SOLUTION: Solve the system. 3x+2y-z=-8 2x-y+7z=33 2x+2y-3z=-17 x= y= z= *Attempted problem more than once, solution still came out incorrect. Please Help!!!!

Algebra ->  Systems-of-equations -> SOLUTION: Solve the system. 3x+2y-z=-8 2x-y+7z=33 2x+2y-3z=-17 x= y= z= *Attempted problem more than once, solution still came out incorrect. Please Help!!!!      Log On


   

Question 186714: Solve the system.
3x+2y-z=-8
2x-y+7z=33
2x+2y-3z=-17
x=
y=
z=
*Attempted problem more than once, solution still came out incorrect. Please Help!!!!
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Solve:
1) 3x%2B2y-z+=+-8
2) 2x-y%2B7z+=+33
3) 2x%2B2y-3z+=+-17
Subtract equation 3) from equation 2).
2x-y%2B7z+=+33
-(2x%2B2y-3z+=+-17)
------------------------
a) highlight%28-3y%2B10z+=+50%29
Now multiply equation 1) by 2 and equation 2) by 3.
2(3x%2B2y-z+=+-8)=6x%2B4y-2z+=+-16
3(2x-y%2B7z+=+33)=6x-3y%2B21z+=+99
Subtract the 1st equation from the 2nd equation.
6x-3y%2B21z+=+99
-(6x%2B4y-2z+=+-16)
-----------------------
b) highlight%28-7y%2B23z+=+115%29
Multiply equation a) by 7 and equation b) by 3.
7(-3y%2B10z+=+50)=-21y%2B70z+=+350
3(-7y%2B23z+=+115)=-21y%2B69z+=+345
Subtract the 2nd equation from the 1st equation.
-21y%2B70z+=+350
-(-21y%2B69z+=+345)
----------------------
highlight%28z+=+5%29 Now substitute this into equation a) and solve for y.
-3y%2B10z+=+50 Substitute z = 5.
-3y%2B10%285%29+=+50 Subtract 50 from both sides.
-3y+=+0 Divide both sides by -3.
highlight%28y+=+0%29 Finally, substitute y = 0 and z = 5 into equation 1) (or 2 or 3) and solve for x.
3x%2B2y-z+=+-8 Substitute y = 0 and z = 5.
3x%2B2%280%29-5+=+-8 Add 5 to both sides.
3x+=+-3 Divide both sides by 3.
highlight%28x+=+-1%29
Now you can check these solutions by substituting x = -1, y = 0, and z = 5 into the three original equations and you'll find that they are correct. I'll leave this as an exercise for you to complete. Good luck!