Question

Solve by elimination, equivalent system or substitution.

x+6z=12
-2x+3y=6
y-z/2=5/2

Answer 1

rearrange the first one in terms of x or z first.

Answer 2

x=-6z+12

-2x+3y=6
-2(-6z+12)+3y=6
12z-24+3y=6
3y+12z=30
y=-4z+10

-4z+10-z/2=5/2
-4z-z/2=-7.5

Do I divide the 2 to both sides?

Answer 3

\[\frac{-4z+10-z}{2} =\frac{5}{2}\]
\[-5z + 10 = \frac{5}{2}\times 2\]
-5z + 10 = 5
-5z = 5-10
-5z=-5
z=-5/-5
z=1

Answer 4

x+6z=12
x+6(1)=12
x=6

-2+3y=6
-2(6)+3y=6
-12+3y=6
3y=18
y=6

Solution: (6,6,1)

Answer 5

Am I correct?