation 2, then add double everything in equation 2 to get -2x-y+z=-3 -4x-2y+2z=-6 then add the equations to get -2x+2y+3z=0 -4x-2y+2z=-6 -------------- -6x+0y+5z = -6 that turns into -6x + 5z = -6 with me so far?
Yeah I think so.
But why isn't the first equation doubled too?
because we have 2y in the first equation we have -y in the second equation we want these two expressions to have equal (yet opposite) coefficients so they add to 0
so we have -y....but we want -2y so we have to double -y to get -2y you have to do this with every term in the equation where -y is (the second equation) to make sure everything is balanced
hopefully that's clearer
Oh okay. Yeah I think I get it.
now let's focus on the 2nd and 3rd equations -2x-y+z=-3 2x-3y+3z=5
in it's current form, if we add -y to -3y, we get -y + (-3y) = -4y that's not 0y like we want, so we have to adjust the -y into +3y to make sure that the two 'y' terms add to 0y ------------------------------------------- so you have to multiply everything in equation 2 by -3 to get -2x-y+z=-3 6x+3y-3z=9 -------------------------------------------- now add this to equation 3 6x+3y-3z=9 2x-3y+3z=5 ------------ 8x+0y+0z=14 that turns into 8x = 14
what's next?
Divide both sides by 8?..
correct
so x = ???
1.75
good, or x = 7/4 as a fraction
now use this value of x to find y and z
I got -3.5+2y+3z=0. For the next step do I subtract 3z from both sides?
pick on an equation that only has x and y or that has x and z
then plug in x = 1.75 or x = 7/4 then solve for the remaining variable
for my x value I got x=(5/2) I solved for x y and z and my values check out when I plug them back in.
I'm confused now..
no it should be x = 7/4 or x = 1.75
x = 5/2 is incorrect
So do I use the -6x+5z=-6?
yes, plug x = 1.75 or x =7/4 into that equation then solve for z
yes you are right....my solutions work for the first 2 equations but not for the 3rd
I got z=-1.09
that is incorrect
-6x+5z=-6 -6(1.75)+5z=-6 -10.5+5z=-6 5z = -6 + 10.5 5z = 4.5 z = ???
@jim_thompson5910 sorry it took so long. had to run somewhere. z=0.9?
Now do I plug that in for z and solve for y?
yes and if you need to you can plug in for both x and z to solve for y. but only do that if you absolutely have to
so x=4?
@killerdime20
you found that x = 1.75 not sure how you got x = 4
but yes z = 0.9
Which equation do I use to find y?
use the first one. it will be the easiest since it is set equal to 0
you know that x = 1.75 and z = 0.9 you now can go to any equation you want that has a y in it
plug those values in, then solve for y
1.2=y?
not quite
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