Question

Solve the system by substitution.

{-x - y - z = -8

{ -4x + 4y + 5z = 7

{ 2x + 2z = 4

Solve the system by elimination.

{-2x + 2y + 3z = 0

{-2x - y + z = -3

{2x+ 3y+ 3z = 5

Answer 1

Using the first equation, -x=y+z-8
and substituting that into the others, you have
4(y+z-8)+4y+5z = 7
-2(y+z-8)+2z = 4
or,
8y+9z = 39
-2y = 12
and the rest is easy

For the elimination, start by subtracting row1 from row2, and adding row1 to row3. That gives you

-2x+2y+3z = 0
0 -3y-2z = -3
0 +5y+6z = 5
Now work with row 2 to get rid of the y in row3, and then you have z, and you can then find y and x.

Answer 2

So what the answer? I’m still confused. How would I do the full equations?

Answer 3

Steve showed you how to do it with elimination. You got z = -6. Now go back and get x and y
for the second one try the Gauss/Jordan link I gave you for the procedure.
http://www.gregthatcher.com/Mathematics/GaussJordan.aspx

Answer 4

I mean he used substitution first. That gave z = -6. Then the link will help with elimination.

Answer 5

I got 6 from that first part not -6.. and I'm confused on what to do after you get the value of z not before ?? Can yall actually help or no cus so far u just confused my g right here and gave us the wrong answer AND didn't actually help with the rest of it

Answer 6

1st question- y=6 x=3 z=-1
2x=4-2z
x=-z+2-(z-2)-y-z=-8
-2-y=-8
y=6-4(-z+2)+4(6)+5z=7
4z-8+24+5z=7
9z+16=7
2x+2(-1)=4
2x-2=4