I'm I doing this problem right Algebra II?

Question

z+3x+y=3

2x+3y=10

2y=8

then I added the top to the bottom and got z+5x+4y=13

and kept the bottom the same 2y=8

if I'm doing this right what do I next and if not what do you do

Answer 1

First multiply the middle by -1 and add the top two, and you get:

z+3x+y=3

-2x-3y=-10

-----------------

z-2y=-7

Then add that and the bottom

z-2y=-7

2y=8

------------

z=1

Now you know that z=1 and plug 1 for z back into z-2y=-7 and you get

1-2y=-7

2y=-6

y=4

Now plug z and y into the original three equations and you should always get x=-2/3

That is wrong though, are you shore that all those equations are really equal?

Answer 2

Start at the bottom.

2y = 8, hence y = 8/2 = 4.

Substitute y = 4 into 2x + 3y = 10

2x + 3(4) = 10

2x + 12 = 10

2x = 10 - 12

2x = -2

x = -2/2

x = -1

Then substitute y = 4 and x = -1 into z + 3x + y = 3.

z + 3(-1) + 4 = 3

z - 3 + 4 = 3

z = 3 - 4 + 3

z = 2.

This means

z + 5x + 4y

= 2 + 5(-1) + 4(4)

= 2 - 5 + 16

=13

Answer 3

The bottom equation only has one variable, perfect! We could solve for this variable.

2y = 8

y = 4

Now, the second equation has two variables ... One of those variables we know, so there's only one unkown variable again, perfect!

2x + 3y = 10

2x + 3*4 = 10

2x + 12 = 10

2x = -2

x = -1

Now, solve for the first equation:

z + 3x + y = 3

z + 3*-1 + 4 = 3

z - 3 + 4 = 3

z = 2

Answer 4

You use the 2y=8 to say that y=4, then plug that in to 2x+3y=10 to find x.   Then you plug the two variables into the top equation to find Z