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Does anyone know how to do these?
5x+4y-z= 1
2x+2y+z= 1
-x-y+z=2
or
3x+y+z=7
x+3y-2z=13
y=2x-1
or
2x+5y+2z=2
x-3y+z=2
3x+2y+5z=0
you have to do it by substitution but im so confused. :(
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i'll help with the first one and then you will probably be able to do the others. we actually just took a test on this in my algebra ii class. okay, the point of substitution is to get one of the equations and make it to where it is either x=..., y=..., or z=. so, going with that, let's take the last equation from that first bit because it looks easier than the rest. let's say you want to isolate x on one side of it (so it will look like x=...).
-x-y+z=2
from both sides, add y, and subtract z so now you have:
-x=y-z+2
that is negative x, and since you want positive x, just change the signs:
x=-y+z-2
so then you can go into another equation (let's take the second one) and plug in the x. so it would be:
2(-y+z-2)+2y+z=1
now, it will really help now if you write that down and make it into a simpler form, which is:
-2y+2z+4+2y+z=1
do you see that there is a -2y and a +2y? those make 0. so, that means that after you do that, all that is left in the equation is z. pretty cool, right? so after that, you can solve for z.
i think that is probably enough info to get you started on that one. just keep going to different equations and isolating variables. now you know what z is, so it will be much easier to solve for x and y. if you have any other math questions, send them to my inbox :)
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you could do it a different way and it comes out the same. 5x+4y-z=1 + -4x-4y-2z=-2 and you get 1x-3z=-1 now 2x+2y+z=1 + -2x-2y+2z=4 and get 3z=4 then add the two equations you have 1x-3z=-1 + 3z=4 and get 1x=-3 then x=-3 thats a part of it now take x and sub. it into one of the two equations u had 1(-3)-3z=-1 then get z alone -3z=2 z=2/3 second part of answer now plug z and x in to an original equation 3-y+2/3=2 makes -y=-1/3 and y=1/3 and the answer is (-3,2/3,1/3).
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Ugh, I answered the wrong question.
Disregard this.
I'll edit later :)
-LM
[15/f]
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