solving systems with 3 variables - where am I wrong

[g187play]

x-3y+6z=-8
3x-2y-10z=11
5x-6y+2z=7
Adding equations (1) and (2) together. Multiplying equation (1) by 5 and equation (2) by 3.. so that I can cancel out the z's.
5(x-3y+6z=-8)
3(3x-2y-10z=11)

5x-15y+30z=-40
9x-6y-30z=33

14x-21y=-7


now adding equations (1) and (3) together. Multiplying equation (3) by -3... so that I can cancel out the z's.
x-3y+6z=-8
-3(5x-6y+2z=7)

x-3y+6z=-8
-15x+18y-6z=-21

-14x+15y=-29

now add the system equations together.
14x-21y=-7
-14x+15y=-29

-6y=-36
divide both sides by -6
y=6

My instruct says this not the correct answer for y. She says it is 7. Thats the only hint she will give me.... HELP!

 

[Loren]

I get y=6 also. Furthermore, it checks.
Possibly your teacher or you have copied the problem wrong. In my opinion, for your simultaneous system, y=6 is correct. Both x and z are fractions.

 

[wjm11]

x-3y+6z=-8
3x-2y-10z=11
5x-6y+2z=7

y=6

My instruct says this not the correct answer for y. She says it is 7.

I am in complete agreement with Loren.

x = 17/2
y = 6
z = ¼

If y does not equal 6, then you have written down the starting equations incorrectly.

 

[g187play]

THANK YOU soooo much. I did copy the problem incorrectly. Why can't teacher's say just say that... errrr. I know it's really my fault. Thanks for the suggestion.